Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Students can Download Chapter 11 Thermal Properties of Matter Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Plus One Physics Thermal Properties of Matter One Mark Questions and Answers

Question 1.
In which of the following processes, convection does nottake place primarily?
(a) Sea and land breeze
(b) Boiling of water
(c) Warming of glass of bulb due to filament
(d) Heating air around a furnace
Answer:
(c) Warming of glass of bulb due to filament
In convection process, the heat is transferred by the bodily motion of the heated particles. It is not so in case of warming of glass bulb due to filament heating. In fact, warming of glass bulb is due to radiation.

Question 2.
\(\frac{\text { Watt }}{\text { Kelvin }}\) a unit of
(a) Stefan’s constant
(b) Wien’s constant
(c) Cooling’s constant
(d) Thermal conductance
Answer:
(d) Thermal resistance
Plus One Physics Thermal Properties of Matter One Mark Questions and Answers 1

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 3.
For measuring temperatures in the range of 2000°C, we should employ
(a) gas thermometer
(b) platinum-rhodium thermometer
(c) barometer
(d) pyrometer
Answer:
(d) pyrometer

Question 4.
There is a hole in metal disc. What happens to the size of metal disc if the metal disc is heated?
Answer:
The size of hole increases.

Question 5.
Which has more specific heat capacity, water, and sand?
Answer:
Water.

Question 6.
Two solid spheres of the same material have the same radius but one is hollow while the other is solid. Both spheres are heated to same temperature. Then
(a) the solid sphere expands more
(b) the hollow sphere expands more
(c) expansion is same for both
(d) nothing can be solid about their relative expansion if their masses are not given
Answer:
(c) expansion is same for both

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 7.
The sprinkling of water reduces slightly the temperature of a closed room because
(a) temperature of water is less than that of the room.
(b) specific heat of water is high
(c) water has large latent heat of vaporisation
(d) water is a bad conductor of heat
Answer:
(c) water has large latent heat of vaporisation
When water is sprinkled over a large area, evaporation takes place. As the latent heat of vaporisation is large cooling takes place.

Question 8.
Why specific heat of gas at constant pressure (Cp) is greater than specific heat at constant volume?
Answer:
More heat is required to raise the temperature of gas at constant pressure than at constant volume.

Question 9.
A body is heated. But there is no change in its temperature. Is it possible?
Answer:
Yes. During change of state, there will be no increase in temperature even when heat is supplied.

Question 10.
When boiling water is put in glass tumbler, the tumbler cracks. Why?
Answer:
Glass is poor conductor of heat. So inner and outer surfaces of tumbler suffer uneven expension. Hence it breaks.

Question 11.
A small space is left between two rails on railway track. Why?
Answer:
If no space is left, the rails would bend due to thermal expansion in summer. So small space is left between two rails to allow thermal expansion.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 12.
The pendulum of clock is made of invar, Why?
Answer:
Invar has low value of coefficient of linear expansion. So length of pendulum remains almost same in all seasons. (The change in length affects time period of pendulum).

Question 13.
Tea gets cooled, when sugar is added to it. Why?
Answer:
When sugar is added, heat content of tea gets shared with sugar & hence’ temperature decreases.

Question 14.
Ice covered in gunny bag does not melt for a long time. Why?
Answer:
A gunny bag is poor conduct of heat & hence it does not allow external heat to enter.

Question 15.
Why two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double the thickness?
Answer:
Because air between two layers of clothes is a bad ‘ conductor of heat.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 16.
On winter nights, we feel warmer when clouds cover the sky than the sky is clear. Why?
Answer:
The clouds are bad conductor of heat. So heat of earth’s atmosphere is not conducted out.

Plus One Physics Thermal Properties of Matter Two Mark Questions and Answers

Question 1.
1. Two bodies at different temperatures T1K and T2K are brought in contact with each other

  • Is the resultant temperature be necessarily (T1 + T2)/2? If not, Why?
  • Should the resultant temperature be between T1 and T1 only? If not, Why?

Answer:
1. Two bodies at different temperatures:

  • The resultant temperature may not be necessarily \(\left(\frac{T_{1}+T_{2}}{2}\right)\). Because specific heat capacity is different for all substances.
  • If heat is not lost to the surroundings, resultant temperature must lie in between T1 and T2.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
A Patient is admitted to hospital. The temperature of the patient is measured by nurse and is found to be 97.6°F

  1. What is meant by temperature?
  2. Convert the temperature (97.6°F) is to centigrade

Answer:
1. Temperature is the degree of hotness.

2.
Plus One Physics Thermal Properties of Matter Two Mark Questions and Answers 2

Question 3.
Why iron rims are heated red hot before being put on the cart wheels?
Answer:
The radius of iron rim is smaller than radius of cart wheel. When iron rim is heated, its radius increases due to thermal expansion. After rim has planted on the wheel, iron rim is allowed to cool. Then it fits tightly on the wheel due to thermal contraction.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 4.
How woolen clothes helps us in winter against cold?
Answer:
Wool is a heat insulator. Moreover, it contains a layer of dry air in it. This air, which is bad conductor of heat does not allow the body heat to escape & it also does not allow external cold to come in. The rough surface of woollen clothes absorbs more and reflects less heat falling on it.

Plus One Physics Thermal Properties of Matter Three Mark Questions and Answers

Question 1.
A Solid material is supplied with heat at constant rate. The temperature of the material is changing with the heat input as shown in figure.

  1. What do the horizontal region AB and CD represent?
  2. What does the slope of DE represent?
  3. The slope of OA is greater than the slope of BC. What does this indicate?

Answer:
Plus One Physics Thermal Properties of Matter Three Mark Questions and Answers 3

  1. Latent heat of fusion, Latent heat of vaporization
  2. It indicates that, the material is in vapour state
  3. It indicates that latent heat of vaporization of the material is greater than the latent heat of fusion.

Plus One Physics Thermal Properties of Matter Four Mark Questions and Answers

Question 1.
A copper block of mass 2.5kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39Jg-1 K-1; latent heat of fusion of water = 335 Jg-1).
Answer:
Mass, m = 2.5kg
= 2.5 × 103 g;
Change in temperature, ∆T = 500°C
Specific heat, c = 0.39 Jg-1K-1;
Latent heat of fusion, L = 335Jg-1
If m’ be the mass of ice melted, then m’L = mc ∆T
or m’ × 335 = 2.5 × 103 × 0.39 × 500 2.5 × 103 × 0.39 × 500
or m’ = \(\frac{2.5 \times 10^{3} \times 0.39 \times 500}{335} \mathrm{g}\)
= 1.5 kg.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
Heat from the sun reaches the earth.

  1. Write the method of transmission of heat.
  2. Thermal conductivity of Aluminium is 205 Jm-1s-1deg-1. What do you mean by this?
  3. Birds swell feathers in winter. Why?

Answer:
1. Radiation, conduction, convenction.

2. The coefficient of thermal conductivity of a substance is defind as the quauitity of heat conducted normally persecond through unit area of the substance per unit temperature gradient when the substance attains steady state.

3. By doing so the birds enclose air between the feathers. Air being a poor conductor, prevents the loss of heat from the body of the bird to the cold surroundings.

Question 3.
Two accidents are happened. The first one with water at 100°C and the second one with steam at 100°C.

  1. Which is dangerous burn due to water at 100°C, and bum due to steam at 100°C? Why?
  2. Latent heat of vapourisation of water to 536 cal/g. Explain the idea of latent heat of vopourisation.
  3. Find the heat required to convert 1g of ice at 0°C to steam at 100°C is

Answer:
1. Burn due to steam is more dangerous, because heat content in steam is very high compared to 100°C water.

2. Latent heat of vapourisation is the amount of heat required to change the state of 1 kg water in to vapour.

3. Q = ML + MC∆Q + ML1
Plus One Physics Thermal Properties of Matter Four Mark Questions and Answers 4
= 716 cal.

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 4.
When a fluid is heated, the particle rises up.

  1. Name the phenomonon behind it.
  2. Explain the formation of land breeze and sea breeze.

Answer:
1. Convection

2. During the day, land heats up more quickly than, water in lake (due to high specific heat capacity of water). The air on the surface of earth gets heated, expands, becomes less dense and rises up. The colder air (wind) replaces the space created by hot air.

It creates a sea breeze. At night the land loses its heat very quickly than water. So water remains more warmer at night.

Plus One Physics Thermal Properties of Matter NCERT Questions and Answers

Question 1.
A 10kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.19Jg-1 K-1.
Answer:
Power, P = 10 kW
= 10 × 103W
= 104W
Mass, m = 8 kg; Time, t = 2.5min = 150s
Specific heat, c = 0.91Jg-1K-1
= 0.91 × 103Jkg-1K-1
Energy, Q = pt = 104 × 150J = 1.5 × 106J
It is given that 50% of energy is lost to the. surroundings. So, energy absorbed by the block is given by
Q = \(\frac{1}{2}\) × 1.5 × 106J = 0.75 × 106J
But Q = mc∆T
∴ ∆T
Plus One Physics Thermal Properties of Matter NCERT Questions and Answers 5

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 2.
A brass boiler has a base area of 0.15m2 and thickness 1.0cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s-1 m-1K-1;
Heat of vaporisation of water = 2256 × 103 J kg-1.
Answer:
A = 0.15m2, d = 10-2m,
\(\frac{m}{t}=\frac{6}{60}\) kgs-1 = 0.1kgs-1,
K = 109J s-1 m-1 K-1, L = 2256 × 103 Jkg-1,
θ1 = ?,
θ2 = 100°C
Plus One Physics Thermal Properties of Matter NCERT Questions and Answers 6

Plus One Physics Chapter Wise Questions and Answers Chapter 11 Thermal Properties of Matter

Question 3.
Explain why?
(a) A body with large reflectivity is a poor emitter.
(b) A brass tumbler feels much colder than a wooden tray on a chilly day.
(c) An optical pyrometer (for measuring high temperature) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
(d) The earth without its. atmosphere would be inhospitably cold?
(e) Heating systems based on circulation of steam are more efficient in warming a building than heating system based on circulation of hot water.
Answer:
(a) a body whose reflectivity is large would naturally absorb less heat. So, a body with large reflectivity is a poor.

(b) The thermal conductivity of brass is high i.e., brass „ is a good conductor of heat. So, when a brass tumbler is touched, heat quickly flows from human body to tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor, so, heat does not flow from the human body to the wooden tray in this case. Thus, it appears comparatively hotter.

(c) Let T the temperature of the hot iron in the furnace. Heat radiated per second per unit area, E = σT4

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Students can Download Chapter 11 The p Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Plus One Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The aqueous solution of borax is
a) Acidic
b) Alkaline
c) Neutral
d) Amphoteric
Answer:
b) Alkaline

Question 2.
Say TRUE or FALSE.
Boron in aqueous solution forms B3+ ion.
Answer:
False

Question 3.
Which of the halide of group 14 does not exist?
a) CF4
b) Cl4
c) SiF4
d) Pbl4
Answer:
d) Pbl4

Question 4.
Orthoboric acid, H3BO3 is a
a) Protonic acid
b) Arrhenius acid
c) Lewis acid
d) Bronsted-Lowery acid
Answer:
c) Lewis acid

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
The zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is _________ .
Answer:
ZSM-5

Question 6.
Dry ice is __________ .
Answer:
Solid CO2

Question 7.
Thermodynamically most stable allotrope of carbon is __________ .
Answer:
Graphite

Question 8.
The alkalimetal used in solar cells is __________ .
Answer:
Cs (Caesium)

Question 9.
Acidity is in the order
Answer:
BBr3 > BCl3 > BF3

Question 10.
AlCl3 fumes in most air because
Answer:
HCl is formed due to hydrolysis in moist air

Plus One Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
Silicon belongs to the carbon family. Graphite is an important allotrope of carbon. But Si does not form an analogue of graphite. What are the possible reasons?
Answer:
Silicon atom is much bigger in size than carbon. Si-Si bond energy is less than C-C bond energy and also Si does not form compounds in sp² hybrid state.

Question 2.
Boron is an element with atomic number 5.
a) Write down the electronic configuration of boron.
b) Mention any two uses of boron.
c) Write down some compounds of boron.
Answer:
a) 1s²2s²2p¹
b) To increase the hardness of the steel.
Used as the semiconductor in electronic devices.
c) Diborane, borax, orthoboric acid, boron trifluoride

Question 3.
Answer the questions, with the help of the following:

  • Hard solid
  • Melting point above 450 K.
  • Four allotropic forms are known.
  • Low electrical conductivity.
  • Mass number! 3

1. Which is the element?
2. Write any two uses of this element.
3. Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 1
4. Write any three compounds of this element.
Answer:
1. Boron
2. To increase hardness of steel
As semiconductor in electric devices
3. 2B
4. Diborane, Borax, Boric acid

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 4.
1. Write any two allotropic forms of carbon.
2. C + ½ O2 → …………….
3. Write any two uses of carbon monoxide.
4. How is carbon dioxide produced?
5. What is the hardest element/form of an element in the world?
Answer:
1. Diamond, Graphite
2. CO (Carbon monoxide)
3. Reducing agent in metallurgy
Used in the manufacture of methanol
4. Carbon dioxide is formed by the complete combustion of carbon.
C + O2 → CO2
5. Diamond

Question 5.
From the compounds of group 14 elements write an appropriate example for each of the following:

No.Type of compoundName/ Formulae of example
1.A strong reducing oxide
2.A giant covalent oxide
3.A strongly reducing chloride
4.A covalent chloride not hydrolysed by water

Answer:
1-CO
2-SiO2
3-SnCl2
4-CCl4

Question 6.
Diborane has an unusual structure. Justify the statement with figure.
Answer:
In diborane, each Batom uses sp³ hybrid orbitals for bonding. Out of the four sp³ hybrid orbitals on each B atom, one is without an electron. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 2

Question 7.
Match the following:

AB
COi. A semi conductor in electronic devices
H2ii. Reducing agent in metallurgy
O3iii. Thermal decomposition of Ammonia
Boroniv. Used as a chemical reagent in organic chemistry

Answer:
CO – ii,
N2 – iii,
O3 – iv,
Boron – i

Question 8.
BCl3 fumes in moist air.
1. Give reason.
2. Write down a balanced equation that can reveal the answer.
Answer:
1. When water of moist air reacts with boron halide hydrogen chloride is formed which causes fumes,

2. BCl3 + 3H2O → H3BO3 + 3HCl

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 9.
‘Generally, non-metal oxides are basic.’
1. Do you agree?
2. What do you meant by oxides?
3. Which are the different types of oxides?
4. Give examples for each type of oxides.
Answer:
1. No. Generally, non metallic oxides are acids.
2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides.
3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide.
4. Acidic oxide → SO2, NO2
Basic oxide → MgO
Amphoteric oxide → Al2O3, ZnO
Neutral oxide → CO, N2O

Question 10.
Match the following:
1. Borane – H3BO3
2. Boric acid – Na2B4O7.10H2O
3. Borax – Amphoteric oxide
4. Al2O3 – Boron hydride
Answer:
1. Borane – Boron hydride
2. Boric acid – H3BO3
3. Borax – Na2B4O7.10H20
4. Al2O3 – Amphoteric oxide

Question 11.
How diborane reacts with
1. Oxygen?
2. Water?
Answer:
1. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.
B2H6 + 3O2 → B2O2 + 3H2O; ∆rHΘ -1976 kJ mol-1

2. Diborane is readily hydrolysed by water to give boric acid.
B2H6(g) + 6H2O(l) → 2B(OH)3(aq) + 6H2(g)

Question 12.
1. CCl4 cannot be hydrolysed. Give reason.
2. Draw the structure of the dimer of AlCl3.
Answer:
1. Carbon has no d orbitals to accommodate the lone pair of electrons from oxygen atom of H20.
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 3

Question 13.
1. Draw the structure of boric acid.
2. Starting from borax how will you prepare boric acid? (Write the chemical equation).
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 4
2. Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3.

Question 14.
CO2 is a gas but SiO2 is a solid. Give reason.
Answer:
CO2 exists as discrete molecules due to the formation of pπ -pπ double bond between carbon and oxygen. But SiO2 has a three dimensional network structure in which each Si atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

Question 15.
1. What are zeolites?
2. What is ZSM-5?
Answer:
1. Zeolites are aluminosilicates with three-dimensional network structure in which Al atoms replace few Si atoms. Cations such as Na+, K+ or Ca2+ balance the negative charge of aluminosilicate anion.

2. ZSM-5 is a zeolite catalyst used in petrochemical industry to convert alcohols directly into gasoline.

Plus One Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Explain the following:
1. Allotropy
2. Coke and Charcoal
Answer:
1. The phenomenon of existence of an element in two or more forms, which have different physical properties but almost similar chemical properties, is called allotropy and the different forms are called allotropes.

2. Coke and Charcoal are amorphous forms of carbon. Coke is formed-by the destructive distillation of coal. It is used as a fuel and also as a reducing agent in metallurgy.

Question 2.
Give justifications.
1. The first ionisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.
2. Graphite is a better lubricant on moon than that on earth.
Answer:
1. This is because carbon has greater nuclear charge. For second ionisation enthalpy, an electron is to be removed from 2p of carbon while in boron the second electron is in 2s orbital. Removal of a 2s electron is more difficult due to the high penetrating power of 2s orbital.

2. Graphite contains hexagonal sheets of carbon held together by weak van der Waals’ forces and low density. In the moon, the different layers experience less weight due to less gravitational force and the density is still reduced. The lighter layers can easily slide over another to make graphite more lubricating on the moon than on earth.

Question 3.
1. Identify the compound with the following structure:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 5
2. Mention the angles a & b.
3. Write down the corresponding elements in the figure.
1……………..
2……………..
3……………..
4……………..
5……………..
6……………..
Answer:
a) Diborane(B2H6).
b) a-97°
b-120°
c) 1-H, 2-H, 3-B, 4-B, 5-H, 6-H

Question 4.
The simplest boron hydride is diborane.
1. Write down the molecular formula of diborane.
2. Distinguish between terminal hydrogen and bridging hydrogen atoms of diboran.
Answer:
1. B2H6

2. In diborane, four hydrogen atoms and two boron atoms are present in a single plane. These hydrogen atoms are called terminal atoms. The terminal B-H bonds are regular two centre-two electron bonds. The remaining two hydrogen atoms above and below this plane are called bridging hydrogen atoms. The two bridge bonds (B-H-B) are three centre-two electron bonds.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
1. How is orthoboric acid prepared?
2. Account for the acidic nature of orthoboric acid.
Answer:
1. Orthoboric acid is prepared by acidifying an
aqueous solution of borax.
Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3

2. Orthoboric acid is a weak nonobasic acid. It acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3 + 2HOH → [B(OH)3] + H3O+

Question 6.
1. How is diborane prepared in the laboratory?
2. BCl3 is a good Lewis acid. Why?
Answer:
1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine.
2NaBH4 + l2 → B2H6 + 2Nal + H2

2. In BCl3, the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

Question 7.
1. Name the allotropes of carbon.
2. Carbon monoxide is highly poisonous. Do you agree? Justify.
Answer:
1. Graphite, diamond and fullerene.

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxyhaemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

Question 8.
1. Diamond is hard and non conducting while graphite is soft and conducting. Why?
2. Explain the action of heat on boric acid.
3. What is inorganic benzene? How is it formed?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. Due to this closely packed arrangement, it is hard. Since, there are no free electrons it is an insulator. But in graphite, the carbon atoms are in sp² hybridised state. It is a good conductor due to the presence of delocalised electrons between the layers. Graphite cleaves easily between the layers. Therefore, It is very soft and slippery.

2. On heating, orthoboric acid above 370 K forms metaboric acid, HBO2 which on further heating yields boric oxide, B2O3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 6

3. B3N3H6 is called Inorganic benzene. It is obtained by heating diborane with ammonia.

Question 9.
1. Explain the difference in properties of diamond and graphite on the basis of their structures.
2. How do you explain the lower atomic radius of gallium as compared to aluminium?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. As a result of this hybridisation, a closely packed arrangement is present in diamond and it explains the hardness of diamond. Due to the absence of electrons, diamond is an insulator.

In graphite, the carbon atoms are in sp² hybridised state. As a result of this hybridisation, graphite has a layered structure with hexagonal rings. Each layer can slide over the other which explains the lubricating property of graphite.

2. This can be explained on the basis of the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 10.
1. Boron resembles silicon in many of its properties. What is this resemblance generally known as?
2. What is dry ice? What is it used for?
3. What are silicones? How do they differ from silicates?
Answer:
1. Diagonal relationship

2. Carbon dioxide can be obtained as a solid in the form of dry ice, by allowing the liquified CO2 to expand rapidly.
Dry ice is used as a refrigerant for ice-cream and frozen food.

3. Silicones are a group of organosilicon polymers containing R2SiO repeating units.
Silicates are minerals with SiO44- as the basic structural unit. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units.

Plus One Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
Match the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 7
Answer:
1 – c – f
2 – a – h
3 – b – g
4 – d – e

Question 2.
a) Elements of group 13 are
1) Al, Cr, Cd, Ga, Ti
2) B, C, Si, Ga, Ti
3) B, Al, Ga, Ti, In
4) B, Al, Cr, Ca, Ti
b) Which of the following is not a mineral of boron? (borax, bauxite, colemanite, tincal)
c) In the reaction between NH3 and BF3 ammonia acts as
(Lewis base, Lewis acid, brownsted base)
d) Diamond and Graphite are carbon’s
(isotropic forms, allotropic forms, isotopic forms, amorphous form)
Answer:
a) 3) B, Al, Ga, TI, In
b) Bauxite
c) Lewis base
d) allotropic forms

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 3.
The hydrides of boron are called boranes.
1. How diborane reacts with ammonia?
2. Account for the exceptional hardness of diamond.
Answer:
1. Diborane reacts with ammonia to give B2H6.2NH3
initially, which can be formulated as [BH2(NH3)2]+[BH4]. Further heating gives borazine, B3N3H6 known as ‘inorganic benzene’ in view of its ring structure with alternate B-H and N-H groups,

2. In diamond, all the carbon atoms are in sp3 hybridised state. As a result of this hybridisation, a rigid three dimensional network of carbon atom is generated with directional covalent bonds throughout the lattice. It is very difficult to break this network structure.

Plus One Chemistry The p Block Elements NCERT Questions and Answers

Question 1.
Why does boron trifluoride behave as a Lewis acid? (2)
Answer:
The B atom in BF3 has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F, NH3, (C2H5)2O, RCH2OH etc. and thus behaves as a Lewis acid.

Question 2.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? (2)
Answer:
Due to increase in atomic size and screening effect the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result, there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 3.
Consider the compounds, BCl3 and CCl4. How will they behave with water? (2)
Answer:
The B atom in BCl3 has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl3 undergoes hydrolysis to form boric acid (H3BO3) and HCl.
BCl3 + 3H2O → H3BO3 + 3HCl

In contrast, C atom in CCl4 has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H2O molecule and hence CCl4 does not undergo hydrolysis in water.

Question 4.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is buddled through. Give reasons. (3)
Answer:
Anhydrous HF is a covalent compound and is strongly H-bonded. Therefore, it does not give free F ions and hence AlF3 does not dissolve in HF. In contrast, NaF is an ionic compound and hence F ions are easily available. As a result, it combines with AlF3 to form the soluble complex.
3NaF + AlF3 → Na3[AlF3]

On bubbling gaseous BF3, AlF3 is precipitated. It is because BF3 is a stronger Lewis acid than AlF3. This is attributed to smaller size and higher electronegativity of Boron. As a result of this B has much higher tending to form complexes than Al. Therefore, when BF3 is added to the above solution, AlF3 gets precipitated.
Na3[AlF6] + 3BF3 → 3Na[BF4] + AlF3(s)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences. (3)
Answer:
Aluminium shows a uniform oxidation state of+3 in its compounds. Like aluminium, thallium also shows +3 oxidation state in some of its compounds like TlCl3, Tl2O3, etc. Al is known to form octahedral complexes like [AlF6]3-. Similarly, Tl also forms octahedral complexes as [TlF6]3-.

Thallium also resembles group 1 metals. Like group 1 metals which show a stable oxidation state of +1 in their compounds Tl, due to inert pair effect, also shows +1 oxidation state in some of its compounds such as Tl2O, TlCl, TlClO4, etc. Similarly, like group 1 oxides, Tl2O is strongly basic.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Students can Download Chapter 10 Mechanical Properties of Fluids Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Plus One Physics Mechanical Properties of Fluids One Mark Questions and Answers

Question 1.
Water is flowing through a very narrow tube. The velocity of water below which the flow remains a streamline flow is known as
(a) relative velocity
(b) terminal velocity
(c) critical velocity
(d) particle velocity
Answer:
(c) critical velocity
Critical velocity is that velocity of liquid flow, upto which the flow of liquid is a streamlined and above which its flow becomes turbulent.

Question 2.
Bernoulli’s equation for steady, non-viscous, imcompressible flow expresses the
(a) conservation of angular momentum
(b) conservation fo density
(c) conservation of momentum
(d) conservation of energy
Answer:
(d) conservation of energy

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 3.
When cooking oil is heated in a frying pan, the oil moves around in the pan more easily when it is hot. The main reason for this is that with rise in temperature, there is a decrease in
(a) surface tension
(b) viscosity
(c) angle of contact
(d) density
Answer:
(d) density

Question 4.
At what temperature density of air is maximum?
Answer:
4°C

Question 5.
A thin glass plate is lying on a wet marble floor, It is difficult to pull the glass plate because of
(i) surface tension
(ii) Viscosity
(iii) friction
(iv) atmosphere
(v) Gravity
Answer:
(i) Viscosity

Question 6.
Why do clouds float in the sky?
Answer:
Zero terminal velocity

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 7.
A spinning cricket ball does not follow parabolic path. Why?
Answer:
Due to the magnus effect.

Question 8.
The deep water runs slow. Why?
Answer:
According to equation of continuity velocity is inversly proportional to velocity (AV = constant). Hence deep water runs slow.

Question 9.
Why dust generally settles down in closed room?
Answer:
The dust particles (tiny spheres) acquire terminal velocity as it fall through air. The terminal velocity is directly proportional to square of radius. Hence terminal velocity of dust particle is very small. So they settle down gradually.

Question 10.
Why more viscous oil is used in summer than in winter in scooters?
Answer:
The viscosity decreases with increase in temperature.

Question 11.
Why is sand drier than clay?
Answer:
Capillary action

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 12.
Why cotton dress is preferred in summer?
Answer:
Cotton dresses have fine pores which act as capillaries for the sweat.

Question 13.
Why oil is poured to calm the sea?
Answer:
When oil is poured in water, the surface tension of water is reduced and water spreads over large area of sea.

Question 14.
How plants draw water from ground?
Answer:
The capillary action.

Question 15.
How do insects run on the surface of water?
Answer:
Because of surface tension, the surface of water behaves like stretched membrane hence it can support weight of small insects.

Question 16.
How ploughing a field helps to retain moisture?
Answer:
When field is ploughed, capillaries are broken and hence water can not rise up and retains moisture.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 17.
Hot soup tastes better than cold soup. Why?
Answer:
The surface tension of hot soap is less compared to cold soap. So hot soap spreads larger area.

Plus One Physics Mechanical Properties of Fluids Two Mark Questions and Answers

Question 1.
Remya found that a piece of metal weighs 210 g in air and 180 g when it is immersed in water. Determine the density of the metal piece.
Answer:
Relative density,
R.D = \(\frac{\text { Weight in air }}{\text { Loss of Weight in water }}\)
= \(\frac{210}{30}\) = 7.

Question 2.
Why is hot soup tastier than cold one?
Answer:
When temperature increases, the surface tension of soap decrease. Hence hot soap can enter into tiny pours of taste buds.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 3.
Why straws are used to drink soft drinks?
Answer:
When we suck the straw, pressure inside straw becomes lower than atmospheric pressure. This pressure difference cause the soft drink to rise through the straw.

Question 4.
Why new earthen pots keeps water more cool than old earthen pots?
Answer:
The capillaries of old earthen pots will get blocked with passage of time. For new earthen pots, water oozes out through capillaries, gets evaporated at the surface and makes it cool.

Plus One Physics Mechanical Properties of Fluids Three Mark Questions and Answers

Question 1.
Air is blown in between two pith balls suspended freely.

  1. What will happen to the balls?
    • They repel each other
    • They attract each other
    • They start oscillating
    • They remain in their initial position They fall on the ground
  2. Give your explanation

Answer:

  1. They attract each other
  2. When air is blown in between two pith balls, the pressure between the balls decreases. Due to this decrease in pressure between the balls, they attract each other.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 2.
A child dipped two identical capillary tubes, one in a beaker containing mercury. He observed that water and mercury have risen through the tubes to a certain heights.

  1. Name this phenomenon.
  2. What difference did he observe in the shape of the meniscus of the two liquids in the tubes?
  3. If he plots a graph connecting the radius of the capillary tube and capillary height, what will be the shape of the graph?

Answer:
1. Capillary rise.

2. The shape of the water meniscus in the tube becomes concave upwards. But the shape of mercury measures in the tube become convex upward.

3. h α \(\frac{1}{r}\), this is in the form y α \(\frac{1}{x}\)
Hence when we draw graph between ‘h’ and ‘r’ we get a graph of hyperbola.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 1

Question 3.
Bernoulli’s theorem is a consequence of energy conservation principle. Using this theorem explain the working of atomiser.
Answer:
Atomiser (application of Bernoulli’s theorem)
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 2
Atomizer is used for getting a fine spray of perfumer or insecticide. It consists of a cylinder with a piston. A small vessel containing liquid to be sprayed is attached to the cylinder. When the piston is moved forward air is blown out through a small opening of the cylinder.

As the velocity of flow of air increases, the pressure at the opening decreases. Due to the lower pressure at the opening, the liquid rises through the narrow tube and gets sprayed out along with air.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 4.
Surface tension is numerically equal to the surface energy.

  1. Difine surface tension.
  2. Derive an expression for the rise of liquid in a capillary tube.

Answer:
1. Liquids acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension.

2. When a drop is split into tiny droplets, the surface area increases. So work has to be done for splitting the drop. Let R be radius of the drop and r the radius of the droplets: R = 1 × 10-3m
surface area of the drop = 4πR2
= 4π × (1 × 10-3)2
= 4π × 10-6m2
Volume of the drop = Volume of 106 droplets
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 3
∴ Surface area of million droplets =106 × 4π2
= 106 × 4π(1 × 10-5)2
= 4π × 10-4 m2
∴ Increase in surface area = 4π × 10-4 – 4π × 10-6
= 3.96π × 10-4m2
∴ Energy expended = 3.967π × 10-4 × S
= 3.96π × 10-4 × 72 × 10-3J
= 8.95 × 10-5J.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 5.
Match the following

1. Pascal’s lawa. \( \sqrt{2 g h}\)
2. Bernoulli’s theoremb. a1v1= a2v2
3. Surface tensionc. Hydraulic jack
4. Velocity of effluxd. Reynolds number
5. Equation of continuitye. Angle of contact
6.  Viscosityf. Ventiurimeter

Answer:
1 – C, 2 – f, 3 – e, 4 – a, 5 – b, 6 – d.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 6.
Give reasons for the following cases.

  1. It is easier to swim in sea water than in river walls.
  2. The passangers are advised to remove ink from pen while going up in an aeroplane.

Answer:
1. The density of sea water is more than that of river water due to the presence of salt. Hence sea water offers more upthrust and only a very small portion of human body will be in sea water compared to river water.

2. In ink pen, ink is filled in atmospheric pressure. As we go higher pressure decreases and hence ink will have a tendancy to come out in order to equalise the pressure.

Plus One Physics Mechanical Properties of Fluids Four Mark Questions and Answers

Question 1.
A large tank containing water has a small hole near the bottom of the tank 1.5 m below the surface of water.

  1. What is the velocity of the water flowing from the hole?
  2. Explain the principle used in deriving the velocity of water flowing from the hole.
  3. Where must a second hole to be drilled so that the velocity of water leaving this hole is half of water flowing through the first hole.

Answer:
1. Velocity of water flowing through the hole
u = \( \sqrt{2 g h}\)
= \(\sqrt{2 \times 10 \times 1.5}\) = 5.47m/s.

2. Bernoulli’s theorem
As we move along a streamline the sum of the pressure (p), the kinetic energy per unit volume \(\frac{\rho v^{2}}{2}\) and the potential energy per unit volume (ρgh) remains a constant.

(OR)

Mathematically Bernoulli’s theorem can be written as
P + \(\frac{1}{2}\)ρv2 + ρgh = constant.

3.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 4
h2 = \(\frac{1.5}{4}\) m = 0.375 m, from the top side of tank.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 2.
Rain drops have an average size of 1 mm when it is formed at the upper atmosphere.

  1. Why the velocity of the rain drop is uniform?
  2. Derive an expression for the terminal velocity of the drop in terms of coefficient of viscosity of air.
  3. If the size of the rain drop become half, then what happens to its terminal speed?

Answer:
1. Due to viscous force acting on the raindrop, it moves with uniform speed.

2. Viscous force, boyancy force and weight of the body
Expression for terminal velocity:
Consider a sphere of radius ‘a’ densitity σ falling through a liquid of density a and viscocity η. The viscous force acting on the sphere can be written as
F = 6πaηv
Where v is the velocity of sphere. This force is acting in upward direction. When the viscous force is equal to the weight of the body in the medium, the net force on the body is zero. It moves with a constant velocity called the terminal velocity.
The weight of a body in a medium,
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 5
When body has terminal velocity, we can write.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 6

Question 3.

  1. Fill in the blanks using the word from the list appended with each statement.
    • Viscosity of gases_____with temperature (increase/decrease)
    • For a fluid in steady flow, the increase in flow speed at a constriction follows from_____ (conservation of mass/Bernoulli’s theorem)
    • The working of a hydraulic lift is based on (Pascal’s Law/ principle of Conservation of Energy)
    • Small insects can walk over the surface of water. It is due to the_____(surface tension of water/viscosity of water)
  2. A girl dips a thin capillary tube in water. Water rises through it.
    • Name the phenomenon.
    • How does this rise vary with the diameter of the tube?

Answer:
1. Fill in the blanks :

  • increases
  • Conservation of mass
  • Pascals law
  • Surface tension

2. A girl dips a thin capillary tube in water:

  • Capillary rise
  • h α \(\frac{1}{r}\) ie. when diameter of tube increases, the. capillary rise decreases.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 4.
The schematic diagram of a sprayer or atomiser is given below.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 7

  1. Name the principle of working of this device from the following:
    • Surface tension
    • Viscosity
    • Bernoulli’s principle
    • Archimedes’ principle
  2. Write its mathematical expression.
  3. Wings of an aeroplane are curved outwards while flattened inwards. Why?

Answer:
1. Bernollis principle

2. P + \(\frac{1}{2}\)ρv2 + ρgh = constant.

3. When the aeroplane moves forward, the air blown in the form of stream lines over the wings of aeroplane is shown figure.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 8
As the upper surface of wing is more curved than its lower surface, the speed of air above the wings is larger than the speed of the air below the wings.

Hence the pressure above the wings becomes less than the pressure below the wings. Due to this pressure difference the aeroplane will get upward force to overcome gravitational force.

Question 5.
During windstorms, roofs of certain houses are blown off without damaging other parts of the houses.

  1. Name the theorem which explains this phenomenon.
  2. State the theorem.
  3. Explain this phenomenon on the basis of this theorem.

Answer:

  1. Bernoulli’s theorem
  2. For a small amount of liquid in stream line flow, between two points, the total energy is constant.
  3. When windstorm blown off, the pressure on the top side of roof decreases. Hence a pressure difference is developed in between roof. Due to this pressure difference, roof of certain houses are blown off.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 6.
Two thin evacuated (one end closed) glass take A and B are carefully immersed in a beaker containing mercury such a way that there is no chance to get air in to the tubes. A is stand vertically and B is making an angle θ with the vertical.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 9

  1. Is any rise of mercury in the tubes?
  2. Is any height difference of mercury levels in tube A and B? Justify your answer.
  3. When the doctors are measuring body pressure, it is advisable to lie on a table. Why?

Answer:

  1. Yes
  2. No. Pressure is same at same level. To get same pressure, height of mercury becomes same.
  3. When we lie on the table, the pressure of our body will be same at all points.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 7.
A small metal sphere is falling through a caster oil.

  1. Name the forces acting on the metal sphere?
  2. Which of these forces change? Why?
  3. Name the velocity of the sphere when the unbalanced force on it is zero?
  4. Write down the expression for this velocity in terms of coefficient of viscosity?

Answer:

  1. forces acting on the metal sphere:
    • Weight of the body (mg)
    • Buoyant force or up thrust
    • Viscous force
  2. Viscous force. Viscous force is the friction offered by the liquid. It is a self adjusting force.
  3. Terminal velocity
  4. Terminal velocity, V = \(\frac{2}{9} a^{2}\left(\frac{f-N}{\eta}\right) g\).

Plus One Physics Mechanical Properties of Fluids Five Mark Questions and Answers

Question 1.
A capillary tube when dipped into water, it is commonly observed that water will rise through the tube.

  1. Which of the following is responsible for this?
    • Gravitational force
    • Viscous force
    • Nuclear force
    • Surface tension
    • Elastic force
  2. Derive an expression for the capillary rise.
  3. If the radius of the tube becomes doubled, then what happens to the height of water column in the tube?

Answer:
1. Surface tension.

2.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 10
Consider a capillary tube of radius ‘a’ dipped in a liquid of density ρ and surface tension S. If the liquid has a concave meniscus it will rise in the capillary tube. Let h be the rise of the liquid in the tube. Let p1 be the pressure on the concave side of the meniscus and p0, that on the other side. The excess pressure on the concave side of the meniscus can be written as
p1 – p0 = \(\frac{2 \mathrm{S}}{\mathrm{R}}\)
Where R is the radius of the concave meniscus. The tangent to the meniscus at the point A makes an angle θ with the wall of the tube.
In the right angled triangle ACO
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 11
substituting the values of R in the equation (1)
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 12
Considering two points M and N in the same horizontal level of a liquid at rest,
pressure at N = pressure at M
But pressure at M = pi, the pressure over the concave meniscus and pressure at N = po + hρg
∴ Pi = Po + hρg
or Pi – Po = hρg ……..(3)
From equations (2) and (3), we get
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 13

3. We know
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 14
The capillary rise decreases to half of the original value.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 2.

  1. Find the odd one out and justify your answer Atomiser, venturi meter, aeroplane, hydraulic lift
  2. Mention one use of venturi meter.
  3. Explain the working of the odd one which you have selected in question (a)

Answer:
1. Hydraulic lift – It is based on pascals law.

2. Venturimeter can be used to find velocity of flow of fluid through a pipe.

3.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 15
A hydraulic lift is used to lift heavy load. Consider a liquid enclosed in a vessel with two cylinders C1 and C2 attached as shown in the figure. The cylinders are provided with two pistons having areas A1 and A2 respectively.
If F1 is the force exerted on the area A1,
pressure P1 = \(\frac{F_{1}}{A_{1}}\).
If F2 is the force exerted on the area A2,
pressure P2 = \(\frac{F_{2}}{A_{2}}\).
According to pascal’s law P1 = P2.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 16
Using this method we can lift heavy load by applying small force.

Question 3.
When a capillary tube of radius ‘r’ is dipped in water, the water rises through it up to height ‘h’.

  1. Which of the following is responsible for the above phenomenon?
    • Viscous force
    • elastic force
    • surface tension
    • gravitational force
    • negative force
  2. To what height will water rise in a glass tube with a bore of radius 0.1 mm (take the angle of contact of glass with 0°, surface tension S = 0.0728 N/m)
  3. If the length of tube is less that the length of capillary rise, will it overflow. Justify your answer.

Answer:
1. Surface tension

2.
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 17

3. The water will never overflow. If the tube is of insufficient length, the radius of curvature of liquid meniscus goes on increasing, making it more and more flat till water is in equilibrium.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 4.
A steel ball of radius 1 mm is falling vertically through a tank of oil at 30°C.

  1. After some time the ball attains a constant velocity called_____
  2. What are the forces acting on the ball and give their directions?
  3. Write down the expression for resultant force acting on the ball?)
  4. If the density of oil is 2 × 103kg/m3, density of steel is 8 × 102 Kg/m3 and ‘η’of oil 2NS/m2, What will be the constant velocity attained by the ball?

Answer:
1. Terminal velocity.

2. Weight of body (down ward), bouyanant force (up ward), Viscous force (upward).

3. Resultant force = weight of body – buoyant force.

4. Terminal velocity,
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 18

Plus One Physics Mechanical Properties of Fluids NCERT Questions and Answers

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Force, F = Weight of girl
= mg = 50 × 9.8N = 490N
Radius, r = 0.5 × 10-2m
Area A = πr2 = \(\frac{22}{7}\)(0.5 × 10-2)2 m2
Pressure
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 19

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
p = hρg, h = \(\frac{p}{\rho g}=\frac{1.01 \times 10^{5}}{984 \times 9.8}\)m = 10.47m.

Question 3.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 of spirit in the other. What is the specific gravity of spirit?
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 20
Answer:
Since the mercury columns in the two arms are at the same level,
∴ pressure due to water column = pressure due to spirit column
∴ hwρwg = hsρsg
or hwρw = hsρ
But hw = 10 cm,
ρw = 1 gcm-3,
hs = 12.5cm
∴ 10 × 1 = 12.5 × ρs
or ρs = \(\frac{10}{12.5}\)gcm-3
= 0.8cm-3
∴ Specific gravity of spirit = 0.8.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 4.
Figs, (a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why?
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 21
Answer:
Fig (a) is incorrect. This is because at a constriction (ie., where the area of cross-section of the tube is smaller), the flow speed is larger due to mass conservation. Consequently, pressure there is smaller according to Bernoulli’s equation. We assume the fluid to be incompressible.

Question 5.
What is the prssure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10-1Nm-1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.
Answer:
Excess pressure = \(\frac{2 \sigma}{R}\)
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 22
Total pressure = 1.01 × 105 + \(\frac{2 \sigma}{R}\)
= 1.01 × 105 + 310
= 1.0131 × 105Pa
Since data is correct upto three significant figures. We should write total pressure inside the drop as 1.1 × 105Pa.

Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids

Question 6.
During blood transfusion, the needle is inserted in a vein where the guage pressure is 2000 Pa, at what height must the blood container be placed so that blood may just enter the vein? Given: density of whole blood = 1.06 × 103kgm-3
Answer:
Guage pressure,
p = hρg, h
Plus One Physics Chapter Wise Questions and Answers Chapter 10 Mechanical Properties of Fluids - 23
= 0.19m.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Students can Download Chapter 8 Arrays Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Plus One Arrays One Mark Questions and Answers

Question 1.
From the following which is not true for an array.
(a) It is easy to represent and manipulate array variable
(b) Array uses a compact memory structure
(c) Readability of program will be increased
(d) Array elements are dissimilar elements
Answer:
(d) Array elements are dissimilar elements

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 2.
Consider the following declaration.
int mark(50) Is it valid? If no give the correct declaration
Answer:
It is not valid. The correct declaration is as follows, int mark[50]. Use square brackets instead of parenthesis

Question 3.
Consider the following declaration.
int mark[200]
The index of the last element is ________.
Answer:
199

Question 4.
Consider the following declaration int mark[200]. The index of the first element is ________.
Answer:
0

Question 5.
Consider the following int age[4] = {15,16,17,18};
From the following which type of initialisation is this.
(a) direct assignment
(b) along with variable declaration
(c) multiple assignment
(d) None of these
Answer:
(b) along with variable declaration

Question 6.
From the following which is used to read and display array elements.
(a) loops
(b) if
(c) switch
(d) if else ladder
Answer:
(a) loops

Question 7.
Write down the corresponding memory consumption in bytes

  1. int age[10] = ——-
  2. charname[10] = ——-
  3. int age[10][10]= ——–

Answer:

  1. 2 × 10 = 20 bytbs (2 bytes for one integer)
  2. 1 × 10 = 10 (one byte for each character)
  3. 2 × 10 ×10 = 200 (2 × (100 elements))

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 8.
A two dimensional array is having ________ number of subscript
(a) one
(b) two
(c) three
(d) none of these
Answer:
(b) two

Question 9.
Consider the following
int age[10][10];
In this array how many elements will contain.
Answer:
10 × 10 = 100 elements

Question 10.
Consider the following int age[4] = {12, 13, 14};
cout<<age[3];
What will be the output?
(а) 14
(b) 12
(c) 13
(d) 0
Answer:
(d) 0

Question 11.
The elements of 2 dimensional array can be read using _________ loop.
Answer:
nested loop

Question 12.
________ is the process of reading/visiting elements of an array.
Answer:
traversal

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 13.
Anjali wants to read the 10 marks that already stored in an array and find the total. This process is known as ________.
(a) insertion
(b) deletion
(c) traversal
(d) linear search
Answer:
(c) traversal

Question 14.
Mani wants to check a number from 100 numbers already stored in an array. This process is called ___________.
(a) insertion
(b) deletion
(c) searching
(d) None of these
Answer:
(c) searching

Question 15.
Divide and conquer method used in ____________ search.
Answer:
binary

Question 16.
“Each element of the array is composed with value to be searched from the beginning of the array”. This method is adopted by _____________ search.
Answer:
linear search

Question 17.
____________ search requires a sorted array as input.
Answer:
binary search

Question 18.
___________ searching is slower for larger array.
Answer:
linear searching

Question 19.
____________ is a simple sorting algorithm.
(a) bubble sort
(b) selection sort
(c) linear search
(d) binary search
Answer:
(a) bubble sort

Question 20.
Adjacent elements are checked and inter changed in _____________ sort.
Answer:
bubble sort

Question 21.
The array is divided into sorted part and unsorted part in ____________ sort.
selection.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 22.
The elements of an array of size ten are numbered from _____ to _______.
Answer:
0 to 9

Question 23.
Element mark[6] is which element of the array?
(a) The sixth
(b) the seventh
(c) the eighth
(d) impossible to tell
Answer:
(b) the seventh

Question 24.
When a multidimensional array is accessed, each array index is ________.
(a) Separated by column.
(b) Surrounded by brackets and separated by commas.
(c) Separated by commas and surrounded by brackets.
(d) Surrounded by brackets,
Answer:
(d) surrounded by brackets

Question 25.
You have used a 2D array with the Name Mat representing a matrix. Write the C++ expression to access the 3rd element in the 2nd row.
Answer:
mat[1] [2];

Question 26.
Write a C++ statement that defines a string variable called ‘name’ that can hold a string of upto 20 characters.
Answer:
char name[21j;

Question 27.
Given some array declaration. Pick the odd man out.
Float a[+40], int num[0-10], double [50], char name[50], amount[20] of float.
Answer:
char name[50]. It is a valid array decalaration the remaining are not valid.

Question 28.
__________ is a collection of elements with same datatype.
Answer:
Array

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 29.
int num[10];
The above C++ statement declares an array named num that can store maximum ________ integer numbers. (SEP-2015 (IMP) (1)
(а) 9
(b) 10
(c) N
(d) none of these
Answer:
(b) 10

Question 30.
Declare a two dimensional array to store the elements of a matrix with order 3 × 5. (MARCH-2016) (1)
Answer:
int m[3] [5]; or float m[3] [5]

Question 31.
_________ search method is an example for ‘divide and conquer method’. (SEP-2016) (1)
Answer:
Binary

Question 32.
Read the following C++ statement:
int AR[10]
How many bytes will be allocated for this array? (1)
Answer:
To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes

Question 33.
Find the value of score [4] based on the following declaration statement. int score [5] = {988,87,92,79,85}; (MARCH-2017) (1)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 1
Hence (158)10 = (9E)16

Plus One Arrays Two Mark Questions and Answers

Question 1.
Whether the statement char text[] = “COMPUTER”; is True/False? Justify.
Answer:
It is a single dimensional array. If the user doesn’t specify the size the operating system allocates the number of characters + one (for null character for text) bytes of memory. So here OS allocates 9 bytes of memory.

Question 2.
Given a word ” Mathematics”. How will you locate this word in a standard dictionary. What are the steps used for this purpose
Answer:
In the standard dictionary the words are arranged in sorted order so binary search method is more suitable to find the word “MATHEMATICS”.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 3.
How can you find out the word “MATHEMATICS” from a text book. Which is the most convenient method? Justify your answer.
Answer:
In a text book the words are not arranged in sorted order so linear search method is convenient.

Question 4.
Consider the statement char str[] = “PROGRAM” What will be stored in last location of this array. Justify.
Answer:
The last location is the null character(\0) because each string must be appedend by a null character.

Question 5.
Explain different array operations in detail.
Answer:

  1. Traversal: All the elements of an array is visited and processed is called traversal
  2. Search: Check whether the given element is present or not
  3. Sorting: Arranging elements in an order is called sorting

Question 6.
Read the following C++ statement: int MAT[5][4];

  1. How many bytes will be allocated for this array?
  2. Suppose MAT[4][4] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the’ sum of all the elements in the array. (MARCH – 2015) (1)

Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 7.
Declare an array of size 5 and intitialize it with the numbers 8, 7, 2, 4 and 6. (MARCH-2016) (2)
Answer:
int n[5] = {8, 7, 2, 4, 6};

Question 8.
Predict the output of the following C++ program.
#include <iostream.h>
int main()
{
int array[] = {1, 2, 4, 6,7,5};
for (int n =1; n<=5; n++)
array [n] = array [n – 1 ];
for (n = 0; n <= 5; n++)
cout<< array [n];
return 0;
}
Answer:
1, 1, 1, 1, 1, 1

Question 9.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (SCERT SAMPLE – II) (2)
Answer:
for (i = 0; i < 5, i++)
for (j = 0; j < 5; j++)
if (i == j)
S = S + M[i][j];

Question 10.

  1. Write C++ program for sorting a list of numbers using Bubble Sort Method. (2)
  2. Write the different passes of sorting the following numbers using Bubble Sort.
    32, 21, 9, 17, 5

Answer:
1. Example for Bubble sort is given below
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 3

2. Bubble sort (ascending order):
Consider the first 2 elements of the numbers and compare it. 32 is greater than 21 then interchange both of them. Then the numbers are
21, 32, 9, 17, 5
Next compare 32 and 9. Here 32 is greater so interchange both of them. Then the numbers are
21, 9, 32, 17, 5
Next compare 32 and 7. Then interchange and the numbers are
21, 9, 17, 32, 5
Next compare 32 and 5. Then interchange and the numbers are
21, 9, 17, 5, 32.
After the first phase the largest number is on the right side. Similarly do the remaining. At last we will get the result as follows.
5, 9, 17, 21, 32

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 11.
Suppose M[5] [5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (MARCH-2017)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 4

Plus One Arrays Three Questions and Answers

Question 1.
Suppose you are given Total mark of 50 students in a class

  1. How will you store these values using ordinary variable?
  2. Is there any other efficient way to store these values? Give reason.

Answer:
We have to declare 50 variables individually to store total marks of 50 students. It is a laborious work. Hence we use array, it is an efficient way to declare 50 variables. With a single variable name we can store multiple elements.
eg: int mark[50]. Here we can store 50 marks. The first mark is in mark[0], second is in mark[1]…. etc the fiftieth mark is in mark[49],

Question 2.
Four friends are in a queue for watching a movie. How can you locate a particular person from that queue? Write all the steps sequentially.
Answer:
Locate a particular person from a queue can be carried out by using two ways.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 3.
Total mark of 50 students in a class are given in an array. A bonus of 10 marks is awarded to all of them. Write the program code for making such a modification.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 5

Question 4.
In School Assembly, try to arrange all the students in your class according to the increasing order of their heights separately for male/female students. Smallest one will be in front and tallest one at the last position. Write sequential steps for doing this, in two different methods.
Answer:
Arranging elements from smallest to largest (ascending) or largest to smallest(descending) is called sorting. There are two sorting methods
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjacent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 5.
Using the concept obtained from algorithms in question number (9), write complete program for the two types of sorting.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 6
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 7
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 8

Question 6.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose these male and female students be seated in two separate benches and you are given a place which is used for sitting these 12 students in linear form. How will you combine and make them sit without mixing male/female students. Write a program for the same.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 9
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 10

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 7.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose you are given the heights according to a sorted order separately for male/female students. How will you combine these group according to the same order in linear form. Write a program for the same.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 11
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 12

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 8.
Write the program code for counting the number of vowels from your school name.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 13
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 14

Question 9.
Write the program code for counting the number of words from the given string.” Directorate of Higher Secondary Examination”
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 15
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 16

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 10.
Given a word like “ECNALUBMA” Write the program code for arranging it in into a meaningful word.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 17

Question 11.
Explain the needs for arrays Array
Answer:
Array is collection of same type of elements. With the same name we can store more elements. The elements are distinguished by using its index or subscript. To store 50 marks of 50 students we have to declare 50 variables, it is a laborious work.

Hence the need for arrays arise. By using array this is very easy as follows int mark[50]. Here the index of first element is 0, then 1, 2, 3, etc upto 49.

Question 12.
Explain different types of arrays.
Answer:
There are two tyes of array
1. Single dimensional:
It contains only one index or subscript. The index starts from 0 and ends with size-1.
eg: int n[50];
char name[10];

2. Multi-dimensional:
It contains more than one index or subscript. The two dimensional array contains two indices, one for rows and another for columns. The row index starts from 0 and end at row size-1 and column index starts at 0 and ends at colunn size-1.
eg: int n[10][10] can store 10 × 10 = 100 elements. The index of the first element is n[0][0] and index of the 100th element is n[9][9],

Question 13.
Write a program to read the 5 marks of a students and display the marks and total
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 18

Question 14.
Write a program to read 3 marks of 5 students and find the total and display it.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 19

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 15.
Differentiate linear search and binary search
Answer:
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is re¬turned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 16.
Differentiate bubble sort and selection sort.
Answer:
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest (in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 17.
Write a program to read a string and a character and find the character by using linear search.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 20

Question 18.
Write a program to read marks of 10 students and read a mark and check whether it is in the array or not using binary search
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 21
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 22
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 23

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 19.
Write a program reverse a string without using a function.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 24

Question 20.
Write a program to read a string and find the no. of vowels .consonents and special characters.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 25

Question 21.
Given a word “COMPUTER”, write a C++ program to reverse the word without using any string functions.
Answer:
#include<iostream>
using namespace std;
int main(()
{
char a[9] = “COMPUTER”;
inti;
for (i=8; i>=0; i- -)
{
cout<<a[i];
}
}

Question 22.
Write a program to accept marks of 10 students and find out the largest and smallest mark from the list.

OR

C++ program to store the scores of 10 batsmen of a school cricket team and find the largest and smallest score.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 26

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 23.
Write the names of two searching methods in arrays. Prepare a chart that shows the comparisons of two searching methods. (MARCH-2015)
Answer:
It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 24.
Write a C++ program to illustrate array traversal. (SAY-2015 (IMP) (3)
Answer:
#include<iostream>
using namespace std;
int main()
{
int i, mark[50];
for(i=0;i<50; i++)
{
cout<<“Enter mark “<<i + 1<<“: cirt>>mark[i];
}
cout<<“\nThe mark of 50 students after adding bonus mark 10 is given below \n”; for(i=0;i<50; i++)
cout<<mark[i] + 10<<endl;
}

Question 25.
Define an array. Also write an algorithm for searching an element in the array using any one method that you are familiar with. (MARCH-2016) (3)
Answer:
An array is a collection of elements with same data type. The index of first element is 0 (zero) and the index of last element is size -1. There are 2 methods linear search and binary search.

Searching: It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.8.3 Two dimensional (2D) arrays.

Some occasions we have to store 6 different marks of 50 students. For this we use 2D arrays. An array with two subscripts is used, eg. int mark[r][c]; Here r is the row and c is the column.

Question 26.

  1. Finding sum of all elements in an array is an example of _________ operation.
  2. If 24, 45, 98, 56, 76, 24, 15 are the elements of an array, illustrate the working of selection sort for arranging the elements in descending order. (SCERT SAMPLE -1) (3)

Answer:
1. Traversal

2. In selection sort the array is divided into two parts the sorted part and unsorted part. First smallest elemant in the unsorted part in searched and exchanged with the first elemant now there is 2 parts sorted part and unsorted part. This process continues.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 27

Question 27.
Read the following C++ statement:

  1. How many bytes will be allocated for this array?
  2. Write algorithm to sort this array using bubble sort method. (SCERT SAMPLE – II) (3)

Answer:
1. To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes.

2. Sorting – Arranging elements of an array in an order(ascending or descending).
(a) Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

(b) Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 28.
Explain ‘Call by Value’ and ‘Call by Reference’ methods of function calling with the help of a suitable example. (SEP-2016) (3)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 28

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 29
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 30

Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Question 29.
What is an array? Write C++ program to declare and use a single dimensional array for storing the computer science marks of all students in your
class. (SEP-2016) (3)
Answer:
An array is a collection of elements with same data type or with the same name we can store many elements, the first or second or third etc can be distinguished by using the index(subscript). The first element’s index is 0, the second elements index is 1, and so on.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 31

Question 30.
Write an algorithm for arranging elements of an array in ascending order using bubble sort. (MARCH-2017) (3)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 32
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 33

Plus One Arrays Five Mark Questions and Answers

Question 1.
Write a program to perform matrix addition, multiplication, subtraction.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 34
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 35
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 36
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 37
Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays 38

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Students can Download Chapter 9 Hydrogen Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Plus One Chemistry Hydrogen One Mark Questions and Answers

Question 1.
In which of the following compounds does hydrogen have an oxidation state of -1?
a) CH4 b) NH3 C) HCl d) CaH2
Answer:
d) CaH2

Question 2.
The radio active isotope of hydrogen is __________ .
Answer:
Tritium

Question 3.
Temporary hardness of water is due to the presence of
a) MgSO4
b)Ca(HCO3)2
c) CaSO4
d) NaHCO3
Answer:
b) Ca(HCO3)2

Question 4.
D2O is used as __________ in nuclear reactors.
Answer:
Moderator

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 5.
30 volumes of H2O2 means
a) 30% H2O2 solution
b) 30 cm³ of the solution contains 1 g of H2O2
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP
d) 30 cm³ of the solution contains 1 mole of H2O2
Answer:
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP

Question 6.
Name the three isotopes of hydrogen.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 1

Question 7.
Dihydrogen is prepared on industrial scale from syngas by
Answer:
Water gas shift reaction

Question 8.
Among the following elements which does not make a hydride is _________
a) Ti
b) Mg
c) Co
d) Pd
Answer:
c)Co

Question 9.
Hydrogen is purified by _________ .
Answer:
Oculusion on pd

Question 10.
H2O2 is _________ .
Answer:
Diamagnetic

Question 11.
Ortho & para hydrogen are __________ .
Answer:
Nuclear spin isomers

Plus One Chemistry Hydrogen Two Mark Questions and Answers

Question 1.
Write one method each for the laboratory preparation of dihydrogen from
i) mineral acid
ii) aqueous alkali.

  1. Which is the catalyst used for the reaction?
  2. Which is the product in this reaction?

Answer:
1.By the reaction of granulated zinc with dilute HCl.
Zn + 2HCl → ZnCl2 + H2

2. By the reaction of zinc with aqueous alkali.
Zn + 2NaOH → Na2ZnO2 + H2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 2.
Classify the following into those causing temporary hardness and permanent hardness:
[Mg(HCO3)2, MgCl2, CaCO3, CaSO4, NaCl, NaHCO3, Ca(HCO3)2]
Answer:
Temporary hardness – Mg (HCO3)2, Ca(HCO3)2
Permanent hardness – MgCl2, CaSO4

Question 3.
The bond angle in water is different from the tetrahedral bond angle.

  1. What is the bond angle and shape of water?
  2. Justify.

Answer:
1. 104.5°

2. There are three types of repulsions in water. They are: Ip – Ip repulsion, Ip – bp repulsion and bp – bp repulsion. In order to minimize the stronger Ip- Ip repulsion, the bond angle is reduced to 104.5° from 109.5°. Thus, the shape of water is distorted tetrahedral or angular.

Question 4.
Among NH3, H2O and HF which would you expect to have highest magnitude of hydrogen bonding? Why?
Answer:
Strength of H-bond depends upon the atomic size and electronegativity of the other atom to which H- atom is covalently bonded. Smaller size and higher electronegativity favour H-bonding. Among N, F and O atoms, F is the smallest and its electronegativity is highest. Hence, HF will have highest magnitude of H-bonding.

Question 5.
1. Soap does not give lather with hard water. Why?
2. What are the disadvantages of hard water?
Answer:
1. Hard water contains Ca+, Mg2+ions in the form of
their bicarbonates, chlorides or sulphates. Hard water forms scum/precipitate with soap. So soap does not give lather with hard water,

2. Hard water is unsuitable for laundry. It is harmful for boilers because of deposition of salts in the form of scale. This reduces the’ efficiency of the boiler.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Write one example each for the oxidising action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium, H2O2 oxidises PbS to PbSO4.
PbS(s) + 4H2O2(aq) → PbSO4(s) + 4H2O(l)
In basic medium, H2O2 oxidises Fe2+ to Fe3+.
2Fe2+ + H2O2 → 2Fe3+ + 2OH

Question 7.
Write one example each for the reducing action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium H2O2 reduces MnO4 to Mn2+.
2MnO4 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
In basic medium H2O2 reduces l2 to l.
l2 + H2O2 + 2OH → 2l + 2H2O + O2

Question 8.
Write two examples for redox reactions involving water.
Answer:
1. Water can be easily reduced to dihydrogen by highly electropositive metals like Na. Here, Na is oxidised to NaOH.
2H2O(I) + 2Na(s) → 2NaOH(aq) + H2(g)

2. With F2, water is oxidised to O2. Here, F2 is reduced to F-.
2F2(g) + 2H2O(l) → 4H+(aq) + 4F(aq) + O2(g)

Question 9.
Distinguish between

  1. Hard and Heavy water
  2. Temporary and permanent hardness of water.

Answer:
1. A sample of water said to be hard water if it does not give lather with soap. Heavy water is the oxide of deuterium (D2O).

2. Temporary hardness is due to the presence of bicarbonate of calcium or magnesium. It can be removed by boiling.

Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium. It is not removed by boiling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 10.
Hydrogen combines with elements to give binary compounds known as hydrides.
1. Name the three categories of hydrides.
2. Classify the given hydrides into different categories: NH3 LiH, TiH, CH4, NaH
Answer:
1. Ionic hydrides, Covalent hydrides and Metallic/ Interstitial hydrides

2. NaH, LiH – Ionic hydrides
NH3, CH4 – Covalent hydrides
TiH – Interstitial hydride

Plus One Chemistry Hydrogen Three Mark Questions and Answers

Question 1.
A chart prepared by a student based on the similarities of hydrogen with alkali metals and halogens is as shown below. Correct the mistakes in it.

Similarities with alkali metalsSimilarities with halogen
-1 oxidation state Reducing agent+ 1 oxidation state High Ionisation energy
Diatomic stateDuring electrolysis, both of them are produced at the cathode
Non metals

Answer:

Similarities with alkali metalsSimilarities with halogen
+1 oxidation state Reducing agent-1 oxidation state High Ionisation energy
During electrolysis, both of them are produced at the cathodeDiatomic state
Non metals

Question 2.
Prepare a short note on different types of hydrides.
Answer:
Hydrogen reacts with metals or non metals to form binary compounds known as hydrides.
Hydrides are mainly classified into the following 3 types:
1. Ionic or salt like hydrides
These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. These are crystalline, non-volatile and non-conducting in solid state. But their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode. They react violently with water producing dihydrogen gas. e.g. NaH, KH.

2. Covalent hydrides or Molecular hydrides
These are formed by the action between dihydrogen and non metals (p-block elements). Covalent hydrides are classified into electron-deficient (e.g. B2H6), electron-precise (e.g. CH4) and electron-rich hydrides (e.g. NH3).

3. Metallic or Non-stoichiometric or Interstitial hydrides
These are formed by the reaction of dihydrogen with many d-block and f-block elements. These hydrides conduct heat and electricity. They are almost always non- stoichiometric, being deficient in hydrogen, e.g.
LaH2.87,YbH2.55,VH0.56 etc.

Question 3.
Consider the chemical equation and fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 2

  1. This method is used for the preparation of ……………
  2. The electrode used is …………
  3.  ……….. is produced at cathode.

Answer:

  1. Hydrogen
  2. Platinum
  3. Hydrogen

Question 4.
Analyse the equation: 2Na + H2 → 2NaH

  1. In this chemical equation, H2 reacts with …………
  2. Hydrogen reacts with metals to form …………
  3. Hydrogen is in ………… oxidation state in NaH.

Answer:

  1. Na
  2. Metal hydrides
  3. -1

Question 5.
a) How is hydrogen of high purity prepared?
b) Dihydrogen is relatively inert at room temperature. Give reason.
c) Write any two uses of hydrogen.
Answer:
a) High purity (>99.95%) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide . solution bewteen nickel electrodes.
b) This is due to high H-H bond enthalpy.
c) 1. In oxy-hydrogen torches.
2. Liquid hydrogen is used as a fuel in rockets.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
A sample of cold river water does not easily give lather with soap, but on boiling it does.

  1. Evaluate and write the chemical equation involved.
  2. In some cases, the water does not give ready lather even if it is boiled. Why?

Answer:
1. If the sample of water has temporary hardness, it is removed during boiling. Here, the bicarbonates of magnesium is precipitated as Mg(OH)2 and that of calcium is precipitated as CaCO3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 3

2. The sample of water might have permanent hardness due to the presence of dissolved chlorides and sulphates of Ca and Mg. This cannot be removed just by boiling.

Question 7.
Hydrogen has three isotopes – Protium, Deuterium and Tritium.
a) Of these which is the radio active one?
b) Name a compound which contains the isotope deuterium.
c) Make a table which shows number of protons, neutrons and electrons in each isotope.
Answer:
a) Tritium
b) Heavy water – D2O
c)

IsotopeNo. of ProtonsNo. of NeutronsNo. of Electrons
Protium101
Deuterium111
Tritium121

Question 8.
What are the three types of hydrates? Give examples for each.
Answer:
1. Hydrates with coordinated water, e.g. [Cr(H2O)6]Cl3.
2. Hydrates with interstitial water, e.g. BaCl2.2H2O
3. Hydrates with hydrogen bonded water, e.g. CuSO4.5H2O.

Question 9.
1. What is meant by amphoteric nature of water?
2. Suggest an example to show this property.
Answer:
1. The amphoteric nature of water means that it has the ability to act as an acid as well as a base.

2. Water acts as an acid with NH3 and a base with H2S.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 4

Question 10.
a) What are the advantages of dihydrogen as a fuel?
b) What are the disadvantages of using dihydrogen as a fuel?
c) What is meant by the term ‘hydrogen economy’?
Answer:
a) 1. It is abundantly available in the combined state as water.
2. Use of dihydrogen as fuel provides pollution free atmosphere because its combustion product is only water.
3. Heat of combustion per gram of dihydrogen is more than twice that of jet fuel.

b) 1. Dihydrogen does not occur in free state in nature.
2. Hydrogen gas has explosive flammability which causes problem to its storage and transportation.
3. A cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol containing the same amount of energy.

c) Hydrogen economy refers to the use of dihydrogen as an alternative source of energy. The basic principle of hydrogen economy is storage and transportation of energy in the form of dihydrogen instead of fossil fuels or electric power.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 11.
What do you mean by temporary and permanent hardness? Explain one chemical method each for removing temporary hardness and permanent hardness.
Answer:
Temporary hardness is due to the presence of bicarbonates of calcium or magnesium. Permanent hardness is due to the presence of chlorides and sulphates of calcium or magnesium ions.
Clark’s method :
This method is used for removing temporary hardness. In this method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 5
Treatment with washing soda: This method is used to remove permanent hardness. Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.

Question 12.
a) Write a method for the preparation of H2O2.
b) What is 100 volume H2O2?
c) Draw the structure of H2O2 in gas phase.
Answer:
a) H2O2 is prepared by acidifying barium peroxide and removing excess water by evaporation under reduced pressure.
BaO2.8H2O(s) + H2SO2(aq) → BaSO4(s) + H2O2(aq) + 8H2O(I)

b) A 30% solution of H2O2 is called 100 volume H2O2. It means that one mL of 30% H2O2 solution will give 100 V of oxygen at STP.

c) H2O2 is has non-planar structure as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 7

Question 13.
lon exchange process is commonly employed for large scale production of soft water.

  1. What is the basic principle involved in ion exchange method of softening water?
  2. What are inorganic cation exchangers? Give example.

Answer:
1. Adsorption

2. These are complex inorganic salts like sodium- aluminium silicate NaAlSiO4 which can exchange cations such as Ca2+ and Mg2+ ions in hard water for Na+ ions. e.g. Zeolite.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 14.
1. What is heavy water? Mention one use of heavy water.
2. Explain why hydrogen peroxide is not stored in glass vessels.
3. What is calgon? What is its use?
Answer:
1. D2O. Used as moderator in nuclear reactors.

2. To prevent its decomposition.

3. Calgon is chemically sodium hexametaphosphate (Na6P6O18). It is used to remove Ca2+ and Mg2+ ions of hard water by converting them into soluble complexes.
M2+ + Na4P6O182- → [Na2MP6O18]2- + 2Na+ (M = Mg,Ca)

Question 15.
1. Name the oxide of isotope of hydrogen used in nuclear reactor.
2. What are cation exchange resins? What is their role in removing permanent hardness of water?
Answer:
1. Heavy water (D2O).

2. Cation exchange resins contain large organic molecule with -SO3H group and are water insoluble. Ion exchange resin (RSO3H) is changed to RNa by treating with NaCl. The resin exchanges Na+ ions with Ca2+ and Mg2+ ions present in hard water to make the water soft.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 16.
1. What is permuitit? How is it useful in removing permanent hardness of water?
2. Compare the structures of water and hydrogen peroxide.
Answer:
1. Hydrated sodium aluminium silicate is called permutit (NaZ). When it is added to hard water it exchanges Ca2+ and Mg2+ ions with Na+ ions.
2NaZ(s) + M2+(aq) —> MZ2(s) + 2Na+(aq) M = Ma, Ca)

2. Water molecule has an angular or bent shape. H2O2 has a non-planar structure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 8

Question 17.
1. Give the industrial method of preparation of H2O2.
2. How is heavy water prepared?
3. How is pure de-mineralised water obtained?
Answer:
1. Industrially H2O2 is prepared by the auto-oxidation of 2-ethylanthraquinol.

2. Heavy water is prepared by exhaustive electrolysis of water or as a byproduct in some fertilizer industries.

3. Pure de-mineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In the cation exchange process, the H+ exchanges for Na+, Ca2+, Mg2+ and other cations present in water. This process results in proton release and makes the water acidic.
2RH(s) + M2+(aq) \(\rightleftharpoons \) MR2(s) + 2H+(aq)

In the anion exchange process, the OH exchanges for anions like Cl, HCO3, SO42- etc. present in water.
RNH3+OH(S) + X(aq) \(\rightleftharpoons \) RNH3+X(s) + OH(aq)

The OH ions, thus liberated neutralise the H+ ions set free in the cation exchange process to get pure de-mineralised water.
H+(aq) + OH(aq) → H2O(l)

Question 18.
1. Explain why hydrogen peroxide is stored in coloured plastic bottles.
2. Write any two uses of H2O2.
Answer:
1. In the presence of metal surfaces or traces of alkali (present in glass conatiners), the decomposition of H2O2 (2H2O2 → 2H2O + O2) is catalysed. It is therefore stored in wax-lined glasses or plastic vessels in dark.

2. 1. As a hair bleach and as a mild disinfectant.
2. In the syntheseis of hydroquinone, tartaric acid and certain food products and pharmaceuticals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 19.
1. A sample of hard water was found to loose its hardness on boiling. Name the type of hardness associated with this sample. 0
2. Write the name and formula of four minerals which cause permanent hardness of water.
3. What are electron deficient hydrides? Whether they behave as Lewis acids or Lewis bases? Why?
Answer:
1. Temporary hardness

2. CaCl2, MgCl2, CaSO4, MgSO4

3. These are covalent hydrides having too few electrons for writing conventional Lewis structure. They act as Lewis acids because they can accept electrons, e.g. B2H6

Plus One Chemistry Hydrogen Four Mark Questions and Answers

Question 1.
Match the following:

AB
Temporary hardnessp-block elements
HydridesReducing agent
Permanent hardnessChloride
Alkali metalsBicarbonate

Answer:

AB
Temporary hardnessBicarbonate
Hydridesp-block elements
Permanent hardnessChloride
Alkali metalsReducing agent

Question 2.
Certain samples of water do no produce easy lather with soap.

  1. What is this condition of water called?
  2. Which are two types of this condition?
  3. Suggest two methods to change this condition of water.
  4. What are the problems caused by this condition of water?

Answer:

  1. Hardwater.
  2. Temporary hardness and permanent hardness.
  3. By boiling water and by adding Na2CO3
  4. Wastage of soap and boiler explosion.

Question 3.
Match the following:

1. D2Oa) Hendry Cavendish
2. Hydrogenb) Water gas
3. Ca + H2c) 31H
4. C2H4 + H2d) Heavy water
5. Tritiume) 21H
6. Deuteriumf) CaH2
7. CO + H2g) C2H6
8. CO + Z2h) Producer gas

Answer:
1) – d)
2) – a)
3) – f)
4) – g)
5) – c)
6) – e)
7) – b)
8) – h)

Question 4.
1. A list of compounds are given below:
H2O, HCl, CH4
Construct chemical reactions to show the preparation of H2 from each of the above compounds.
2. What is syn gas? How is it prepared?
3. What is coal gasification?
4. Explain water gas shift reaction.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 9
2. Syn gas is a mixture of CO and H2. It is prepared by passing steam over red hot coke.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 10

3. The process of production of syngas from coal is called coal gasification.

4. The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst. This is called water gas shift reaction.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 11

The CO2 is removed by scrubbing with sodium arsenite solution.

Plus One Chemistry Hydrogen NCERT Questions and Answers

Question 1.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? (2)
Answer:
The various isotopes of hydrogen are:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 12

Question 2.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions? (2)
Answer:
Hydrogen atom has only one electron and thus, to achieve stable inert gas configuration of helium, it shares its single electron with electron of another hydrogen atom to form a stable diatomic molecule. The stability of H2 is further confirmed by the fact, that formation of one mole of gaseous H2 molecules results in the release of 435.8 kJ of energy.
H(g) + H(g) H2(g); ∆H = – 435.8 kJ mol-1

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 3.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain. (2)
Answer:
In some of the transition metal hydrides, hydrogen is absorbed as H-atoms. Due to the inclusion of H- atoms, the metal lattice expands and thus becomes less stable. Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely divided metal. The hydrogen evolved in this manner can be used as a fuel. Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

Question 4.
What causes the temporary and permanent hardness of water? (2)
Answer:
Temporary hardness is caused by presence of bicarbonates of calcium and magnesium, i.e., Ca(HCO3)2 and Mg(HCO3)2 in water whereas permanent hardness is caused by presence of soluble chlorides and sulphates of calcium and magnesium, i.e., CaCl2, CaSO4, MgCl2 and MgSO4 in water.

Question 5.
Write chemical reactions to show amphoteric nature of water. (2)
Answer:
Water is amphoteric in character. It means that it can act as proton donor as well as proton acceptor. When it reacts with acids, (stronger than itself), it behaves as a base. When it reacts with bases (stronger than itself) it acts as an acid.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? (2)
Answer:
Demineralised or distilled water is not useful for drinking purposes because it does not contain even useful minerals. Therefore, to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised or distilled water.

Question 7.
How does H2O2 behave as a bleaching agent? (2)
Answer:
The bleaching action of H2O2 is due to the nascent oxygen which is liberates on decomposition.
H2O2 → H2O + [O]
The nascent oxygen oxidise the colouring matter to colourless products. Hence, H2O2 is used for the bleaching of delicate maerials like ivory, feather, silk, wool, etc.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Students can Download Chapter 10 Functions Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Plus One Functions One Mark Questions and Answers

Question 1.
The process of dividing big programs into small programs are called __________.
Answer:
Modularization

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 2.
The big programs are divided into smaller programs. This smaller programs are called __________.
Answer:
Functions

Question 3.
The execution of the program begins at __________ function.
Answer:
main function

Question 4.
One of the following is not involved in the creation and usage of a user defined function.
(a) Define a function
(b) Declare a function
(c) invoke a function
(d) None of these
Answer:
(d) None of these

Question 5.
The default data type returned by a function is ________.
(a) float
(b) double
(c) int
(d) char
Answer:
(c) int

Question 6.
After the execution of a function, it is returned back to the main function by executing _________ keyword.
Answer:
return

Question 7.
Supplying data to a function from the called func-tion by using _________.
Answer:
parameters (arguments)

Question 8.
_________ keyword is used to give a value back to the called function.
Answer:
return.

Question 9.
_________ key word is used to specify a function returns nothing.
Answer:
void

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 10.
One of the following is not necessary in the function declaration. What is it?
(a) name of the function
(b) return type
(c) number and type of arguments
(d) name of the parameters
Answer:
(d) name of the parameters

Question 11.
A function declaration is also called _________.
Answer:
prototype

Question 12.
Consider the following declaration
int sum(int a , int b)
{
return a+b;
}
From the following which is the valid function call.
(a) n = sum(10)
(b) n = sum(10,20)
(c) n = sum(10,20,30)
(d) n = sum()
Answer:
(b) n = sum(10,20)

Question 13.
The ability to access a variable or a function from some where in a program is called ________.
Answer:
scope

Question 14.
A variable or a function declared within a function is have ______ scope.
Answer:
local

Question 15.
A variable or a function declared out side of all the functions is have __________ scope.
Answer:
global

Question 16.
State True/False

  1. A local variable exist till the end of the program
  2. A global variable destroyed when the sub function terminates

Answer:

  1. False
  2. False

Question 17.
consider the following declaration
int x;
void main()
{
……..
}
Here x is a _____ variable.
Answer:
global

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 18.
consider the following declaration
void main()
{
int x;
}
Here x is a _______ variable
Answer:
local

Question 19.
__________ parameter is used when the function call does not supply a value for parameters
Answer:
default

Question 20.
The parameter used to call a function is called ____________.
Answer:
Actual parameter

Question 21.
The parameters appear in a function definition are ___________.
Answer:
formal parameters

Question 22.
After the distribution of answer scripts, the teacher gives the Photostat copy of the mark list to the students to check the marks. If the students make any change that do not affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by address
(d) none of these
Answer:
(a) call by value

Question 23.
Your class teacher gives you the original mark list to check the mark. If you make any change it will affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by function
(d) none of these
Answer:
(b) call by reference

Question 24.
Consider the following function declaration
int sum(int,a, int b)
{
Body
}
Here the arguments are passed by __________.
Answer:
call by value method

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 25.
Consider the following function declaration
int sum(int &a, int &b)
{
Body
}
Here the arguments are passed by __________.
Answer:
call by reference method

Question 26.
A function calls it self is known as ____________.
Answer:
recursive function

Question 27.
Varun wants to copy a string by using strcpy() function. From the following which header file is used for this?
(a) cstdio
(b) cmath
(c) cstring
(d) cctype
Answer:
(c) cstring

Question 28.
__________ is a named group of statements to perform a job/task and returns a value.
Answer:
Function

Question 29.
In his C++ program Ajith wants to accept a lengthy text of more than one line. Which function in C++ can be used in this situation.
Answer:
gets() function can be used to accept a lengthy text.

Question 30.
A function can call itself for many times and return a result. What is the name given to such a function? (MARCH-2016)
Answer:
Recursive function

Question 31.
How many values can a C++ function return? (SCERT SAMPLE-I) (1)
Answer:
One Value

Question 32.
A function may require data to perform the task assigned to it. Which is the component of function prototype that serves this purpose? (SCERT SAMPLE – II) (1)
Answer:
Arguments (Parameters)

Question 33.
“Arguments used in call statement are formal arguments”. State true or false. (SAY-2016) (1)
Answer:
False

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 34.
Name the built-in function to check whether a character is alphanumeric or not. (MARCH-2017)
Answer:
pisalnum()

Plus One Functions Two Mark Questions and Answers

Question 1.
How does C++ support modularity in programming C++
Answer:
The process of converting big programs into smaller programs is known as modularisation. This small programs are called modules or sub programs or functions. C++ supports modularity in programming called functions.

Question 2.
Consider the following code snippet
charch;
cout << “Enter an alphabet”;
cin>>ch;
cout<<toupper (ch);
What is the output of the above code? Give a sample output.
If the above code is used in a computer that has no cctype file, how will you modify the code to get the same output?
Answer:
It reads a character and convert it into upper case.
eg:
Enter an alphabet: a
The output is A.
If a computer has no cctype header file the code is as follows,
char ch;
cout<< “Enter an alphabet”;
cin»ch;
if (ch >= 97 && ch<<122)
cout<<ch – 32;

Question 3.
The following assignment statement will generate a compilation error.
char str[20];
str = “Computer”
Write a correct C++ statement to perform the same task
Answer:
char str[20] = “Computer”;

OR

char str[20];
strcpy(str,”Computer”); (The header file <string.h> should be included)

Question 4.
float area(const float pi = 3.1415, const float r)
{
r = 10;
return pi*r;
}
Is there any problem? If yes what is it?
Answer:
There is an error. The error is ‘r’ is a constant V must be initialised and cannot be changed during execution.

Question 5.
Match the following

a. strcmp()1. cctype
b. tolower()2. iomanip
c. sqrt()3. iomanip
d. abort()4. cstdlib
e. setw()5. cmath

Answer:
a. 3
b. 1
c. 5
d. 4
e. 2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 6.
What are the jobs of a return statement in a program?
Answer:
In the case of a sub function a return statement helps to terminate the sub function and return back to the main function or called function. But in the case of a main function it terminates the program.

Question 7.
How to invoke a function in C++ program?
Answer:
function can be called or invoked by providing the name of the function followed by the arguments in parenthesis.
eg: sum(m,n);

Question 8.
Briefly explain constant arguments
Answer:
By using the keyword const we can make argument (parameter) of a function as a constant argument. The value of the const argument cannot be modified within the function.

Question 9.
Short notes on header files
Answer:
A header file is a prestored file that helps to use some operators and functions. To write C++ pro-grams the header files are must. Following are the header files
alloc
iostream
iomanip
cstdio
cctype
cmath
cstring
The syntax for including a header file is as follows
#include<name of the header file>
eg: #include<iostream>

Question 10.
Identify the appropriate header files for the following.

  1. setw()
  2. cout
  3. toupper()
  4. exit()
  5. strcpy()

Answer:

  1. setw() – iomanip.h
  2. cout – iostream
  3. toupper() – cstring
  4. exit() – process
  5. strcpy() – cstring

Question 11.
Construct the function prototypes for the following functions.

  1. The function Display() accepts one argument of type double and does not return any value.
  2. Total () accepts two arguments of type int, float respectively and returns a float type value. (MARCH-2015) (2)

Answer:

  1. void Display(double);
  2. float Total(int, float);

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 12.
Explain two types of variable according to its scope and life. (SAY-2015) (2)
Answer:
Scope and life of variables and functions
1. Local scope:
A variable declared inside a block can be used only in the block. It cannot be used any other block.
eg:
#include<iostream>
using namespace std;
int sum(int n1 ,int n2)
{
int s;
s = n1 + n2;
return (s);
}
int main()
{
int n1 ,n2;
cout<<“Enter 2 numbers cin>>n1>>n2;
cout<<“The sum is “<<sum(n1 ,n2);
}
Here the variable s is declared inside the function sum and has local scope;

2. Global scope:
A variable declared outside of all blocks can be used any where in the program.
#include<iostream>
using namespace std;
int s;
int sum(int n1,int n2)
{
s = n1 + n2;
return(s);
}
int main()
{
int n1 ,n2;
cout<<“Enter 2 numbers cin>>n1>>n2;
cout<<“The sum is “<<sum(n1,n2);
}
Here the variable s is declared out side of all functions and we can use variable s any where in the program.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 13.
There is a way to pass more than one value to the calling function from the called function. Explain how? (SCERT SAMPLE – I) (2)
Answer:
Methods of calling functions
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example
#include<iostream.h>
#include<conio.h> void
swap(int a, int b)
{
int temp;
temp = a;
a = b;
b = temp;
}
main()
{
clrscr();
int a,b;
cout<<“Enter values for a and b:-“;
cin>>a>>b;
cout<<“The values before swap a =”<<a<<“ and
b =”<<b;
swap(a,b);
cout<<“\nThe values before swap a =”<<a«“ and
b =”<<b;
getch();
}

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
#include<iostream.h>
#include<conio.h>
void swap(int &a, int &b)
{
int temp;
temp = a;
a = b;
b = temp;
}
main()
{
clrscrO;
int a,b;
cout<<“Enter values for a and b:- “; cin>>a>>b;
cout<<“The values before swap a =”<<a<<“ and
b=”<<b;
swap(a,b);
cout<<“\nThe values before swap a =”<<a<<“ and
b =”<<b;
getch();
}

Question 14.
A
Answer:
int sum (int N, int Start_No) .
{
int i, S = 0;
for (i = Start_No; i < = N + Start_No; i++)
S = S + i;
reurn S;
}

Question 15.
Differentiate between the string functions strcmp() and strcmpi(). (SAY-2016) (2)
Answer:
strcmp(): It is used to compare two strings and returns an integer.
Syntax:
strcmp(string1,string2)

  • if it is 0 both strings are equal.
  • if it is greater than 0(i.e. +ve) stringl is greater than string2
  • if it is less than 0(i.e. -ve) string2 is greater than string1

strcmpi():
It is same as strcmp() but it is not case sensitive. That means uppercase and lowercase are treated as same.
eg: “ANDREA” and “Andrea” and “andrea” these are same.

Question 16.
Read the function definition given below. Predict the output, if the function is called as convert (7). (MARCH-2017) (2)
void convert (int n)
{
if (n>1)
convert (n/2);
cout<<n%2;
}
Answer:
The out put is 111. convert() is a recursive function.

Plus One Functions Three Mark Questions and Answers

Question 1.
Consider the following function declaration with optional (default) arguments and state legal or illegal and give the reasons

  1. int sum(int x = 10, int y, int z)
  2. int sum(int x = 10, int y = 20, int z)
  3. int sum(int x = 10, int y = 20, int z = 30)
  4. int sum(int x, int y = 20, int z)
  5. int sum(int x, int y = 20, int z = 30)
  6. inf sum(int x, int y, int z = 30)
  7. int sum(int x = 10, int y, int z = 30)

Answer:
There is a rule to make an argument as default argument,i.e., to set an argument with a value that must be in the order from right to left. All the arguments in the right side of an argument must be set first to make an argument as a default argument.

  1. illegal, because y and z are not have values
  2. illegal, because z has no value
  3. legal
  4. illegal, because z has no value
  5. legal
  6. legal
  7. illegal, because x has a value but y has no value

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 2.
A list of C++ built-in functions are given. Classify them based on the usage and prepare a table with proper group names.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 1
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 2

Question 3.
Read the following program
#include<iostream>
using namespace std;
int main()
{
cout<<sum(2,3);
}
int sum(int x, int y)
{return (x + y);}
On compilation on the program, an error will be displayed. Identify and explain the reason. How can you rectify the problem
Answer:
The compilation of the program starts from the first line arid next line and so on( i.e. line by line). While compiling the line cout<<sum(2,3); The compiler does not understand the word sum(2,3) because it is not declared yet hence the error prototype required. To rectify this problem there are two methods

First method:
Give the function definition just before the main function as follows.
#include<iostream>
using namespace std;
int sum(intx, int y)
{return (x + y);}
int main()
{
cout<<sum(2,3);
}

Second Method:
Give the function declaration(prototype only) in the main function as follows.
#include<iostream>
using namespace std;
int main()
{
int sum(int.int);
cout<<sum(2,3);
}
int sum(int x, int y)
{return (x + y);}

Question 4.
Considering the following function definition.
{
for (int f = 1; n>0; n-)
f = f*n;
return f;
}
void main()
{
int a = 5, ans;
ans = fact (a);
cout<< a << “! = ” << ans;
}
The expected, desired output is 5! = 120
What will be the actual output of the program? It is not the same as above, why? What modification are required in the program to get the desired output.
Answer:
The output is 0! = 120
Because the address of variable ’a’ is given to the variable ‘n’ of the function fact(call by reference method). So the function changes its value (i.e. n-) to 0. Hence the result.
To get the desired result call the function as call by value method in this method the copy of the value of the variable ‘a’ is given to the function. So the actual value of ‘a’ will not changed. So instead of int fact(int &n) just write int fact(int n), i.e., no need of & symbol.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 5.
A function is defined as follows
int sum (int a, int b = 2)
{return (a + b);}
Check whether each of the following function calls is correct or wrong, Justify your answer

  1. cout << sum (2,3);
  2. cout << sum( 2) ;
  3. cout << sum();

Answer:
Here the function is declared with one optional argument. So the function call with minimum one argument is compulsory.

  1. It is valid. Here a becomes 2 and b becomes 3.
  2. It is also valid . Here a becomes 2 and b takes the default value 2.
  3. It is not a valid call. One argument is compulsory.

Question 6.
The factorial of a number, say N is the product of first N natural numbers. Thus, factorial of 5 can be obtained by taking the product of 5 and factorial of 4. Similarly factorial of 4 be found out by taking the product of 4 and factorial of 3. At last the factorial of 1 is 1 itself.

Which technique is applicable to find the factorial of a number in this fashion? Write a C++ function to implement this technique. Also explain the working of the function by giving the number 5 as input.
Answer:
A function calls itself is known as recursion.
#include<iostream>
using namespace std;
int fac(int);
int main()
{
int n;
cout<<“enter a number”;
cin>>n;
cout<<fac(n);
}
int fac(int n)
{
if (n == 1) return (1);
else
return (n × fac(n – 1));
}
The working of this program is as follows If the value of n is 5 then it calls the function as fa. The function returns value 5 × fac(4), That means this function calls the function again and returns 5 ×4 × fac(3). This process continues until the value n = 1. So the result is 5 × 4 × 3 × 2 × 1 = 120.

Question 7.
How do two functions exchange data between them? Compare the two methods of data transfer from calling function to called function
Answer:
There are two methods they are call by value and call by reference
1. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

2. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Question 8.
Read the following program
#include<iostream>
using namespace std;
int a = 0;
int main()
{
int showval(int);
cout<< a; a++;
cout << showval (a);
cout<< a;
}
int showval(int x)
{
int a = 5;
return (a + x);
}
Write down the value displayed by the output of the above program with suitable explanation. What are the inferences drawn regarding the scope of variables?
Answer:
The output is 061.
Global variable: A variable declared out side of all functions it is known as global variable.
Local variable: A variable declared inside of a function it is known as local variable.
If a variable declared inside a function(main or other) with the same name of a global variable. The function uses the value of local variable and does not use the value of the global variable.

Here int a = 0 is a global variable. In the main function the global variable ‘a’ is used. There is no local variable so the value of ‘a’, 0 is displayed. The statement ‘a++’ makes the value of ‘a’ is 1. It calls the function showval with argument ‘a = T.

The argument ‘x’ will get this value i.e. ‘x = 1 But in the function showval there is a local variable ‘a’ its value is 5 is used. So this function returns 6 and it will be displayed. After this the value 1 of the global variable ‘a’ will be displayed. Hence the result 061.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 9.
The following are function calling statements. Some of them will be executed, while some other generate compilation error. Write down your opinion on each of them with proper justification Function

  1. har ch =
  2. sqrt(25);
  3. strcat (“Computer”, “Program”);
  4. double num = pow(2,3,5)
  5. putchar(getchar());

Answer:

  1. getch get a character from the console(keyboard) but does not echo to the screen. So we can’t read a character from the console.
  2. It returns the square root of 25.
  3. It concatenates Program to computer, i.e. we will get a string “computer program”
  4. The function pow should contains only two arguments. But here it contains 3 arguments so it is an error. We can write this function as follows Double num = pow(pow(2,3),5)
  5. It reads a character from the console and display it on the screen.

Question 10.
Write down the operation performed by the following statements.

  1. int l = strlen(“Computer Program”);
  2. char ch[ ] = tolower(“My School”);
  3. cout <<(strcmp(“High”, “Low”)>0 ? toupper(“High”):tolower(“Low”));

Answer:

  1. The built in function strlen find the length of the string i.e. 16 and assigns it to the variable I.
  2. This is an error because tolower is a character function.
  3. This is also an error because tolower and toupper are character functions.

Question 11.
A line of given length with a particular character is to be displayed. For example, ********** js a line with ten asterisks (*). Define a function to achieve this output.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 3

Question 12.
Read the following function
int fib(int n)
{
if (n<3)
return 1;
else
return (fib(n – 1) + fib(n – 2));
}

  1. What is the speciality of this function
  2. How does it work?
  3. What will be the output of the following code?

Answer:

  1. This function is a recursive function. That means the function calls itself.
  2. It works as follows
    • if i = 1, The function fib calls with value 1. i.e. fib(1) returns 1
    • if i = 2, The function fib calls with value 2. i.e. fib(2) returns 1
    • if i = 3, The function fib calls with value 3. i.e. fib(3) returns fib(2) + fib(1) i.e. it calls the function again.
      So the result is 1 + 1 = 2
    • if i = 4, The function fib calls with value 4. i.e. fib(4) returns fib(3) + fib(2) i.e. it calls the function again.
      So the result is 2 + 1 = 3
  3. The output will be as follows 1 1 2 3

Question 13.
Explain scope rules of functions and variables in a C++ program
Answer:
Local variable or function:
A variable or function declared inside a function is called local variable or function. This cannot be accessed by the out-side of the function.
eg:
main()
{
int k; // local variable
cout<<sum(a,b); // local function
}

Global variable or function:
A variable or function declared out side of a function is called global variable or function. This can be accessed by any statements.
eg:
int k; // global variable
int sum(int a, int b); // global function
main()
{
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 14.
Briefly explain default arguments Default arguments.
Answer:
A default value can be set for a parameter (argument) of a function. When the user does not give a value the function will take the default value. An important thing remember,is an argument cannot have a default value unless all arguments on its right side must have default value.
Functions with valid default arguments are given below

  • float area(int x, int y, int z = 30);
  • float area(int x, int y = 20, int z = 30);
  • float area(int x = 10, int y = 20, int z = 30);

Functions with invalid default arguments are given below

  • float area(int x = 10, int y, int z);
  • float area(int x, int y = 20, int z);
  • float area(int x = 10, int y = 20, int z);

Question 15.
How to pass an array to function.
Answer:
By using call by reference method we can send an array as argument. We have to send only the array name and its data type. The array name holds the starting address and will access the subsequent elements of an array.
The following example shows this
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 4

Question 16.
How to pass a structure to a function.
Answer:
A structure can be passed as argument to a function either call by value or call by reference method. An example is given below to read name and age of a structure and pass it to a function.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 5

Question 17.
Write a program to read a character and check whether it is alphabet or not. If it is an alphabet check whether it is upper case or lower case?
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 6

Question 18.
Write a program to read 2 strings and join them 2 string.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 7

Question 19.
Write a program to read 2 strings and compare it 2 string.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 8
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 9

Question 20.
Write a program to read a string and display the number of alphabets and digits and special characters.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 10

Question 21.
Write a program to perform the following opera-tions on a string

  1. Length of a string
  2. Search a character
  3. Display the string

Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 11
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 13

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 22.
Write short notes about conversion functions.
Answer:
The header file stdlib.h is used for conversion functions.Following are the important conversion functions
1. itoa():
It is used to convert an integer into a string.
eg:
int n = 100;
charstr[20];
itoa(n,str,10);

2. Itoa():
It is used to convert an long into a string.
eg:
long n = 1002345L
char str[20];
ltoa(n,str,10);

3. atoi():
It is used to convert a string into integer
eg:
int n;
str[] = “123”
n = atoi(str);

4. atoll():
It is used to convert a string into long
eg:
long n;
str[] = “123456”
n = atol(str);

Question 23.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 14

  1. Predict the output of both programs.
  2. Justify your predictions.

Answer:
1. A. Output
value = 40

B. output
value = 0

2. In the first case (A) the argument x is passed by reference method. So the changes made in the function reflects in main()

In the second case (B) the argument x is passed by value method. So the changes made in the function will not reflect in main()

Question 24.
Consider the following C++ function.
int recfact(int x)
{
if(x == 0)
return 1;
else
return x * recfact(x – 1);
}

  1. What type of function is this? Explain. (2 Scores)
  2. What is the output of the above function when x = 5

Answer:

  1. recursive function. The abilty of a function to call itself is called recursion. A function is said to be recursive if a statement in the body of the function calls itself.
  2. Output 120

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 25.
void initialise()
{
int k = 10;
}
int main()
{
int a, b;
float marks;
a = 20;
cout<<“First value =”<< a;
initialise();
b = k + a;
cout<<“New value =”<<b;
}

  1. Identify the error in the above code and explain its reasons.
  2. Correct the errors.

Answer:

  1. K is a local variable in the function initialize 0. It is not accessible in main()
  2. Making the variable K as global we can correct the error.

Question 26.
Write a program to read 2 strings and join them using string function.
Answer:
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
char str1 [50],str2[50],str[100];
cout<<“Enter a string:”; gets(strl);
cout<<“Enter another string:”; gets(str2);
strcat(str,str1);
strcat(str,str2);
cout<<“The concatenated string is”<<str;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 27.
Consider the following code.
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
char str1 [] = “123”,str2[] = “45”;
int a,b;
a = atoi(str1);
b = atoi(str2);
cout<<a + b;
}
What will be the output and give the reason?
Answer:
The output is 123 + 45 = 168. Here the conversion function atoi is used to convert the string “123” into integer 123 and “45” into integer 45 so the result.

Question 28.
Name the different methods used for passing arguments to a function. Write the difference between them with examples. (MARCH-2015)
Answer:
Methods of calling functions:
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 15
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 16

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 17

Question 29.
Write a C++ program to display the simple interest using function. (SAY-2015) (3)
Answer:
include < iostream>
using namespace std;
float Simplelnt (float p, int n, float r)
{
return (p*n*r/100);
}
int main()
{
float p, r, SI;
int n;
cout<<“Enter values for p,n,r:”; cin>>p>>n>>r;
SI = Simplelnt (p,n,r); cout<<“simple interest is”<<SI;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 30.
Write a function’s definition of the above type to find the sum of matural numbers from 1 to N. (Hint: If the value of N is 5, the answer will be 1 + 2 + 3 + 4 + 5 = 15). (MARCH – 2016) (3)
Answer:
int sum (int n)
{
if (n == 0)
return 0;
else
return (n + sum (n – 1));
}

Question 31.
A program requires functions for adding 2 numbers, 3 numbers and 4 numbers, How can you provide a solution by writing a single functio0n? (SCERT SAMPLE -1) (3)
Answer:
It can be solved by writing function with default values.
eg: int add (int n1, int n2, int n3 = 0, int n4 = 0)
{
return (n1 + n2 + n3 + n4);
}
This function can be invoked by the following function calling
1. add (5,2);
2. add (5,2,7);
3. add (5,2,7,10);

The 1. Call returns 5 + 2 = 7
2. Call returns 5 + 2 + 7 = 14
3. Call returns 5 + 2 + 7 + 10 = 24

Question 32.
Explain the difference between call-by-value method and call-by-reference method with the help of examples. (SCERT SAMPLE – II) (3)
Answer:
Methods of calling functions:
Two types call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the ’ function makes any change will not affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 18
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 19

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 20

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 33.
Explain recursive functions with the help of a suitable example. (SAY-2016) (3)
Answer:
A function calls itself is called recursive function.
#include<iostream>
using namespace std;
void convert(int n)
{
if(n>1)
convert(n/2);
cout<<n % 2;
}
int main()
{
convert(7);
}
Here the function convert is a recursive function, that means it calls itself and output of the above program is 111.

Question 34.
Name the built-in function to check whether a character is alphanumeric or not. (MARCH-2017) (3)
Answer:
isalnum()

Question 35
Briefly explain the three components in the structure of a C++ program. (MARCH-2017) (3)
Answer:
1. Preprocessor directives:
A C++ program starts with the preprocessor directive i.e., # .include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
eg: #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

2. Header files:
A header file is a pre stored file that helps to use some operators and functions. To write C++ programs the header files are must. Following are the header files iostream

  • iomanip
  • cstdio
  • cctype
  • cmath
  • cstring

The syntax for including a header file is as follows
#include<name of the header file>
eg: #include<iostream>

3. The main function:
The main function is the first function which is invoked at the time of execution and the program ends within main(). The other functions are invoke from main().

Question 36.
Explain the difference between call by value method and call by reference method with the help of examples. (MARCH-2017) (3)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change wilt not affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 21

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 22
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 23

Plus One Functions Five Mark Questions and Answers

Question 1.
Short notes about character functions and string functions
Answer:
A. Character functions:
1. isalnum():
t is used to check whether a character is alphabet or digit. It returns a non zero value if it is an alphabet or digit otherwise it returns zero.

2. isalpha():
It is used to check whether a character is alphabet or not. It returns a non zero value if it is an alphabet otherwise it returns zero.

3. isdigit():
It is used to check whether a character is digit or not. It returns a non zero value if it is digit otherwise it returns zero.

4. islower():
It is used to check whether a character is lower case alphabet or not. It returns a non zero value if it is a lowercase alphabet otherwise it returns zero.

5. isupper():
It is used to check whether a character is upper case alphabet or not. It returns a non zero value if it is an upper case alphabet otherwise it returns zero.

6. tolower():
It is used to convert the alphabet into lower case

7. toupper():
It is used to convert the alphabet into upper case

B. String functions:

1. strcpy():
This function is used to copy one string into another

2. strcat():
This function is used to concatenate(join) second string into first string

3. strlen():
This function is used to find the length of a string.

4. strcmp():
This function is used to compare 2 strings. If the first string is less than second string then it returns a negative value. If the first string is equal to the second string then it returns a value zero and if the first string is greater than the second string then it returns a positive value.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 2.
Write functions to perform the following operations

  1. sqrt()
  2. power of 2 numbers
  3. sin
  4. cos

Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 24
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 25

Question 3.
Write a C++ program to display the roots of quadratic equation. (SAY-2015) (5)
Answer:
#include<iostream>
#include<cmath>
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 26

Question 4.
Explain ‘Call by Value’ and ‘Call by Reference’ methods of function calling with the help of a suitable example. (SAY-2016)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 27
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 29

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example
Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions 30

Plus One Computer Science Chapter Wise Questions and Answers Chapter 10 Functions

Question 5.
Define network topology. Explain any four network topologies in detail. (SAY-2016)
Answer:
Network topologies:
Physical or logical arrangement of computers on a network is called structure or topology. It is the geometrical arrangement of computers in a network. The major topologies developed are star, bus, ring, tree and mesh

1. Star Topology:
A star topology has a server all other computers are connected to it. If computer A wants to transmit a message to computer B. Then computer A first transmit the message to the server then the server retransmits the message to the computer B.

That means all the messages are transmitted through the server. Advantages are add or remove workstations to a star network is easy and the failure of a workstation will not effect the other. The disadvantage is that if the server fails the entire network will fail.

2. Bus Topology:
Here all the computers are attached to a single cable called bus. Here one computer transmits all other computers listen. Therefore it is called broadcast bus. The transmission from any station will travel in both the direction. The connected computers can hear the message and check whether it is for them or not.

Advantages are add or remove computer is very easy. It requires less cable length and the installation cost is less. Disadvantage is fault detection is very difficult because of no central computer.

3. Ring Topology:
Here all the computers are connected in the shape of a ring and it is a closed loop. Here also there is no central computer. Here a computer transmits a message, which is tagged along with its destination computer’s address. The message travels in one direction and each node check

whether the message is for them. If not, it passes to the next node. FIt requires only short cable length. If a single node fails, at least a portion of the network will fail. To add a node is very difficult.

4. Hybrid Topology:
It is a combination of any two or more network topologies. Tree topology and mesh topology can be considered as hybrid topology.
(a) Tree Topology:
The structure of a tree topology is the shape of an inverted tree with a central node and branches as nodes. It is a variation of bus topology. The data transmission takes place in the way as in bus topology. The disadvantage is that if one node fails, the entire portion will fail.

(b) Mesh Topology:
In this topology each node is connected to more than one node. It is just like a mesh (net). There are multiple paths between computers. If one path fails, we can transmit data through another path.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Students can Download Chapter 8 Redox Reactions Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Plus One Chemistry Redox Reactions One Mark Questions and Answers

Question 1.
In which of the following, oxidation number of chlorine is +5?
a) Cl
b) ClO
c) ClO2
d) ClO3
Answer:
d) ClO3

Question 2.
An oxidising agent is a substance which can
a) Gain electrons
b) Lose an electronegative radical
c) Undergo decrease in the oxidation number of one of its atoms
d) Undergo any one of the above changes
Answer:
d) Undergo any one of the above changes

Question 3.
The arrangement of metals in the order of decreasing tendency to lose electrons is called _________ .
Answer:
Activity series

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 4.
When KMnO4, reacts with acidified FeSO4
a) Only FeSO4 is oxidised
b) Only KMnO4 is oxidised
c) FeSO4 is oxidised and KMnO4 is reduced
d) KMnO4 is oxidised and FeSO4 is reduced
Answer:
c) FeSO4 is oxidised and KMnO4 is reduced

Question 5.
In the disproportionation reaction, which of the following statements is not true?
a) The same species is simultaneously oxidised as well as reduced
b) The reacting species must contain an element having at least three oxidation states
c) The element in the reacting species is present in the lowest oxidation state
d) The element in the reacting species is present in the intermediate oxidation state
Answer:
c) The element in the reacting species is present in the lowest oxidation state

Question 6.
Find the oxidation state of oxygen in OF2.
Answer:
The oxidation number of fluorine in its compounds is always taken as -1.
In OF2
X+ (-1 × 2) = 0
X = +2

Question 7.
The oxidation numbers of chlorine atoms in bleaching powder is _________ .
Answer:
-1

Question 8
SO2 can act as
a) Oxidising agent only
b) Reducing agent only
c) Both oxidising and reducing agents
d) Acid and a reducing agent only
Answer:
c) Both oxidising and reducing agents

Question 9.
In the reaction
2KMnO4 +16HCl → 5Cl2 + MnCl2 + 2KCl + 8H2O the reduction product is _________ .
Answer:
MnCl2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 10.
The strongest reducing agent is .
a) K
b) Ba
c) Li
d) Na
Answer:
c) Li

Question 11.
Oxidation state of oxygen in H2O2 is _________ .
Answer:
-1

Plus One Chemistry Redox Reactions Two Mark Questions and Answers

Question 1.
Balance the following equation using oxidation number method:
MnO2 + Cl → Mn2+ + Cl2
Answer:
Assigning oxidation numbers:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 1
Equating the increase and decrease in oxidation number:
MnO2 + 2Cl → Mn2+ + Cl2
Balancing hydrogen and oxygen atoms:
MnO2 + 2Cl + 4H+ → Mn2+ + Cl2 + 2H2O

Question 2.
Balance the following equation using half reaction method:
Cu + NO3 → Cu2+ + NO2
Answer:
Separating into half reactions:
Oxidation half: Cu → Cu2+
Reduction half: NO3 → NO2
Balancing oxygen and hydrogen atoms:
NO3 + 2H+ → NO2 + H2O
Balancing charge by adding electrons and making the number of electrons equal in the two half reactions:
Cu → Cu2+ + 2e
2NO3 + 4H+ + 2e → 2NO2 + 2H2O
Adding the two half reactions to achieve the overall reaction:
Cu + 2NO3 + 4H+ → Cu2+ + 2NO2 + 2H2O

Question 3.
Complete the following ionic equations:

  1. Al3+ + 3e → …………….
  2. MnO42- → + e
  3. K → K+ + ……………
  4. Fe2+ → Fe3+ +

Answer:

  1. Al3+ + 3e → Al
  2. MnO42- → MnO4+ e
  3. K → K+ + e
  4. Fe2+ → Fe3+ + e

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 4.
Find the oxidation number of P in the following compounds:

  1. Na2PO4
  2. H3P2O7
  3. PH3
  4. H3PO4

Answer:

  1. Na2PO4, Oxidation state of P = +6
  2. H4P2O7, Oxidation state of P = +5
  3. PH3, Oxidation state of P = -3
  4. H3PO4, Oxidation state of P = +5

Question 5.
Choose the correct oxidation number of sulphur in the compounds in column Afrom column B.

Column AColumn B
Na2SO4-2
H2sO3+7
H2S+6
H2S2O7+4

Answer:

Column AColumn B
Na2SO4+6
H2SO3+4
H2S-2
H2S2O8+7

Question 6.
Explain oxidation number and valency.
Answer:
Valency of an atom is its combining capacity and is denoted by a number without sign. The valency of an element is always a whole number.

Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.

Question 7.
Some rules related to oxidation number are given below. Correct the mistakes.

  • Oxidation number of alkali metals and alkaline earth metals is +2.
  • Oxidation number of hydrogen is always +1.
  • Algebraicsum of oxidation number of all the atoms in an ion is not equal to the charge on the ion.

Answer:

  • Oxidation number of alkali metals is +1.
  • Oxidation number of alkaline earth metals is +2.
  • Oxidation number of H is +1 except in metallic hydrides.

Question 8.
Match the following:

Oxidation number of Cl in Cl2O7Cu
OxidantZn
Stannous Chloride, SnCl2+7
Oxidation number of C in diamondGet reduced easily
The metal which can’t displace H from dil.HClZero
Reducing agent for mercuric chloride

Answer:

Oxidation number of Cl in Cl2O7+7
OxidantGet reduced I easily
Stannous Chloride, SnCl2Reducing agent for mercuric chloride
Oxidation number of C in diamondZero
The metal which can’t displace H from dil.HClCu

Question 9.
1. Calculate the oxidation number of oxygen in OF2 and KO2.
2. When Zn rod is dipped in blue CuSO4 solution ‘ the blue colour of CuSO4 fades due to displacement reaction. Write the reaction and identify the following:
i) The substance oxidised and the substance reduced.
ii) The oxidant and the reductant.
Answer:
1. OF2: x + (-1 × 2) = 0
x – 2 = 0
x = +2
KO2: (+1 × 1) + 2x = 0
1 + 2x = 0
2x = -1
x = –\(\frac{1}{2}\)

2. Zn(s) + CuSO4 (aq) → ZnSO4(aq) + Cu(s)
i) Substance oxidised – Zn
Substance reduced – Cu
ii) Oxidant-Cu
Reductant – Zn

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 10.
a) Calculate the oxidation number of C in CH4 and in CH3Cl.
b) The sum of oxidation numbers of all atoms in a molecule is …………
Answer:
a) CH4:
x + (1 × 4) = 0
x + 4 = 0
x = -4
Oxidation number of C in CH4 is -4.

CH3Cl:
x + (3 × 1) + -1 = 0
x + 3 – 1= 0
x + 2 = 0
x = -2
Oxidation number of C in CH3Cl is -2.

b) Zero

Question 11.
1. Write the oxidation state of each element and identify the oxidising agent and reducing agent in the following reaction:
H2S(g) + Cl2(g) → 2HCl(g) + S(s)
2. Fill in the blanks and classify the following reactions into oxidation and reduction:
i) Mn7+ + 5e → ……………
ii) Sn4+ + …………… → Sn2+
iii) Na → Na+ + ……………
iv) Fe3+ +…………… → Fe2+
Answer:
1.Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 2
Reducing agent – H2S
Oxidising agent -Cl2

2. i) Mn7+ + 5e → Mn2+
ii) Sn4+ + 2e → Sn2+
iii) Na → Na+ + e
iv) Fe3+ + e → Fe2+
Oxidation: Reaction (iii)
Reduction: Reactions (i), (ii) and (iv)

Question 12.
Dihydrogen undergoes redox reactions with many metals at high temperature.
a) Write the reaction between hydrogen with sodium.
b) Comment, whether the product formed, is covalent compound or ionic compound.
c) Which is the reducing agent in this reaction?
Answer:
1. 2Na + H2 → 2NaH
2. Ionic compound is formed. When alkali metals react with hydrogen ionic hydrides are formed.
3. Na is the reducing agent.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 13.
1. Is it possible to keep copper sulphate solution in zinc pot? Why?
2. Assign oxidation numbers of the underlined elements.
i) NaH2\(\underline { P } \)O4
ii) NaH\(\underline { S } \)O4
Answer:
1. No. Zn being more reactive will displace Cu from CuSO4. Thus Cu will be deposited on the vessel.

2. i) NaH2\(\underline { P } \)O4
+1 +(+1 × 2) + x +(-2 × 4) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x = +5
ii) NaH\(\underline { S } \)O4
+1 + 1 + x +(-2 × 4) = 0
+1 + 1 + x – 8 = 0
x – 6 = 0
x = +6

Question 14.
Identify the substance oxidised, reduced, oxidising agent and reducing agent in the reaction:
2Cu2O + Cu2S → 6Cu + SO2
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 3
In this reaction, Cu is reduced from +1 state to zero. oxidation state and S is oxidised from -2 state to +4 state. Cu2O helps S in Cu2S to increase its oxidation number. Therefore, Cu(l) is the oxidising agent. S of Cu2S helps Cu both in Cu2S itself and Cu2O to decrease its oxidation number. Therefore, S of Cu2S is the reducing agent.

Question 15.
Explain the following in terms of electron transfer concept:

  1. Oxidation
  2. Reduction
  3. Oxidising agent
  4. Reducing agent

Answer:

  1. Oxidation: Loss of electron(s) by any species.
  2. Reduction: Gain of electron(s) by any species.
  3. Oxidising agent: Any species which accepts electrons).
  4. Reducing agent: Any species which donates electron^).

Question 16.
Represent the following compounds using Stock notation:
Cu2O, SnCl4, MnO, Fe2O3, V2O5
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 4

Question 17.
In a redox reaction, oxidation and reduction occur simultaneously.
a) Write the classical concept of oxidation and reduction.
b) Identify the species undergoing oxidation and reduction in the following reaction:
H2S(S) + Cl2(g) → 2HCl(g) + S(s)
Answer:
1. Oxidation:
addition of oxygen/electronegative element to a substance or removal of hydrogen/ electropositive element from a substance.

Reduction:
removal of oxygen/electronegative element from a substance or addition of hydrogen/ electropositive element to a substance.

2.Oxidised species:
H2S. This is because a more electronegative element, Cl is added to H or a more electro positive element, H has been removed from S.

Reduced species:
Cl. This is due to addition of more electropositive element H to it.

Plus One Chemistry Redox Reactions Three Mark Questions and Answers

Question 1.
An equation is given below:
HNO3+ l2 → HlO3 + NO2 + H2O

  • Find the oxidising agent and reducing agent.
  • Balance the equation using half reaction method.

Answer:
Oxidising agent = HNO3
Reducing agent = l2
Skeletal equation:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 5
Balancing the charge on the half reactions by adding electrons and equalising the number of electrons:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 6

Question 2.
1. Define redox reactions.
2. Predict whether the following reaction is a redox reaction or not? Justify.
Cr2O72- + H2O → 2CrO42- + 2H+
Answer:
1. Redox reactions are those reactions are those reactions in which reduction and oxidation occur simultaneously. These reactions involve change in oxidation state of the interacting species.

2. No.
Because no element undergoes change in oxidation number.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 3.
a) Find out the oxidising agent and reducing agent in the following reaction:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
b) Balance the following redox reaction in acid medium using oxidation number method.
Cr2O72- + Fe2+ → Cr3+ + Fe3+
Answer:
1. Oxidising agent – Ag
ReducingAgent – Cu

2. Assigning oxidation number:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 7

Question 4.
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO4, Cr2O72- and NO3.
Answer:
H2SO4
(2 × +1) + x+ (4 × -2) = 0
+2 + x – 8 = 0
x – 6 = 0
x = +6

Cr2O72-
2x + 7 ×-2 = -2
2x = -2 + 14
2x = 12
∴ x = +6

NO3
x + 3 × -2 = -1
x = -1 + 5
x = +4

Question 5.
1. Assign oxidation numbers
(i) P in NaH2PO4
(ii) Mn in KMnO4
(iii) B in NaBH4
(iv) S in H2SO4
2. Identify the oxidising and reducing agents in the following reaction:
CuO + H2 → Cu + H2O
Answer:
1. i) NaH2PO4
Na+1H2+1PO4-2
1+2 + x- 8 = 0
3 + x – 8 = 0
x – 5 = 0
x =+ 5

ii) K+1MnO4-2
1+ x – 8 = 0
x – 7 = 0
x = +7

iii) Na+1BH4+1
1 + x + 1 × 4 = 0
x + 5 = 0
x = -5

iv) H2+1SO4-2
2 + x – 8 = 0
x – 6 = 0
x = +6

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 8

Question 6.
A copper rod is dipped in silver nitrate solution.

  1. What are the observations?
  2. Write the displacement reaction.
  3. Identify the species getting oxidised and reduced.

Answer:

  1. The colour of the solution changes to blue. Silver is deposited on the copper rod.
  2. Cu(s) +2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
  3. Oxidised species – Cu Reduced species – Ag+

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 7.
1. Identify the oxidising and reducing agent in the reaction:
CuS + O2 → Cu + SO2
2. Determine the oxidation number of the underlined element in the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 9
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 10

Question 8.
1. Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the reaction:
Cl2 + 2l → 2Cl +l2

2. Calculate the oxidation number of underlined elements in the following compounds:
i) K2\(\underline { Cr } \)2O7
ii) H\(\underline { H } \)O3
Answer:
1. Cl2 is reduced, therefore Cl2 is the oxidising agent. I’ is oxidised, therefore I” is the reducing agent.

2. i) K2\(\underline { Cr } \)2O7
(+1 × 2) + 2x +(-2 × 7) = 0
+2 + 2x – 14 =0
2x – 12 =0
2x = 12
x = +6
ii) H\(\underline { H } \)O3
(+1 × 1) + x + (-2 × 3) = 0
1+ x – 6 = 0
x – 5 = 0
x = +5

Question 9.
Determine oxidation number of the elements underlined in each of the following.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 12
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 13

Plus One Chemistry Redox Reactions Four Mark Questions and Answers

Question 1.
Permanganate ion (MnO4) reacts with bromide ion (Br) in basic medium to give manganese dioxide
(MnO2) and bromate ion (BrO3).
a) Write the balanced ionic equation for this reaction.
b) Identify the oxidising agent and reducing agent in this reaction.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 14

Question 2.
A redox reaction involves oxidation and reduction.
a) What do you understand by electrode potential?
b) Define a redox couple.
c) Explain the set-up for Daniell cell with a diagram.
d) Write the electrode reactions and overall cell reaction which occur in the Daniel cell.
Answer:
a) The potential difference between metal and its own ion is called electrode potential.

b) A redox couple is defined as the combination of oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

c) Take copper sulphate solution in a beaker and put a copper strip. Take zinc sulphate solution in another beaker and put a zinc rod. The two redox couples are represented as Zn2+/Zn and Cu2+/Cu. Put the beaker containing copper sulphate solution and beaker containing zinc sulphate side by side. Connect two solution by a salt bridge. The Zn and Cu rods are connected by a metalic wire with a provision for ammeter and switch. Transfer of electrons now does not take place directly from Zn to Cu2+, but through metallic wire. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 15

Question 3.
Redox reactions are those in which oxidation and reduction takes place. Explain the different types of redox reactions with suitable examples.
Answer:
Combination Reactions: The reactions in which two substances combine together to form a new compound are called combination reactions. These can be denoted as A+ B → C where either A or B or both A and B should be in the elemental form.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 16
Decomposition reactions:
The reactions in which a compound breaks up into two or more substances at least one of which is in elemental form are called decompositions reactions.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 17

Displacement reactions:
The reactions of the type X + YZ → XZ + Y in which an atom or ion in a compound is displaced by an ion (atom) of another element, such that X and Y are in elemental form are called displacement reactions. They are of two categories:
1. Metal displacement reactions: Reactions in which a more electropositive metal displaces a less electropositive metal from its compound.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 18

2. Non-metal displacement reactions: These are reactions in which a non-metal is displaced by another metal or non-metal.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 19
Disproportionation reactions: These are special type of redox reactions in which an element in one oxidation state is simultaneously oxidised and reduced. Here one of the reactants should contain an element that should exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
e.g. The decomposition of hydrogen peroxide.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 20
Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O2 and to -2 state in H2O.

Plus One Chemistry Redox Reactions NCERT Questions and Answers

Question 1.
Fluorine reacts with ice and results in the change :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 21
Justify that this reaction is a redox reaction.
Answer:
In the given reaction O.N. of F2 changes from zero to -1 in HF and HOF whereas O.N. of oxygen change from -2 in H2O to zero in HOF. Thus, F2 is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.

Question 2.
Write formulas for the following compounds:

  1. Mercury (II) chloride
  2. Nickel (II) sulphate
  3. Tin (IV) oxide
  4.  Thallium (I) sulphate
  5. Iron (III) sulphate
  6. Chromium (III) oxide

Answer:

  1. Hg(II)Cl2
  2. Ni(II)SO4
  3. Sn(IV)O2
  4. Tl2(I)SO4
  5. Fe2(III)(SO4)3
  6. Cr2(III)O3

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 3.
The compound AgF2 is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?
Answer:
In AgF2, oxidation state of Ag is + 2 which is very unstable. Since Ag can exist in a stable state of + 1 it quickly accepts an electron to form the more stable + 1 oxidation state.
Ag2+ + e → Ag+

Question 4.
Consider the reactions:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 22
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average O.N. of S in S2O32- is + 2 while in S4O62- it is + 2.5. The O.N. of S in SO42- is+6. Since Br2 is a stronger oxidising agent that l2, it oxidises S of S2O32- to a higher oxidation state of + 6 and hence forms SO42- ion. l2, however, being a weaker oxidising agent oxidises S of S2O32- ion to a lower oxidation of + 2.5 in S4O62- ion.

Question 5.
Why does the following reaction occur?
XeO64-(aq) + 2F(aq) + 6H+(aq) → XeO3(s) + F2(g) + 3H2O(I)
What conclusion about the compound Na4XeO6 (of which XeO64- is a part) can be drawn from the reaction?
Answer:
The balanced equation along with O.N. of the elements above their symbols will be as:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 23
In the equation the, O.N. of Xe decreases from + 8 in XeO64- to + 6 in XeO3 while that of F increases from – 1 in F to 0 in F2. Therefore, XeO64- is reduced while F is oxidised. This reaction occurs because Na4XeO6 (0r XeO64-) is stronger oxidising agent than F2.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Students can Download Chapter 9 String Handling and I/O Functions Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Plus One String Handling and I/O Functions One Mark Questions and Answers

Question 1.
To read a single character for gender i.e. ‘m’ or ‘f’ ________ function is used.
Answer:
(a) getch()
(b) getchar()
(c) gets()
(d) getline()
Answer:
(b) getchar()

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
To use getchar(), putchar(), gets() and puts(), which header file is used?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(b) cstdio

Question 3.
To use cin and cout, which header file is needed?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(a) iostream

Question 4.
Predict the output of the following code snippet.
#include<cstdio>
int mainO
{
char name[ ] = “ADELINE”;
for(int i=0; name[i]!=’\0′;i++)
putchar(name[i]);
}
Answer:
The output is “ADELINE”.

Question 5.
From the following which is equivalent to the function getc(stdin).
(a) putchar()
(b) gets()
(c) getchar()
(d) puts()
Answer:
(c) getchar()

Question 6.
From the following which is equivalent to the function putc(ch, stdout).
(a) putchar(ch)
(b) ch = gets()
(c) ch = getchar()
(d) puts(ch )
Answer:
(a) putchar(ch)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 7.
To print a single character at a time which function is used?
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) putchar()

Question 8.
To read a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) gets()

Question 9.
To print a string _______ function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) puts()

Question 10.
Consider the following code snippet.
main()
{
char str[80];
gets(str);
for(int i=0. len=0;str[il!=’\0′;i++.len++);
cout<<“The length of the string is ” <<len;
}
Select the equivalent forthe under lined statement from the following
(a) int len = strlen(str)
(b) int len = strcmp(str)
(c) int len = strcount(str)
(d) None of these
Answer:
(a) int len = strlen(str)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 11.
Arjun wants to read a string with spaces from the following which is suitable.
(а) cin>>
(b) cin.getline(str,80)
(c) str = getc(stdin)
(d) none of these
Answer:
(b) cin.getline(str,80)

Question 12.
State whether the following statement is true or false. The ‘<<‘ insertion operator stops reading a string when it encounters a space.
Answer:
True

Question 13.
_________ function is used to copy a string to another variable. (SAY-2016) (1)
Answer:
strcpy();

Question 14.

  1. Write the declaration statement for a variable ‘name’ in C++ to store a string of maximum length 30.
  2. Differentiate between the statement cin>>name and gets (name) for reading data to the variable ‘name’. (SAY-2016)

Answer:
1. char name[31];(One for null(\0) character).

OR

cin>> does not allows space. It will take characters up to the space and characters after space will be truncated . Here space is the delimiter. Consider the following code snippet that will take the input upto the space.
#include<iostream>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
cin>>name;
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis. The string after space is truncated.

2. gets(): This function is used to get a string from the keyboard including spaces. Considerthe following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis Emerin” then the output will be Hello Alvis Emerin.

Question 15.
What is the advantage of using gets() function in the C++ program to input string data? Explain with an example.
Answer:
gets() function is used to get a string from the keyboard including spaces. Consider the following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}
If you input a name “Alvis” then the output is Hello Alvis.

Plus One String Handling and I/O Functions Two Mark Questions and Answers

Question 1.
In a C++ program, you forgot to include the header file iostream. What are the possible errors occur in that Program? Explain?
Answer:
Prototype error. To use cin and cout the header file iostream is a must.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
Categorise the following into three according to their relationship
iostream, cstdio, gets(), puts(), getchar(), putchar(), getline(), write(), cin, cout.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and IO Functions 1

Question 3.
Pick the odd one out from the following and give reason.
gets(), getline(), getch() getchar().
Answer:
getline() – It is a stream function whereas the others are console functions.

Question 4.
My_name is a variable contains a string. Write two different C++ statements to display the string. (SAY-2016) (2)
Answer:

  1. cout<<my_name;
  2. puts(my_name);

Question 5.
Suggest most suitable built-in function in C++ to perform the following tasks: (MARCH-2016) (2)

  1. To find the answer for 53
  2. To find the number of characters in the string “KERALA” “HAPPY NEW YEAR”
  3. To get back the number 10 if the argument is 100.

Answer:

  1. pow(5,3);
  2. strlen(“KERALA”)
  3. tolower(‘M’)
  4. sqrt(100);

Question 6.
Read the following C++ statements:
charstr[50];
cin>>str;
cout<<str;
During execution, if the string given as input is “GREEN COMPUTING”, the output will be only the word “GREEN”. Give reason for this. What modification is required to get the original string as output? (SCERT SAMPLE -1) (2)
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets() function.

Because gets() function reads character upto the enter key.
Hence gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Question 7.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (2)
Answer:
gets() function is used to get a string from the keyboard including spaces. To use gets () function the header file cstdio must be included. It reads the characters upto the enter key pressed by the user.
eg:
char name[20];
cout << “Enter your name”;
gets(name);
cout<< “Hello”<< name;
When the user gives Alvis Emerin. It displays as “Hello Alvis Emerin”.

Plus One String Handling and I/O Functions Three Mark Questions and Answers

Question 1.
Suresh wants to print his name and native place using a C++ program. The program should accept name and native place first.
Name is: Suresh Kumar
Address is: Alappuzha
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20],place[20];
cout<<“Enter your name”;
cin.getline(name,80);
cout<<“Enter your place”;
cin.getline(place,80);
cout<<“Your name is puts(name);
cout<<“Your place is puts(place);
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 2.
“Programming is Fun”. Write a C++ program to read a string like this in lower case and print it in UPPER CASE. With out using toupper() library function.
Answer:
using namespace std;
#include<cstdio>
int main()
{
char line[80];
int i;
puts(Enter the string to convert”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
if (Iine[i]>=97 && line[i]<=122)
line[i]=line[i] – 32;
puts(line);
}

Question 3.
An assignment Kumar has written a C++ program which reads a line of text and print the number of vowels in it. What will be his program code?
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’:
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is “<<vowel;

Question 4.
What will be the output of the following code if the user enter the value “GOOD MORNING”.
1. char string [80];
gets(string);
cout<<string;

2. char string [80];
cin>>string;
cout<<string;

3. charch;
ch = getchar();
cout<<ch;

4. char string [80];
cin.getline(string,9);
cout<<string;
Answer:

  1. GOOD MORNING
  2. GOOD
  3. G
  4. GOOD MORN

Question 5.
Consider the following code snippet.
int main()
{
int n;
cout<<“Enter a number”;
cin>>n;
cout<<“The number is “<<n;
}
Write down the names of the header files that must be included in this program
Answer:
Here cin and cout are used so the header file iostream must be included.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 6.
Write a program to display the following output.
A
BB
CCC
#include<iostream>
using namespace std;
int main()
{
char str[]=”ABC”;
int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<=i;j++)
cout<<str[i];
cout<<endl;
}
}

Question 7.
Distinguish getchar and gets.
Answer:
getchar is a character function but gets is a string function. The header file cstdio.h must be included. It reads a character from the keyboard.
Eg.
char ch;
ch = getchar();
cout<<ch;
gets is used to read a string from the keyboard. It reads the characters upto enter key. The header file cstdio must be included.
char str[80J;
cout<<“Enter a string”;
gets(str);

Question 8.
Distinguish putch and puts.
Answer:
putch is a character function but puts is a string function. The header file cstdio must be included. It prints a character to the monitor.
Eg:
char ch;
ch = getc(stdin);
putch(ch);
puts is used to print a string. The header file stdio.h must be included.
charstr[80];
puts(“Entera string”);
gets(str);
puts(str);

Question 9.
Write a program to check whether a string is palindrome or not. (A string is said to be palindrome if it is the same as the string constituted by reversing the characters of the original string. eg: “MALAYALAM”, “MADAM”, “ARORA”, “DAD”, etc.)
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40];
int len,i,j;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!-\0′;len++);
for(i=0,j=len-1;i<len/2;i++,j–)
if(str[i]!=str[j])
break;
if(i==len/2) .
cout<<str<<” is palindrome”;
else
cout<<str<<” is not palindrome”;
}

Question 10.
Explain multi-character function.
Answer:
getline() and write() functions are multi character functions:
1. getline() It reads a line of text that ends with a newline character. It reads white spaces also.
eg:
char line[80];
cin.getline(line,80);

2. write() It is used to display a string.
Eg.
char line[80];
cin.getline(line,80);
cout.write(line,80);

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 11.
Read a string and print the number of vowels.
Answer:
#include<cstdio>
#include<cctype>
#include<iostream>
using namespace std;
int main()
{
char line[80];
int i,vowel=0;
puts(“Enter a string”);
gets(line);
for(i=0;line[i]!=’\0′;i++)
switch(tolower(line[i]))
{
case ‘a’:
case ‘e’:
case ‘i’;
case ‘o’:
case ‘u’:
vowel++;
}
cout<<“The number of vowels is in the string is “<< vowel;
}

Question 12.
Distinguish between get() and put() functions.
Answer:
get() function:
get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.
Syntax is cin.get(variable);
eg: char ch;
cin.get(ch);

put() function:
put() is an output function. It is used to print a character.
Syntax is cout.put(variable);
eg:
charch;
cin.get(ch);
cout.put(ch);

Question 13.
Write a program to read a string and print the number of consonants.
Answer:
#include<iostream>
#include<cstdio>
#include<cctype>
using namespace std;
int main()
{
char str[40],ch;
int consonent = 0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
{
ch = toupper(str[i]);
if(ch>=’B’ && ch<=’Z’)
if(ch!=’E’&& ch!=’I’&& ch!=’0’&& ch!=’U’)
consonent++;
}
cout<<“The number of consonents is “<<consonent;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 14.
Write a program to read a string and print the number of spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[40];
int space=0,i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
space++;
cout<<“The number of spaces is “<<space;
}

Question 15.
Describe in detail about the unformatted console I/O functions.
Answer:
1. Single character functions: This function is used to read or print a character at a time,
(i) getchar():
It reads a character from the keyboard and store it in a character variable.
eg:
char ch;
ch=getchar();

(ii) putchar():
This function is used to print a character on the screen.
eg:
char ch;
ch = getchar();
putchar(ch);

2. String functions This function is used to read or print a string.
(i) gets():
This function is used to read a string from the keyboard and store it in a character variable.
eg:
charstr[80];
gets(str);

(ii) puts():
This function is used to display a string on the screen.
eg:
char str[80];
gets(str);
puts(str);

Question 16.
Write a program to input a string and find the number of uppercase letters, lowercase letters, digits, special characters and white spaces.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i,digit=0, Ualpha=0, Lalpha=0, special=0, wspace=0;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]>=48 && str[i]<=57)
digit++;
else if(str[i]>=65 && str[i]<=90)
Ualpha++;
else if(str[i]>=97 && str[i]<=122)
Lalpha++;
else if(str[i]==’ ‘ || str[i]==’\t’)
wspace++;
else
special++;
cout<<“The number of alphabets is “<<Ualpha+Lalpha<<
” the number of Uppercase letters is “<<Ualpha<< ” the number of Lowercase letters is “<<Lalpha<<” the number of digits is “<<digit<<” the special characters is “<<special<<” and the number of white spaces is “<<wspace;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 17.
Write a program to count the number of words in a sentence.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i,words=1;
char str[80];
cout<<“Enter a string\n”;
gets(str);
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32)
words++;
cout<<“The number of words is “<<words;
}

Question 18.
Write a program to input a string and replace all lowercase vowels by the corresponding uppercase letters.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[100];
int i;
cout<<“Enter a string:”;
gets(str);
for(i=0;str[i]!=\0′;i++)
if(str[i]>=65 && str[i]<=90 || str[i]> = 97 && str[i]<=122)
switch(str[i])
{
case ‘a’:
str[i] = str[i]-32;
break;
case ‘e’:
str[i] = str[i]-32;
break;
case ‘i’:
str[i] = str[i]-32;
break;
case ‘o’: .
str[i] = str[i]-32;
break;
case ‘u’:
str[i] = str[i]-32;
}
cout<<str;
}

Question 19.
Write a program to input a string and display its reversed string using console I/O functions only. For example if the input is “AND” the output should “DNA”.
Answer:
#include<iostream>
using namespace std;
int main()
{
char str[40],rev[40];
int len.ij;
cout<<“Enter a string:”;
cin>>str;
for(len=0;str[len]!=’\0′;len++);
for(i=0,j=len-1 ;i<len;i++,j–)
rev[il=str[j];
rev[i]=’\0′;
cout<<“The reversed string is “<<rev;
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 9 String Handling and I/O Functions

Question 20.
Write a program to input a word(say COMPUTER) and create a triangle as follows.
C
C O
C O M
C O M P
C O M P U
C O M P U T
C O M P U T E
C O M P U T E R
Answer:
#include<iostream>
#include<cstring>//for strlen()
using namespace std;
int main()
{
charstr[20];
cout<<“enter a word(eg.COMPUTER):”;
cin>>str;
int ij;
for(i=0;i<strlen(str);i++)
{
for(j=0;j<=i;j++)
cout<<str[j]<<“\t”;
cout<<endl;
}
}

Question 21.
Write a program to input a line of text and display the first characters of each word. Use only console I/O functions. For example, if the input is “Save Water, save Nature”, the output should be “SWSN”.
Answer:
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i;
charstr[80];
cout<<“Enter a string\n”;
gets(str);
if(str[0]!=32)
cout<<str[0];
for(i=0;str[i]!=’\0′;i++)
if(str[i]==32 && str[i+1]!=32)
cout<<str[i+1];
}

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Students can Download Chapter 6 Data Types and Operators Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Plus One Data Types and Operators One Mark Questions and Answers

Question 1.
___________ is the main activity carried out in computers.
Answer:
Data processing

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 2.
The data used in computers are different. To differentiate the nature and size of data ___________ is used.
Answer:
Data types

Question 3.
Classify the following data types,
int, array, function, char, pointer, void, float, double, structure
Answer:

Fundamental data typesDerived data types
intarray
floatfunction
doublepointer
voidstructure
char

Question 4.
Sheela wants to store her age. From the following which is the exact data type.
(a) void
(b) char
(c) int
(d) double
Answer:
(c) int

Question 5
Integer data type uses ________ bytes of memory.
(a) 5
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 6.
char data type ________uses bytes of memory.
(a) 1
(b) 3
(c) 7
(d) 8
Answer:
(a) 1

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 7.
From the following which data type uses 4 bytes of memory.
(a) float
(b) short
(c) char
(d) double
Answer:
(a) float

Question 8.
Full form of ASCII is ___________
Answer:
American Standard Code for Information Interchange

Question 9.
Ramu wants to store the value of From the following which is correct declaration.
(a) char pi = 3.14157
(b) int pi = 3.14157
(c) float pi = 3.14157
(d) long pi = 3.14157
Answer:
(c) float pi = 3.14157

Question 10.
From the following which is not true, to give a variable name.
(a) Starting letter must be an alphabet
(b) contains digits
(c) Cannot be a keyword
(d) special characters can be used
Answer:
(d) special characters can be used

Question 11.
Pick a valid variable name from the following.
(а) 9a
(b) float
(c) age
(d) date of birth
Answer:
(c) age.

Question 12.
To perform a unary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(c) 1 (Unary means one)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 13.
To perform a binary operation how many number of operands needed?
(а) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(a) 2 (binary means two)

Question 14.
To perform a ternary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
Answer:
(b) 3 (eg: ternary means three)

Question 15.
In C++ 13 % 26 =
(a) 26
(b) 13
(c) 0
(d) None of these
Answer:
(b) % is a mod operator i.e. it gives the remainder. Here the remainder is 13.

Question 16.
In C++ 41/2 =
(a) 20.5
(b) 20
(c) 1
(d) None of these
Answer:
(b) 20. (The actual result is 20.5 but both 41 and 2 are integers so .5 must be truncated)

Question 17.
++ is a __________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(a) Unary.

Question 18.
Conditional operator is _________ operator.
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(c) Ternary

Question 19.
% is a _______ operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
Answer:
(b) Binary

Question 20.
State True/False

  1. Multiplication, division, modulus have equal priority
  2. Logical and (&&) has less priority than logical or ()

Answer:

  1. True
  2. False

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 21.
_______is composed of operators and operands.
(a) expression
(b) Keywords
(c) Identifier
(d) Punctuators
Answer:
(a) expression

Question 22.
Supply value to a variable at the time of declaration is known as __________.
Answer:
Initialisation

Question 23.
From the following which is initialisation.
(a) int k;
(b) int k = 100;
(c) int k[10];
(d) None of these
Answer:
(b) int k = 100;

Question 24.
State True/False
In an expression all the operands having lower size are converted(promoted) to the data type of the highest sized operand.
Answer:
True

Question 25.
Classify the following as arithmetic/Logical expression.
(a) x + y * z
(b) x < y && y > z
(c) x / y
(d) x > 89 || y < 80
Answer:
(a) and (c) are Arithmetic, (b) and (d) are Logical

Question 26.
Suppose x = 5 and y = 2 then what will be cout<<(float)x/y
Answer:
2.5 The integer x is converted to float hence the result.

Question 27.
Consider the following.
a = 10;
a* =10;
Then a =
(a) a = 100
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(a) a = 100. This short hand means a = a*10

Question 28.
Consider the following.
a = 10;
a+ = 10;
Then a =
(a) a = 30
(b) a = 50
(c) a = 10
(d) a = 20
Answer:
(d) a = 20. This short hand means a = a + 10

Question 29.
Pick the odd one out
(a) structure
(b) Array
(c) Pointer
(d) int
Answer:
(d) int, it is fundamental data type the others are derived data types

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 30.
From the following select not a character of C++ language
(a) A
(b) 9
(c) \
(d) @
Answer:
(d) @

Question 31.
Consider the following
float x = 25.56;
cout<<(int)x;
Here the data type of the variable is converted. What type of conversion is this?
(a) type promotion
(b) type casting
(c) implicit conversion
(d) None of these
Answer:
(b) type casting (explicit conversion);

Question 32.
Identify the error in the following C++ statement and correct it.
short population = 68000;
Answer:
The maximum number that can store in short type is less than 32767. So to store 68000 we have to use long data type.

Question 33.
Consider the following statements in C++ if(mark>=18)
cout<<“Passed”;
else
cout<<“Failed”;
Suggest an operator in C++ using which the same output can be produced.
Answer:
Conditional operator (?:)

Plus One Data Types and Operators Two Mark Questions and Answers

Question 1.
Analyses the following statements and write True or False. Justify

  1. There is an Operator in C++ having no special character in it
  2. An operator cannot have more than 2 operands
  3. Comma operator has the lowest precedence
  4. All logical operators are binary in nature
  5. It is not possible to assign the constant 5 to 10 different variables using a single C++ expression
  6. In type promotion the operands with lower data type will be converted to the highest data type in expression.

Answer:

  1. True (sizeof operator)
  2. False(conditional operator can have 3 operands)
  3. True
  4. False
  5. False(Multiple assignment is possible)
    eg: a = b = c ==5
  6. True

Question 2.
Consider the following declaration.
const int bp;
bp = 100;
Is it valid? Explain it?
Answer:
This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration is as follows, const int bp = 100;

Question 3.
Consider the following statements in C++

  1. cout<<41/2;
  2. cout<<41/2.0;

Are this two statements give same result? Explain?
Answer:
This two statements do not give same results. The first statement 41/2 gives 20 instead of 20.5. The reason is 41 and 2 are integers. If two operands are integers the result must be integer, the real part must be truncated.

To get floating result either one of the operand must be float. So the second statement gives 20.5. The reason 41 is integer but 2.0 is a float.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 4.
If mark = 70 then what will be the value of variable result in the following
result = mark > 50? ’P’: ’F’;
Answer:
The syntax of the conditional operator is given below
Condition ? Value if true : Value if false;
Here the conditional operator first checks the condition i.e., 70 > 50 it is true. So ’P’ is assigned to the variable result.
So the result is ’P’;

Question 5.
Is it possible to initialise a variable at the time of execution. What kind of initialisation is this? Give an example
Answer:
Yes it is possible. This is known as Dynamic initialisation. The example is given below
eg: int a = 10, b = 5; int c = a*b; here the variable c is declared and initialised with the value 10*5.

Question 6.
Boolean data type is used to store True/False in C++. Is it true? Is there any data type called Boolean in C++?
Answer:
No there is no data type for storing boolean value true/false. But in C++ non -zero (either negative or positive) is treated as true and zero is treated as false

Question 7.
Consider the following
n=-15;
if (n)
cout<<“Hello”;
else
cout<<“hai”;
What will be the output of the above code?
Answer:
The output is Hello, because n = -15 a non zero number and it is treated as true hence the result.

Question 8.
Is it possible to declare a variable in between the program as and when the need arise? Then what is it?
Answer:
Yes it is possible to declare a variable in between the program as and when the need arise. It is known as dynamic initialisation.
eg. int x = 10, y = 20;
int z = x*y;

Question 9.
char ch;
cout<<“Enter a character”;
cin>>ch;
Consider the above code, a user gives 9 to the variable ‘ch’. Is there any problem? Is it valid?
Answer:
There is no problem and it is valid since 9 is a character. Any symbol from the key board is treated as a character.

Question 10.
“With the same size we can change the sign and range of data”. Comment on this statement.
Answer:
With the help of type modifiers we can change the sign and range of data with same size. The important modifiers are signed, unsigned, long and short.

Question 11.
Write short notes about C++ short hands?
Answer:
x = x + 10 can be represented as x + = 10, It is called shorthands in C++. It is faster. This is used with all the arithmetic operators as follows.

Arithmetic Assignment ExpressionEquivalent Arithmetic Expression
x+ = 10x = x + 10
x- = 10x = x -10
X* = 10x = x * 10
x/ = 10x = x /10
x% = 10x = x % 10

Question 12.
What is the role of ‘const’ modifier?
Answer:
This ‘const’ keyword is used to declare a constant.
eg: const int bp=100;
By this the variable bp is treated as constant and cannot be possible to change its value during execution.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 13.
Specify the most appropriate data type for handling the following data.

  1. Roll no. of a student.
  2. Name of an employee.
  3. Price of an article.
  4. Marks of 12 subjects

Answer:

  1. short Roll no;
  2. char name[20];
  3. float price;
  4. short marks[12];

Question 14.
Write C++ statement for the following,

  1. The result obtained when 5 is divided by 2.
  2. The remainder obtained when 5 is divided by 2.

Answer:

  1. 5/2
  2. 5 % 2

Question 15.
Predict the output of the following code. Justify.
int k = 5;
b = 0;
b = k++ + ++k;
cout<<b;
Answer:
Output is 12. In this statement first it take the value of k in 5 then increment it K++. So first operand for + is 5. Then it becomes 6. Then ++k makes it 7. This is the second operand. Hence the result is 12.

Question 16.
Predict the output.
1. int sum = 10, ctr = 5;
sum = sum + ctr–;
cout<<sum;

2. int sum = 10, ctr = 5;
sum = sum + ++ctr;
cout<<sum;
Answer:

  1. 15
  2. 16

Question 17.
Predict the output.
int a;
float b;
a = 5;
cout<<sizeof(a + b/2);
Answer:
Output is 4. Result will be the memory size of floating point number

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 18.
Predict the output.
int a, b, c;
a = 5;
b = 2;
c = a/b;
cout<<c;
Answer:
Output is 2. Both operands are integers. So the result will be an integer

Question 19.
Explain cascading of i/o operations
Answer:
The multiple use of input or output operators in a single statement is called cascading of i/o operators.
eg: To take three numbers by using one statement is as follows
cin>>x>y>>z;
To print three numbers by using one statement is as follows
cout<<x<<y<<z;

Question 20.
Trace out and correct the errors in the following code fragments

  1. cout<<“Mark =”45;
  2. cin<<“Hellow World!”;
  3. cout>>”X + Y;
  4. Cout<<‘Good,<<‘Morning’

Answer:

  1. cout<<“Mark = 45”;
  2. cout<<“Hellow World!”;
  3. cout<<X + Y;
  4. Cout<<“Good Morning”;

Question 21.
Raju wants to add value 1 to the variable ‘p’ and store the new value in ‘p’ itself. Write four different statements in C++ to do the task.
Answer:

  1. P=P+1;
  2. p++;(post increment)
  3. ++p;(pre increment)
  4. p+=1;(short hand in C++)

Question 22.
Read the following code
char str [30];
cin>>str;
cout<<str;
If we give the input “Green Computing”, we get the output “Green”. Why is it so? How can you correct that? (2)
Answer:
The input statement cin>> cannot read the space. It reads the text up to the space, i.e. the delimiter is space. To read the text up to the enter key gets() or getline() is used.

Question 23.

NameSymbol
(a) Modulus operator(i) ++
(b) Logical Operator(ii) ==
(c) Relational Operator(iii) =
(d) Assignment operator(iv) ?:
(e) Increment operator(v) &&
(f) Conditional Operator(vi) %

Answer:

NameSymbol
(a) Modulus operator(vi) %
(b) Logical Operator(v) &&
(c) Relational Operator(ii) ==
(d) Assignment operator(iii) =
(e) Increment operator(i) ++
(f) Conditional Operator(iv) ?:

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 24.
Write a C++ expression to. calculate the value of the following equation.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 1
Answer:
x = (-b + sqrt(b*b – 4*a*c)/(2*a)

Question 25.
A student wants to insert his name and school address in the C++ program that he has written. But this should not affect the compilation or execution of the program. How is it possible? Give an example.
Answer:
He can use comments to write this information. In C++ comments are used to write information such as programmer’s name, address, objective of the codes etc. in between the actual codes. This is not the part of the programme. There are two types of comments

  1. Single line (//) and
  2. Multi-line (/* and *f)

1. Single line (if):
Comment is used to make a single line as a comment. It starts with //.
eg: /./programme starts here.

2. Multi-line (/* and */):
To make multiple lines as a comment. It starts with /* and ends with */.
Eg: /* this programme is used to find sum of two numbers */

Question 26.
Consider the following C++ statements:
char word [20];
cin>>word;
cout<<word;
gets(word);
puts(word);
If the string entered is “HAPPY NEW YEAR”, predict the output and jsutify your answer.
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets () function. Because gets () function reads character upto the enter key.
Hence
gets(word);
puts(word);
Displays “HAPPY NEW YEAR”

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 27.
Write the difference between x = 5 and x == 5 in C++.
Answer:
x = 5 means the value 5 of the RHS is assigned to the LHS variable x . Here = is the assignment operator. But x == 5, == this is the relational (comparison) operator. Here it checks whether the value of RHS is equal to the value of LHS and this expression returns a boolean value as a result. It is the equality operation.

Question 28.
1. What is the output of the following program?
# include <iostream.h>
void main ()
{
int a;
a = 5 + 3*5;
cout << a;
}

2. How do 9, ‘9’ and “9” differ in C++ program?
Answer:
Here multiplication operation has more priority than addition.
hence
1. a = 5 + 15 = 20

2. Here 9 is an interger
‘9’ is a character
“9” is a string

Question 29.
Read the following C++ program and predict the output by explaining the operations performed.
#include<iostream.h>
void main ()
{
int a = 5, b = 3;
cout<<a++ /–b;
cout<<a/ (float) b;
}
Answer:
Here a = 5 and b = 3
a++ /– b = 5/2 = 2
That is a++ uses the value 5 and next it changes its value to 6
So a/(float) b = 6/(float)2
= 6/2.0
= 3
So the output is 2 and 3

Question 30.
What is the preprocessor directive statement? Explain with an example.
Answer:
A C++ program starts with the preprocessor directive i.e., # include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
eg. #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 31.
The following C++ code segment is a part of a program written by Smitha to find the average of 3 numbers.
int a, b, c;
float avg;
cin>>a>>b>>c;
avg = (a + b + c)/3;
cout<<avg; .
What will be the output if she inputs 1, 4 and 5? How can you correct it?
Answer:
= (1 + 4 + 5)/3
= 10/3
= 3.3333
Instead of this 3.3333 the output will be 3. This is because if both operands are integers an integer division will be occurred, that is the fractional part will be truncated. To get the correct output do as follows
case 1: int a,b,c; is replaced by float a,b,c;

OR

case 2: Replace (a + b + c)/3 by (a + b + c)/3.0;

OR

case 3: Type casting.
Replace avg = (a + b + c)/3;
by avg = (float)(a + b + c)/3;

Plus One Data Types and Operators Three Mark Questions and Answers

Question 1.
In a panchayath or municipality all the houses have a house number, house name and members. Similar situation is in the case of memory. Explain
Answer:
The named memory locations are called variable. A variable has three important things

  1. variable name: A variable should have a name
  2. Memory address: Each and every byte of memory has an address. It is also called location (L) value
  3. Content: The value stored in a variable is called content.lt is also called Read(R) value.

Question 2.
Briefly explain constants.
Answer:
A constant or a literal is a data item its value doe not change during execution. The keyword const is used to declare a constant. Its declaration is as follows
const data type variable name = value;
eg.const int bp = 100;
\const float pi = 3.14157;
const char ch = ‘a’;
const char[]=”Alvis”;

1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
eg: For decimal 100, 150, etc.
For octal 0100, 0240, etc.
For hexadecimal 0x100, 0x1A, etc.

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
eg: 3.14157, 79.78, etc

3. Character literal:
A valid C++ character enclosed in single quotes, its value does not change during execution.
eg: ‘m’, ‘f, etc.

4. String literal:
One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’)
eg: “Mary’s”, “India”, etc.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 3.
Consider the following statements
int a = 10, x = 20;
float b = 45000.34, y = 56.78;
1. a = b;
2. y = x;
Is there any problem for the above statements? What do you mean by type compatibility?
Answer:
Assignment operator is used to assign the value of RHS to LHS. Following are the two chances
(a) The size of RHS is less than LHS. So there is no problem and RHS data type is promoted to LHS. Here it is compatible.

(b) The size of RHS is higher than LHS. Here comes the problem sometimes LHS cannot possible to assign RHS. There may be a chance of wrong answer. Here it is not compatible.
Here
1. a = b; There is an error since the size of LHS is 2 but the size of RHS is 4.
2. y = x; There is no problem because the size of LHS is 4 and RHS is 2.

Question 4.
A company has decided to give incentives to their salesman as perthe sales. The criteria is given below.
If the total sales exceeds 10,000 the incentive is 10%

  1. If the total sales >= 5,000 and total sales <10,000, the incentive is 6 %
  2. If the total sales >= 1,000 and total sales <5,000, the incentive is 3 %

Write a C++ program to solve the above problem and print the incentive after accepting the total sales of a salesman. The program code should not make use of ‘if’ statement.
Answer:
#include<iostream>
using namespace std;
int mainO
{
float sales,incentive;
cout<<“enter the sales”;
cin>>sales;
incentive = (sales>10000 ? sales*.10: (sales > =5000 ? sales * .06 : (sales >= 1000 ? sales * -03: 0)));
cout<<“\nThe incentive is ” << incentive;
}

Question 5.
A C++ program code is given below to find the value of X using the expression
\(x=\frac{a^{2}+b^{2}}{2 a}\)
where a and b are variables

#include<iostream>
using namespace std;
int main()
{
int a;b;
float x
cout<<“Enter the values of a and b;
cin>a>b;
x = a*a + b*b/2*a;
cout>>x;
}
Predict the type of errors during compilation, execution and verification of the output. Also write the output of two sets of input values

  1. a = 4, b = 8
  2. a = 0, b = 2

Answer:
This program contains some errors and the correct program is as follows.
#include<iostream>
using namespace std;
int main()
{
int a,b;
float x; .
cout<<“Enterthe values of a and b”;
cin>>ab;
x=(a*a + b*b)/(2*a);
cout<<x;
}
The output is as follows

  1. a = 4 and b = 8 then the output is 10
  2. a = 0 and b = 2 then the output is an error divide by zero error(run time error)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 6.
A list of data items are given below
45, 8.432, M, 0.124,8 , 0, 8.1 × 1031, 1010, a, 0.00025, 9.2 × 10120, 0471,-846, 342.123E03

  1. Categorise the given data under proper headings of fundamental data types in C++.
  2. Explain the specific features of each data type. Also mention any other fundamental data type for which sample

data is not given
Answer:
1.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 2

2. The specific features of each data type.
(i) int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It con¬sumes 4 bytes (32 bits) of memory.i.e. 232 . numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve ) So a total of 232 numbers. We can store a number in between -231 to + 231-1.

(ii) char data type:
Any symbol from the keyboard, eg. ‘A’ , ‘?’, ‘9’,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 7.
Write valid reasons after reading the following statements in C++ and comment on their correctness by give reasons.

  1. char num = 66;
    char num = B’;
  2. 35 and 35L are different
  3. The number 14, 016 and OxE are one and the same
  4. Char data type is often said to be an integer type
  5. To store the value 4.15 float data type is preferred over double

Answer:

  1. The ASCII number of B is 66. So it is equivalent.
  2. 35 is of integer type but 35L is Long
  3. The decimal number 14 is represented in octal is 016 and in hexadecimal is OXE.
  4. Internally char data type stores ASCII numbers.
  5. To store the value 4.15 float data type is better because float requires only 4 bytes while double needs 8 bytes hence we can save the memory.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 8.
Suggest most suitable derived data types in C++ for storing the following data items or statements

  1. Age of 50 students in a class
  2. Address of a memory variable
  3. A set of instructions to find out the factorial of a number
  4. An alternate name of a previously defined variable.
  5. Price of 100 products in a consumer store
  6. Name of a student

Answer:

  1. Integer array of size 50
  2. Pointer variable
  3. Function
  4. Reference
  5. Float array of size 100
  6. Character array

Question 9.
Considering the following C++ statements. Fill up the blanks

  1. If p = 5 and q = 3 then q%p is _______
  2. If E1 is true and E2 is False then E1 && E2 will be _______
  3. If k = 8, ++k < = 8 will be ________
  4. If x = 2 then (10* ++x) % 7 will be ________
  5. If t = 8 and m = (n=3,t-n), the value of m will be ______
  6. If i = 12 the value i after execution of the expres¬sion i+ = i- – + – -i will be ______

Answer:

  1. 3
  2. False
  3. False(++k makes k = 9. So 9<=8 is false)
  4. 2(++x becomes 3 ,so 10 * 3 = 30%7 = 2)
  5. 5( here m = (n = 3,8-3) = (n = 3,5), so m = 5, The maximum value will take)
  6. Here i = 12

i + = i- – + – -i
here post decrement has more priority than pre decrement. So “i- -” will be evaluated first. Here first uses the value then change so it uses the value 12 and i becomes 11
i + = 12 + – -i
now i = 11.
Here the value of i will be changed and used so “i- -” becomes 10
i + = 12 + 10 = 22
So, i = 22 +10
i = 32
So the result is 32.

Question 10.
The Maths teacher gives the following problem to Riya and Raju.
x = 5 + 3 * 6.
Riya got x = 48 and Raju got x = 23. Who is right and why it is happened? Write down the operator precedence in detail?
Answer:
Here the answer is x = 23. It is because of precedence of operators. The order of precedence of operators are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 8
Here multiplication has more priority than addition

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 11.
Explain the data types’ in C++. (3)
Answer:
Fundamental data types:
It is also called built-in data type. They are int, char, float, double and void
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory. i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type Any symbol from the keyboard, eg. ‘A’,‘9’, …. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having an ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
Void means nothing. It is used to represent a function returns nothing.

  • User defined Data types: C++ allows programmers to define their own data type. They are Structure(struct), enumeration (enum), union, class, etc.
  • Derived data types: The data types derived from fundamental data types are called Derived data types. They are Arrays, pointers, functions, etc

Question 12.
Predict the output of the following C++ statements.
int a = -5, b = 3, c = 4;
C+ = a++ + –b;
cout<<a<<b<<c;
Answer:
a = -4, b = 2 and c = 1.

Question 13.
Match the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 9
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 10

Question 14.
Write any five unary operators of C++. Why are they called so?
Answer:
A unary operator is an operator that need only one operand to perform the operation. The five unary operators of C++ are given below.
Unary +, Unary -, ++, – – and ! (not)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 15.
Write C++ examples for the following:

  1. Declaration statement
  2. Assignment statement
  3. Type casting

Answer:

  1. int age;
  2. age = 16;
  3. avg = (float)a + b + c/3;

Plus One Data Types and Operators Five Mark Questions and Answers

Question 1.

  • Name: Jose
  • Roil no: 20
  • Age: 17
  • Weight: 45.650

Consider the above data, we know that there are different types of data are used in the computer. Explain different data types used in C++.
Answer:
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory, i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

2. char data type:
Any symbol from the keyboard, eg. A’,’9′,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e, the numbers with decimal point. It uses 4 bytes(32 bits) of memory.
eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators

Question 2.
Define an operator and explain operator in detail.
Answer:
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators
1. lnput(>>) and output(<<):
These operators are used to perform input and output operation.
eg: cin>>n;
cout<<n;

2. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction(-), division (/), multiplication (*) and modulus (%- gives the remainder) operations.
eg: If x = 10 and y = 3 then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 3
x/y = 3, because both operands are integer. To get the floating point result one of the operand must be float.

3. Relational operator:
It is also a binary operator. It is used to perform comparison or relational operation between two values and it gives either true(1) or false(O). The operators are <,<=,>,>=,== (equality)and !=(not equal to)
eg: If x = 10 and y = 3 then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 4

4. Logical operators:
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true(1) or false(O). If x=True and y=False then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 5
Both operands must be true to get a true value in the case of AND(&&) operation If x = True and y = False then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 6
Either one of the operands must be true to get a true value in the case of OR(||) operation If x = True and y = False then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 6 Data Types and Operators 7

5. Conditional operator:
It is a ternary operator hence it needs three operands. The operator is ?:
Syntax: expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.
eg: If x = 10 and y = 3 then x>y ? cout<<x : cout<<y;. Here the output is 10

6. sizeof():
This operator is used to find the size used by each data type.
eg. sizeof(int) gives 2.

7. Increment and decrement operator:
These are unary operators.
(a) Increment operator (++): It is used to incre¬ment the value of a variable by one i.e., x++ is equivalent to x = x + 1;
(b) Decrement operator (–): It is used to decre¬ment the value of a variable by one i.e., x-is equivalent to x = x – 1.

8. Assignment operator (=):
lt is used to assign the value of a right side to the left side variable.
eg: x = 5; Here the value 5 is assigned to the variable x.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Students can Download Chapter 6 Thermodynamics Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Plus One Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
Hot coffee in a thermos flask is an example of system.
Answer:
Isolated

Question 2.
Which of the following statements is incorrect about internal energy?
a) The absolute value of internal energy cannot be determined
b) The internal energy of one mole of a substance is same at any temperature or pressure
c) The measurement of heat change during a reaction by bomb calorimeter is equal to the internal energy change
d) Internal energy is an extensive property
Answer:
b) The internal energy of one mole of a substance is same at any temperature or pressure

Question 3.
For which of the following the standard enthalpy is not zero?
a) C (Diamond)
b) C (Graphite)
c) Liquid mercury
d) Rhombicsulphur
Answer:
a) C (Diamond)

Question 4.
Say TRUE or FALSE?
Any spontaneous process must lead to a net increase in entropy of the universe.
Answer:
TRUE

Question 5.
The ∆H fora reaction is-30 kJ. On the basis of this fact, we can conclude that the reaction
a) Gives off thermal energy
b) Is fast
c) Is slow
d) Is spontaneous
Answer:
a) Gives off thermal energy

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 6.
Write the type of system in each of the following:

  1. Hot water taken in an open vessel
  2. Hot water taken in a closed metallic vessel
  3. Hot water taken in a thermos flask

Answer:

  1. Open system
  2. Closed system
  3. Isolated system

Question 7.
In a reversible process the total change in entropy is ∆s(universe) is
Answer:
Zero

Question 8.
For the reaction Ag2O \(\rightleftharpoons \) 2Ag + \(\frac{1}{2}\)O2(g) ∆S and ∆H are 66J K-1mol-1 and 30.56 Kg mol respectively. The reaction will not be spontaneous at.
Answer:
463K

Question 9.
One mole of methane undergoes combustion to form CO2 and water at 25°C. The difference between ∆U & ∆H will be
Answer:
-2RT

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
A gas expands from 1 l to 6 l against a constant pressure of 1 atm and it absorbs 500J of heat ∆μ is
Answer:
-6.5J

Question 11.
Born Haber cycle is to find out __________
Answer:
lattice energy

Plus One Chemistry Thermodynamics Two Mark Questions and Answers

Question 1.
1. Explain enthalpy of fusion.
2. Give illustration of fusion of ice.
Answer:
1. It is the enthalpy change when one mole of a solid is converted into its liquid at its melting point.

2. Enthalpy of fusion of ice is 6 kJ/mol. From this it is clear that 6 kJ of energy is required to convert one mole of ice (18 g) into water at 0°C.

Question 2.
a) What do you meant by enthalpy of vapourisation?
b) Explain enthalpy of sublimation.
Answer:
a) It is the enthalpy change when one mole of a liquid is converted into its vapour at its boiling point.
b) It is the enthalpy change when one mole of a solid is converted into its vapour at its transition temperature.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
One equivalent of an acid reacts completely with one equivalent of a base in dilute solution.
1. Which type reaction is this?
2. HCl + NaOH → NaCl + H2O
On the basis of above equation, explain enthalpy of neutralisation.
Answer:
1. Nneutralisation.

2. When one equivalent of HCl (36.5 g) reacts completely with one equivalent of NaOH (40 g), 57.1 kJ energy is liberated.

Question 4.
1. What is the difference between system and surroundings?
2. There are different types of systems. What are they? Explain.
3. Give example for different types of systems.
Answer:
1. A system in thermodynamics refers to that part of universe in which observations are made. The remaining part of the universe other than the system constitutes the surroundings.

2. System is classified into the following three types. Open system: This is a system in which there is exchange of energy and matter between system and surroundings.
Closed system:
This is a system in which there is no exchange of matter, but exchange of energy is possible between system and the surroundings.

Isolated system:
This is a system in which there is no exchange of energy or matter between the • system and the surroundings.

3. Open system
Presence of reactants in an open beaker
Closed system
Presence of reactants in a closed vessel made of conducting material
Isolated system
Presence of reactants in a • thermos flask or any other closed insulated vessel

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 5.
Match the following:

AB
1. IsothermalTemperature varies
2. AdiabaticTemperature constant
3. IsobaricVolume constant
4. IsochoricPressure constant

Answer:

AB
1. IsothermalTemperature constant
 2.AdiabaticTemperature varies
3. IsobaricPressure constant
4. IsochoricVolume constant

Question 6.
1. What is meant by enthalpy?
2. Derive an equation for enthalpy change.
3. What is enthalpy change?
Answer:
1. Enthalpy is the sum of internal energy and pressure volume energy.
i.e.,H = U + pV

2. ∆H = ∆U + ∆pV)
∆H = ∆U + p∆V + V∆p
At constant pressure, ∆p=0
∆H = ∆U+p∆V
But ∆U=q+w
∆H=q+w+p∆V
w = -p∆V
i.e; ∆H= q – p∆V + p∆V
∆H = qp

3. Enthalpy change is heat absorbed or released at constant pressure.

Question 7.
1. Find the enthalpy of the reaction,
C(graphite) + O2(g) → CO2(g)
Given,
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =-110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?
Answer:
1. Considerthe reaction,
C(grahite) + O2(g) → CO2 (g); ∆H = x
CO2 can also be prepared through the following two steps:
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
Then by Hess’s law, x = (-110.5+-283.0) kJ=-393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.
For a spontaneous process ∆STotal is +ve.
For a nonspontaneous process ∆STotal is -ve.

Question 8.
Explain the following:

  1. Enthalpy of atomization
  2. Enthalpy of solution at infinite dilution

Answer:

  1. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
  2. It is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions or solute molecules are negligible.

Question 9.
The enthalpy change for the reaction,
N2(g) + 3H2(g) → 2NH3(g) is -92.38 kJ at 298 K.
What is ∆U at 298 K?
Answer:
∆U = ∆H — ∆ngRT
= -92.38 × 10³ J – [-2 × 8.314J K-1 mol-1 × 293 K)]
= -92.38 × 10³J +4.872 × 10³J
= -87.51 × 10³J
= – 87.51 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
What are the two types of heat capacities? How they are related?
Answer:
The two types of heat capacities are heat capacity at constant pressure (Cp) and heat capacity at constant volume (Cv). These two are related as Cp – Cv = R, where R is the universal gas constant.

Question 11.
Enthalpy and Entropy changes of two reactions are given below: Find out whether they are spontaneous or not at 27°C. Justify.
1. ∆H = 26 kJ/mole, ∆S = 8.3 J/K/mole
2. ∆H = -393.4 kJ/mole, ∆S = 6 J/K/mole
Answer:
1. ∆G = ∆H -T∆S
= 26000 – 300 × 8.3 = 23.510
Since ∆G is positive, the process is non-spontaneous.

2. ∆G = ∆H -T∆S
= -393400 – 300 × 6 = -391600
Since ∆G is negative, the process is spontaneous.

Question 12.
1. What is enthalpy of solution?
2. What is enthalpy of dilution?
Answer:
1. Enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure.

2. Enthalpy of dilution is the heat withdrawn from the surroundings when additional solvent is added to the solution. It is dependent on the original concentration of the solution and the amount of solvent added.

Question 13.
What is the significance of the second law of thermodynamics in the spontaneity of exothermic and endothermic reactions?
Answer:
The second law of thermodynamics provides explanation for the spontaneity of chemical reactions. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall 1 entropy change is positive which makes the reaction spontaneous.

In the case of endothermic reactions, since heat is ‘ absorbed by the system from the surroundings, the entropy change of the surroundings becomes negative (∆Ssurr < 0). In this case the process will be spontaneous only if the entropy change of the reacting system is postive (∆Ssys > 0) and is also greater than ASsurr in magnitude so that the overall entropy change (∆Stotal) is positive.

Question 14.
Explain the importance of third law of thermodynamics.
Answer:
The importance of third law of thermodynamics lies in the fact that it permits the calculation of absolute values of entropy of pure substances from thermal data alone. For a pure substance, this can be done by summing \(\frac { { q }_{ rev } }{ T } \) increments from 0 K to 298 K.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 1

Question 15.
C12H22O11 + 12O2 → 12CO2 + 11H2O
Consider this equation and answer the following questions.
a) Thermodynamically, which type reaction is this?
b) What is enthalpy of combustion?
c) Give another example.
Answer:
a) Combustion.
b) It is the enthalpy change when one mole of a substance undergoes complete combustion in excess of air or oxygen.
c) C6H12O6 + 6O2 → 6CO2 + 6H2O

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 16.
Bond dissociation energies of hydrogen and nitrogen are 430 kJ and 41.8 kJ respectively and the enthalpy of formation of NH3 is – 46 kJ. What is the bond energy of N-Hbond?
Answer:
3H2 + N2 → 2NH3; AH = -46 kJ
3 × ½H2 + ½N2 → 2 × ½NH3
[(3× ½ × 430) + (½ × 941 .8)] – (3N-H) = – 46
[(3 × 215) + (470.9) + (46)] → [3N-H]
[645 + 470.9 + 46] = 3N-H
N-H = 387.3 kJ

Plus One Chemistry Thermodynamics Three Mark Questions and Answers

Question 1.
In 1840, G.H.Hess (a Russian chemist) proposed an important generalisation of thermochemistry which is known after his name as Hess’s law.
1. State Hess’s law.
2. Give illustration of Hess’s law.
Answer:
1. Enthalpy change in a chemical reaction is same whether it takes place in one step or in more than one step.

2. Considerthe formation of CO2.
C + O2 → CO2; ∆H = x
CO2 can be prepared through the following two steps:
C + ½O2 → CO ; ∆H = y
CO + ½O2 → CO2; ∆H = Z
Then by Hess’s law,
x = y + z

Question 2.
∆U = q-p∆V. If the process is carried out at constant
volume, then ∆V=0. Answer the following questions.
1. Give the equation for ∆U.
2. 1000J was supplied to a system at constant volume. It resulted in the increase of temperature of the system from 45 °C to 50 °C. Calculate the change in internal energy.
Answer:
1. ∆U = qv

2. Since the volume kept constant, ∆V=0
∴ ∆U = qv = 1000J

Question 3.
Thermodynamics deals with macroscopic properties.
1. What is the difference between extensive and intensive properties?
2. Classify the following properties into extensive and intensive.
Pressure, Mass, Volume, Temperature, Density, Heat capacity, Viscosity, Surface tension, Internal • energy, Molar heat capacity, Refractive index, Enthalpy, Specific heat capacity
Answer:
1. Extensive properties are those properties whose value depend on the quantity or size of matter present in the system.
Intensive properties are those properties which do not dependent on the quantity or size of matter present in the system.

2. Extensive properties: Mass, Volume, Heat capacity, Internal energy, Enthalpy Intensive: Pressure, Temperature, Density, Viscosity, Surface tension, Molar heat capacity, Refractive index, Specific heat capacity.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 4.
1. What is meant by state of the system and state variables?
2. Give any four examples for state variables/state functions.
Answer:
1. The state of a system refers to the conditions of existence of a system when its macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. The measurable properties required to describe the state of a system are called state variables or state functions. A state function is a property of a system whose value depends only upon the initial and final states of the system and is independent of the path by which this state has been reached. Properties whose values depend on the path followed are called path functions.

2. State variables/State functions – Temperature, Pressure, Enthalpy, Entropy

Question 5.
1. Explain the Zeroth law of thermodynamics.
2. What are the important modes of transference of energy. Explain.
Answer:
1. The Zeroth law of thermodynamics states that if two bodies say, ‘A’ and ‘B’ are in thermal equilibrium with another body say, ‘C’, then the bodies A’ and ‘B’will also be in thermal equilibrium with each other. It provides the basis for the measurement of temperature.

2. The two important modes of transference of energy are heat and work.
Heat:
The exchange of energy, which is a result of temperature difference between system and surroundings is called heat (q).

Work:
The exchange of energy between system and surroundings can occur in the form of work which can be mechanical work, electrical work or pressure-volume work. The exchange of energy as pressure-volume work can occur if system consists of gaseous substance and there is a difference of pressure between system and surroundings.

Question 6.
1. Explain the symbols and sign conventions of heat and work.
2. Explain internal energy.
Answer:
1. Heat is represented by the symbol ‘q’. The ‘q’ is positive, when heat is transferred from the surroundings to the system and ‘q’ is negative when heat is transferred from system to the surroundings.
Work is represented by the symbol ‘w’. The ‘w’ is positive when work is done on the system and ‘w’ is negative when work is done by the system.

2. Every substance is associated with a definite amount of energy due to its physical and chemical constitution. This is called internal energy. It is the sum of the different types of energies such as chemical, electrical, mechanical etc.

Question 7.
Fill in the blanks.

  1. If heat is released, ‘q’ is ……………
  2. For exothermic process ‘∆H’is ………………
  3. If work is done on the system, ‘w’ is ………………
  4. For endothermic process ‘∆H’ is ………………
  5. If work is done by the system, ‘w’ is ………………

Answer:

  1. Negative
  2. Negative
  3. Positive
  4. Positive
  5. Negative

Question 8.
1. What is meant by enthalpy of formation?
2. What is the value of standard enthalpy of formation (∆fH) of an element?
Answer:
1. Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable states of aggregation (i.e., reference state).

2. Zeno

Question 9.
First Law of thermodynamics is the law of conservation of energy.

  1. Give the mathematical form of the first law.
  2. Write the Gibb’s equation.
  3. What is the sign for ∆G for a spontaneous process?

Answer:

  1. ∆U=q+w w = work done
    q = heat absorbed
  2. G = H- TS or ∆G= ∆H – T∆S
  3.  In the case of spontaneous process ∆G = -ve

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
1. Predict the sign of ∆S for the reaction,
NH3(g) + HCl(g) → NH4Cl(s)
2. The reaction between gaseous hydrogen and chlorine is
H2(g) + Cl2(g) → 2HCl(g); ∆rH = -1840 kJ
i) What is the enthalpy of formation of HCl?
ii) How much heat will be liberated at 298 K and 1 atm for the formation of 365 g of HCl?
Answer:
1. ∆S is negative

2. i) ∆fH = \(\frac{-1840}{2}\) = -920 kJ mol-1
ii) Heat liberated during the formation of 1 mole (36.5 g) of HCl = 920 kJ
∴ Heat liberated during the formation of 365 g of HCl = 9200 kJ

Question 11.
Derive the Meyer’s relationship.
Answer:
We have, q = C × ∆T
At constant volume, qv = Cv × ∆T = ∆U
At constant pressure, qp = Cp × ∆T = ∆H
For 1 mole of an ideal gas,
∆H = ∆U + ∆(pV) = ∆U + ∆(RT) = ∆U + R∆T
∴ ∆H = ∆U + R∆T
On putting the values of ∆H and ∆U,
Cp∆T = Cv∆T + R∆T
Cp = Cv + R
Cp – Cv = R, which is the Meyer’s relationship.

Question 12.
1. In a process 701J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
2. What is free expansion? What is the work done during free expansion of an ideal gas?
Answer:
1. ∆U=q+w = 9 + w = 701J – 394 J = 307J

2. Expansion of a gas in vacuum is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.

Question 13.
1. Name the instrument used for measuring the ∆U of a process.
2. What is the value of ∆G for a reaction at equilibrium?
3. ∆H and ∆S of a reaction are 30.56 and 0.666 kJ/ mol respectively at 1 atm pressure. Calculate the temperature at which the reaction is in equilibrium.
Answer:
1. Bomb calorimeter
2. Zeno
3. ∆H-T∆S = 0 or ∆H = T∆S
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 2

Question 14.
Thermodynamic process differ based on the manner
in which it is carried out.
1. Distinguish between reversible and irreversible processes.
2. Calculate the amount of work done when 2 moles of a gas expands from a volume of 2 L to 6 L isothermally and irreversibly against a constant external pressure of 1 atm.
Answer:
1.

Reversible processIrreversible process
1) Which can be reversed1) Which cannot be reversed spontaneous process
2) Takes place infinitesimally slowly2) Takes place spontaneous
3) Work done maximum3) Work done minimum

2. w = -p∆V = -1 × (6 – 2)
= – 4 L atm

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 15.
1. What are thermochemical equations?
2. Give an example for a thermochemical equation.
Answer:
1. A balanced chem ical equation together with the value of its ∆rH is called a thermochemical equation.
2.

Question 16.
1. Define lattice enthalpy of an ionic compound.
2. What is Born-Haber cycle?
Answer:
1. The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

2. It is a simplified method developed by Max Born and Fritz Haberto correlate lattice enthanpies of ionic compounds to otherthermodynamic data.

Question 17.
Predict what happens to entropy in the following changes:

  1. Metal is converted into alloy.
  2. Solute crystallizes from solution.
  3. Hydrogen molecule dissociates.

Answer:

  1. The entropy will increase.
  2. The entropy will decrease.
  3. The entropy will increase.

Question 18.
1. Give the relation between change in enthalpy and change in free energy.
2. Name the above relation.
3. What is the significance of the above relation?
Answer:
1. ∆G = ∆H – T∆S

2. Gibbs equation orGibbs-Helmholtz equation.

3. This relation is used to predict the spontaneity of a process based on the value of ∆G . If ∆G is negatve, the process is spontaneous. If ∆G is positive, the process is non-spontaneous.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 19.
1. Predict in each of the following whether entropy increases or decreases.
i) Sublimation of camphor
ii) 4Fe(s) + 3O2(g) → 2Fe2O3(g)
2. The equilibrium constant for a reaction at 30 °C
2.5 x 10-29. What will be the value of ∆G?
Answer:
1. i) entropy increases
ii) entropy increases
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 4

Question 20.
1. Explain the effect of temperature on the spontaneity of a process based on Gibbs equation.
2. For a reaction 2A(g) + B(g) → 2D(g), enthalpy and entropy changes are – 20.5 kJ mol-1 and – 50.4 J K-1mol-1 respectively. Predict whether the reaction occurs at 25 °C.
Answer:
1. If ∆H is -ve and ∆S is +ve, ∆G would certainly be -ve and the process will be spontaneous at all temperatures.
If both ∆H and ∆S are – ve ∆G would be -ve if ∆H > T∆S
If both ∆H and ∆S are + ve ∆G would be -ve if T∆S > ∆H
If ∆H is +ve and AS is -ve, ∆G would certainly be +ve and the process will be non-spontaneous at all temperatures.

2. According to Gibbs equation, ∆G = ∆H – T∆S
∆G = (-20.5 × 10³)-(298 ×-50.4)
= – 20500 + 15019.2 = – 5480.8 J mol-1
Since ∆G is -ve, the process is spontaneous.

Plus One Chemistry Thermodynamics Four Mark Questions and Answers

Question 1.
1. Explain the first, second and third laws of thermodynamics.
2. What do you meant by entropy?
3. Explain the spontaneous process.
Answer:
1. First law:
Energy can neither be created nor be destroyed. Energy in one form can be converted into another form without any loss or gain.

Second law:
Entropy of the universe increases during a spontaneous process.

Third law:
Entropy of a perfect crystalline substance is zero at absolute zero of temperature.

2. Entropy:
Entropy is a measure of randomness or disorder of a system.

3. Spontaneous process:
A spontaneous process is defined as an irreversible process which has a natural tendency to occur either of its own or after proper initiation under the given set of conditions.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 2.
U1, q, w, U2 are given. U1 is internal energy, q is absorbed heat, w is work done and U2 is final energy.
a) Derive an equation for ∆U.
b) Give the equation for w.
c) Calculate the change in internal energy of a system which absorbs 200 J of heat and 315 J of work is done by the system.
Answer:
a) U2 = U1 + q +w
U2 – U1 = q + w
or ∆U = q + w
b) w= -p∆V
c) q = 200J
w = -315J
∆U = ?
∆U = q + w
= 200 + {315}
= 200-315 = -115J

Question 3.
a) Predict whether entropy increases or decreases in the following changes:
i) l2(s) → l2(g)
ii) Temperature of a crystalline solid is raised from 0 Kand 115 K.
iii) Freezing of water
b) Calculate the enthalpy of combustion of methane. Given that standard enthalpies of formation of CH4, CO2 and H2O are -75.2, -394 and -285.6 kJ/mol respectively.
Answer:
a) i) Entropy increases
ii) Entropy increases
iii) Entropy decreases
b) The required equation is,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 5

Plus One Chemistry Thermodynamics NCERT Questions and Answers

Question 1.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? (2)
Answer:
Heat absorbed by the system, (q) = + 701 J
Work done by the system (w) = – 304 J
Change in internal energy (∆U) = q + w
= 701 – 394
= 307 J

Question 2.
The reaction of cyanamide, NH2CN (s) with oxygen was carried out in a bomb calorimeter and AU was found to be – 742.7 kJ mol1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. (2)
Answer:
NH2CN(S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
∆U = 742.7 kJ mol-1;
∆n(g) =2 – 3/2= + 0.5
R = 8.314 × 10-3kJ K-1 mol-1;
T = 298K
According to the relation, ∆H = ∆U + ∆ngRT
∆H = -742.7 kJ + 0.5 mol × 8.314 × 10-3kJ K-1 mol-1 × 298 K
=-742.7 kJ + 1.239kJ
=-741.5 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol-1 K-1. (2)
Answer:
Moles of Al (n) = \(\frac { 60{ g } }{ 27{ g }{ mol }^{ -1 } } \) = 2.22 mol
Molar heat capacity (Cm) = 24 J mol-1 K-1
∆T = 55 °C – 35 C° = 20C° or 20 K
Now, q = Cm × n × ∆T
= 24.0 J mol-1 K-1 × 2.22 mol × 20 K
= 1065.6 J
= 1.067 kJ

Question 4.
The enthalpy of formation of CO(g), CO2 (g), N2O (g), N2O4 (g) are -110, – 393, 81 and 9.7 kJ mol-1 respectively. Find the value of ∆rH for the reaction:
N2O4 (g) + 3CO(g) → N2O(g) + 3CO2(g)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 6

Question 5.
The equilibrium constant for the reaction is 10. Calculate the value of ∆G ; Given R = 8 J K-1 mol-1; T = 300 K.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 7

Question 6
Calculate the entropy change in surroundings when 1.0 mol of HzO (I) is formed under standard conditions. Given ∆fH = – 286 kJ mol-1.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 8

Question 7.
Comment on the thermodynamic stability of NO(g) and NO2 (g) given :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 9
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 10

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Students can Download Chapter 5 States of Matter Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Plus One Chemistry States of Matter One Mark Questions and Answers

Question 1.
When a gas is compressed at constant temperature
a) The speeds of the molecules increase
b) The collisions between the molecules increase
c) The speed of the molecules decrease
d) The collisions between the molecules decrease
Answer:
b) The collisions between the molecules increase

Question 2.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called ___________
Answer:
Boyle temperature or Boyle point

Question 3.
The compressibility factor is given by the expression
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 1
Answer:
a) \(\frac{p V}{n R T}\)

Question 4.
Two flasks of equal volume contain CO2 and SO2 respectively at 298 K and 1.5 atm pressure. Which of the following is equal in them?
a) Masses of the two gases
b) Rates of effusion
c) Number of molecules
d) Molecular structures
Answer:
c) Number of molecules

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 5.
With rise in temperature, viscosity of a liquid
a) Increases
b) Decreases
c) Remains constant
d) May increase or decrease
Answer:
b) Decreases

Question 6.
The unit of‘b’ in VanderWaals equation of state.
Answer:
l mol-1

Question 7.
Most probable velocity, average velocity, and root mean square velocity are related by
Answer:
1 : 1.128 : 1.224

Question 8.
The volume of 2.8g of CO at 27°C and 0.821 atm pressure is (R = 0.0821 l atm Km-1 ol-1)
Answer:
3L

Question 9.
The density of gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temp will its density become 0.75d?
Answer:
400K

Question 10.
The rms velocity of an ideal gas at 27°C is 0.3ms-1. ‘ Its rms velocity at 927°C in (ms-1) is
Answer:
0.6m/s

Plus One Chemistry States of Matter Two Mark Questions and Answers

Question 1.
Find out the relation between the first pair and complete the second pair.
a) Boyle’s law: Temperature
Charles’ law : ……………….
b) Avagadro’slaw: V α n
Ideal gas equation: ……………..
Answer:
a) Pressure
b) pV=nRT

Question 2.
The graphs of Boyle’s law as plotted by Student 1 (Graph 1) and Student 2 (Graph 2) are given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 2

  1. Which is the correct graph?
  2. Justify your answer.

Answer:

  1. Both the graphs are correct.
  2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 3.
The rate of diffusion of hydrogen is less than that of oxygen.

  1. Do you agree?
  2. Which law is applied here?
  3. State the law.

Answer:

  1. Yes.
  2. Graham’s law of diffusion.
  3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Question 4.
The ideal gas equation has been modified for real gases by applying pressure and volume corrections.

  1. What is the corrected equation known as?
  2. Write the equation and explain the terms.

Answer:

  1. The van derWaals’equation.
  2. \(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V — nb) = nRT
    where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Question 5.
‘Moist soil grains are pulled together.’

  1. Name the related phenomenon.
  2. Justify.

Answer:

  1. Surface tension.
  2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

Question 6.
1. What is aqueous tension?
2. What is its significance in the determination of pressure of a dry gas?
Answer:
1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature.

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total presssure of the moist gas.
Pdry gas=PTotal-Aqueous tensi0n

Question 7.
A balloon filled with air, when kept in sunlight bursts after some time.

  1. Name the related law.
  2. Justify.

Answer:

  1. Charles’ law
  2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the baloon expanded more than that the balloon could afford, it bursted.

Question 8.
Define surface energy. What is its SI unit?
Answer:
Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m-2.

Question 9.
a) Based on Boyle’s law how will you show that at a constant temperature, pressure is directly proportional to the density of a fixed mass of the gas?
b) Give the relation between density and molar mass of a gaseous substance.
Answer:
a) According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 3

Question 10.
The isotherm of carbon dioxide at various temperatures is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 4
1. What is the significance of the shaded area?
2. Identify the pressure at which liquid CO2 appears for the first time when cooled form 30.98 °C. What is this pressure called?
Answer:

  1. At any point in the dome shaped shaded area liquid and gaseous CO2 exists in equilibrium.
  2. 73 atm. Critical pressure (pc).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 11.
Certain properties of liquids are given below: Classify them on the basis of effect of temperature on them,

  1. Evaporation
  2. Vapour pressure
  3. Surface tension
  4. Viscosity

Answer:
Properties which increase with increase in temperature: Evaporation & Vapour pressure Properties which decrease with increase in temperature: Surface tension & Viscosity

Question 12.
The size of the water bubbles increases on moving to the surface.
1. Name the law responsible for this.
2. What is your justification?
Answer:
1. Boyle’s law.

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Question 13.
What are the properties of liquid state?
Answer:
Vapour pressure, Boiling point, Viscosity and Surface tension.

Plus One Chemistry States of Matter Three Mark Questions and Answers

Question 1
Analyse the following graph :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 5
1. Name the gas law associated with this graph.
2. State the law.
3. Give the mathematical expression of this law.
Answer:
1. Boyle’s law.

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversly with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n)
Or p = k × \(\frac{1}{V}\) where k is the proportionality
constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 2.
1. What are van der Waals’ forces?
2. Which are the different types of van der Waals’ forces?
3. Arrange the van der Waal’s forces in the increasing order of their strength.
Answer:
1. The attractive intermolecular forces are known as van der Waals’ forces.

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding

Question 3.
A graphical representation of Charles’ law is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 6

  1. What is the temperature corresponding to the point‘A’called? .
  2. What will be the temperature at that point A’ in degree Celsius?
  3. What is the significance of this temperature?

Answer:

  1. Absolute zero temperature
  2. -273.15 °C
  3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 4.
Assume that two gases X and Y at the same temperature and pressure have the same volume.
1. Which of the following is correct?
No. of moles of X= No.of moles of Y
No. of moles of X ≠ No.of moles of Y
2. Which law helped you to find the answer?
3. State the law.
Answer:
1. No. of moles of X= No.of moles of Y

2. Avogadro’s law

3. Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 5.
During a seminar session in the class, the presenter argued that equal amounts of both H2and N2 on heating at constant pressure will expand in the same rate. Another student objected this argument by saying that they will expand differently since their molecular masses are different.

  1. Who is correct in your opinion?
  2. Which law helped you to reach the answer?
  3. State the law and give its mathematical expression.

Answer:
1. The argument of the presenter is correct.

2. Charles’ law

3. Charles’ law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically, V ∝ T (at constant n and P) Or \(\frac{V}{T}\) = K, where K is the proportionality constant which depends on the pressure of the gas, its amount and the unit in which V is expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 6.
1. What is an ideal gas?
2. Give the ideal gas equation and explain the terms.
3. Derive the ideal gas equation.
Answer:
1. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly at all conditions is called an ideal gas.

2. The ideal gas equation is pV = nRT where p = pressure, V = volume, n = number of moles, R = universal gas constant and T = absolute temperature.

3. According to Boyle’s law:
V ∝ \(\frac{1}{p}\) at constant T and n.
According to Charles’ law:
V ∝ T. at constant n and p.
According to Avogadro law,
V ∝ n , at constant p and T Combining the above three equations,
V ∝ \(\frac{nT}{p}\)
⇒ V = R\(\frac{nT}{p}\), where R is the proportionality constant known as universal gas constant.
Or pV = nRT, the ideal gas equation.

Question 7.
Partial pressure of a vessel containing Cl2, CO2 and CO is the sum of the partial pressures of Cl2, O2 and CO.
1. If so, is it correct to say partial pressure of a vessel containing NH3 and HCl gases is the sum of their partial
pressures? Justify.
2. Which law helped you to answer this?
3. State the law.
Answer:
1. No. NH3 reacts with HCl to form NH4CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure.

2. Dalton’s law of partial pressures.

3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 8.
The average kinetic kenergy of the gas molecules is directly proportional to the absolute temperature.

  1. Which theory is related to this assumption?
  2. Write the other postulates of this theory.

Answer:
1. Kinetic moleculartheory of gases.

2.

  • The volume of a gas molecule is negligible when compared to the whole volume of the gas.
  • There is no force of attraction between the particles of a gas at ordinary temperature and pressure.
  • The gas molecules are in random motion.
  • During motion, they collide with each other and also with the walls of the container.
  • Gravity has no influence in the movement of gas molecules.
  • Pressure of a gas is due to the collision of gas molecules with the walls of the container.
  • At any particular time, different particles in the gas have different speeds and hence different kinetic energies.

Question 9.
Three mateorological baloons filled with equal amount of helium, rising in the atmosphere are shown below: (Assume that temperature remains constant).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 7

  1. Which of the baloons will be at the lowest altitude?
  2. Which law helped you to find the answer?
  3. State the law.

Answer:

  1. C –
  2. Boyle’s law.
  3. At low altitudes pressure is high. According to Boyle’s law for a given mass of a gas, greater the pressure lower is the volume at constant temperature.

Question 10.
‘All the postulates of the kinetic molecular theory of gases are correct.’

  1. Do you agree with the statement?
  2. If no, write the wrong postulates of this theory.
  3. Give justification.

Answer:
1. No.

2.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  •  There is no force of attraction between the molecules of a gas.

3. If assumption i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens.
If assumption ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

Question 11.
Two gases with equal molecular mass will have the same rate of diffusion.’

  1. Do you agree?
  2. Explain.
  3. Substantiate your answer with an example.

Answer:
1. Yes.

2. According to Graham’s law of diffusion the rate of diffusion depends only on the molecular mass. So if the molecular masses are the same, their rate of diffusion is same.

3. Both CO and N2 have the same molecular mass (28 g mol-1)
Rate of diffusion of CO = Rate of diffusion of N2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 12.
Water can be boiled more quickly on the top of a mountain.

  1. Do you agree?
  2. What is the reason?
  3. What is called boiling point of a liquid?
  4. How normal boiling point and standard boiling point differ?

Answer:
1. Yes.

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly.

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure.

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

Question 13.
Ethanol flows faster than honey.

  1. Name the related phenomenon.
  2. Explain this phenomenon.
  3. What is the effect of temperature on this?

Answer:
1. Viscosity.

2. Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.

3. Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

Question 14.
Liquid drops attain spherical shape.

  1. Which property of liquids is responsible for this?
  2. Explain the phenomenon and justify.
  3. Suggest another consequence of this phenomenon.

Answer:
1. Surface tension.

2. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. The lowest energy state of the liquid will be when surface area is minimum. Spherical shape satisfies this condition.

3. Fire polishing of glass – On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges due to surface tension, which makes the edges smooth.

Question 15.
Vapour pressure is an important property of liquids.

  1. What is vapour pressure?
  2. How boiling point and vapour pressure are related?
  3. Pressure cooker is used for cooking food at higher altitudes. Give reason.

Answer:
1. Vapour pressure of a liquid is the pressure exerted by the vapour which is in equilibrium with liquid at a given temperature.

2. Boiling point of a liquid is the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. Thus, lower the vapour pressure of a liquid higher will be its boiling point and vice-versa.

3. At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. In a pressure cooker, the internal pressure is greater than atmospheric pressure. Hence, in a pressure cooker water boils at a temperature higher than its normal boiling point of 100 °C. Thus, cooking becomes more effective.

Question 16.
Assume that ‘A’, ‘B’ and ‘C’ are three non-reacting gases kept in a vessel at a constant temperature.
Then, PTotal= PA + PB + PC
1. Name the related law.
2. How can you explain the above law on the basis of kinetic molecular theory of gases?
Answer:
1. Dalton’s law of partial pressures.

2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases.

Question 17.
1. Write the general equation which relates the different variables of a gas used to describe the state of any ideal gas.
2. A flask at 295 K contains a gaseous mixture of N2 and O2 at a total pressure of 1.8 atm. If 0.2 moels of N2 and 0.6 moles of O2 are present, find the partial pressures of N2 and O2.
3. What is meant by Boyle temperature or Boyle point?
Answer:
1. PV = nRT
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 8

3. It is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.

Question 18.
1. Liquid tries to rise or fall in the capillary. Name the related phenomenon.
2. What is the effect of temperature on the above phenomenon?
3. WhatistheSI unit of the above phenomenon.
Answer:
1. Surface tension.

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

3. N m-1.

Question 19.
1. Define critical temperature (Tc ).
2. CO2 cannot be liquified above 31.1°C. Why?
3. The critical temperatures of ammonia and carbon dioxide are 405.5 Kand 304.10 K respectively. On cooling, which of these gases will liquify first? Justify.
Answer:
1. It is the highest temperature at which a gas can
be liquified by applying external pressure.

2. The critical temperature (Tc ) of CO2 is 30.98°C. This is the highest temperature at which liquid CO2 is observed. Above this temperature it is gas.

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 20.
a) Will water boils at higher temperature at sea level or at top of a mountain. Explain.
b) A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer:
a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 9

Question 21.
Real gases deviate from ideal behaviour.
1. What are the two wrong postulates of kinetic theory of gases, responsible for deviation of real gases from ideal behaviour?.
2. When do real gases deviate from ideal behaviour?
Answer:
1.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  • There is no force of attraction between the molecules of a gas.

2. Real gases deviate from ideal behaviour at high pressure and low temperature, when the gas molecules are very close to each other.

Question 22.
1. What is meant by compressibility factor, Z?
2. What is the significance of compressibility factor?
3. A plot of pV/nRT of oxygen gas against p is as follows:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 10
a) Is the gas ideal or real?
b) Write the equation of state of the above gas.
Answer:
1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(z=\frac{p V}{n R T}\).

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor.

3. a) Real gas.
\(\left[p+\frac{a n^{2}}{V^{2}}\right]\)[V-nb] = nRT (van der Waals’ equation)

Question 23.
a) What is the difference between gas and vapour?
b) Analyse the vapour pressure vs temperature curve shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 11
Arrange the compounds shown in the graph in the decreasing order of their normal boiling points,

c) The density of a gas was found to be 2.92 g L1 at 27 °C and 2.0 atm. Calculate the molar mass of the gas.
Answer:
a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance.

b) water > ethyl alcohol > carbon tetrachloride > diethyl ether
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 12

Question 24.
1. How will you account for the observation that automobile tyre is inflated with lesser air in summer than in winter?
2. A sample of gas occupies 250 mLat27 °C. What volume will it occupy at 35 °C if there is no change in pressure?
Answer:
1. This can be explained on the basis of Gay Lussac’s law, according to which at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. In summer season the temperature will be higher. Hence, pressure will increase and the tyre may burst if filled with more air. But during winter temperature is low and hence pressure will below.

2. According to Charles’ law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 13

Question 25.
Real gases behave ideally at low temperature and high pressure.

  1. Is the above statement correct or not?
  2. Justify.
  3. Write the van der Waals’ equation for 1 mole of a real gas.

Answer:
1. The statement is wrong.

2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart.

3. \(\left(p+\frac{a}{V^{2}}\right)\)(V – b) = RT

Question 26.
1. Distinguish between real gas and ideal gas.
2. Explain the deviation of the following gases from ideal behaviouron the basis of the pV vs. p plot. CO, CH4, H2 and He.
Answer:
1. Real gas do not follow, Boyle’s law, Charles law, and Avagadro law perfectly under all conditions. Ideal gas follow, Boyle’s law, Charles’ Law and Avagadro law strictly under all conditions.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 14
It can be seen that at constant temperature pV vs p plots for these gases are not straight lines. Two types of curves are seen. In the curves for H2 and He, as the pressure increases the value of pV also increases. These gases show positive deviation from ideal behaviour at all pressures. The second type of plot is seen in the case of CO and CH4. For these gases the pV value decreases with increase in pressure and reaches to a minimum value characteristics of the gas. After that PV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously.

Question 27.
1. What is meant by laminar flow?
2. Derive the expression for the force responsible for flow of layers of a liquid.
Answer:
1. When a liquid flows overa fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

2. Consider three layers of a flowing liquid as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 15
For any layer, the layer above it accelerates its flow and the layer below this retards its flow. If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by \(\frac{du}{dz}\). A force is required to maintain the
flow of layers. This force is proportional to the area of contact (A) of layers and velocity gradient.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 16
where η is a proportionality constant called coefficient of viscosity.

Question 28.
1. What are London forces?
2. What is the relation between London forces and the distance between the particles?
Answer:
1. The attractive force between two temporary dipoles is known as London forces or Dispersion forces.

2. London forces are always attractive and the interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.

Question 29.
1. What is meant by thermal energy and thermal motion?
2. Can oxygen exist as a gas at -273.15°C? Write the significance of this temperature.
Answer:
1. Thermal energy is the energy of a body arising from motion of its atoms or molecules. The movement of particles due to thermal energy is called thermal motion.

2. At – 273.15 °C oxygen will not exist as a gas. In fact all the gases get liquified before this temperature is reached. It is the absolute zero of temperature, which is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 30.
Molecues of a gas are in a state of continuous motion
1. What is most probable speed?
2. Give the equation for average speed of molecules.
Answer:
1. Most probable speed is the speed possessed by the maximum fraction of molecules of the gas at a given temperature.

2. If there are ‘n’ molecules in a sample and their individual speeds are u1, u2 ……… un, then, average speed of molecules, uav is given by the equation:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots+u_{n}}{n}\)

Plus One Chemistry States of Matter Four Mark Questions and Answers

Question 1.
1. State the Avogadro law.
2. Give the mathematical expression of this law.
3. What is the value of molar volume of an ideal gas at 273.15 K and 1 bar?
4. Show that, at constant temperature and pressure, the density of an ideal gas is proportional to its molar volume.
Answer:
1. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

2. v ∝ n (at constant P and T)
⇒ V = k × n where k is a proportionality constant.

3. 22.71098 L

4. According to Avogadro law, for n moles of an ideal gas,
V = k × n
But n = \(\frac{m}{M}\) where m is the mass of the gas and M is its molar mass.
Thus, V = k × \(\frac{m}{M}\)
On rearranging the above equation,
M = k × \(\frac{m}{V}\) = k × d
⇒ M ∝ d

Question 2.
The speed of molecules is a measure of their average kinetic energy.
a) What is root mean square speed?
b) Give the equation for root mean square speed.
c) Calculate the following:
i) Root mean square speed of methane molecule at 27°C.
ii) Most probable speed of nitrogen molecule at 25°C
Answer:
a) It is the square root of the mean of the squares of speeds of various molecules of the gas at a given temperature.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 17

Question 3.
The graph A is drawn at high temperature and low pressure and graph B is drawn at low temperature and high pressure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 18
a) Which graph represents ideal behaviour?
b) Give the equation for combined gas law.
c) A baloon occupies volume of 700 mL at 25°C and 760 mm of pressure. What will be its volume at higher attitude when temperature is 15°C and pressure is 600 mm Hg.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 19

Question 4.
1. Give the relationship among the three types of molecular speeds.
2. Drawthe Maxwell-Boltzmann distribution showing all the molecularspeeds.
3. Which of the following molecules will have the higher value of most probable speed at the same temperature, N2 or Cl2? Justify.
Answer:
1. Root mean square speed, average speed and the most probable speed have the following relationship:
Urms > Uav > Ump
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 20
3. N2. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules.

Plus One Chemistry States of Matter NCERT Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C? (2)
Answer:
p1 = 1 bar p2 = ?
V1 = 500 dm³ V2 = 200 dm³
Temperature remains constant.
According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 21

Question 2.
Using the equation of state pV=nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. (2)
Answer:
According to ideal gas equation :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 22

Question 3.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP. (3)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 23

Question 4.
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure (R = 0.083 bar L K-1 mol-1). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 24

Question 5.
Explain the significance of van der Waal parameters. (2)
Answer:
The van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effectve size of the gas molecules.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Students can Download Chapter 5 Introduction to C++ Programming Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Plus One Introduction to C++ Programming One Mark Questions and Answers

Question 1.
IDE means _____________
Answer:
Integrated Development Environment

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 2.
We know that C++ is a high level language. From the following which statement is true.
(a) C++ contains English like statements.
(b) C++ contains mnemonics
(c) C++ contains only 0 and 1
(d) None of these
Answer:
(a) C++ contains English like statements.

Question 3.
C++ is a ______ language.
(a) High level
(b) Low level
(c) Middle level
(d) None of these
Answer:
(a) High level

Question 4.
C++ was developed at ___________
(a) AT & T Bell Laboratory
(b) Sanjose Laboratory
(c) Kansas University Lab
(d) None of these
Answer:
(a) AT & T Bell Laboratory

Question 5.
C++ is a successor of ___________ language
(a) C#
(b) C
(c) java
(d) None of these
Answer:
(b) C

Question 6.
The most adopted and popular approach to write programs is __________
Answer:
Structured programming

Question 7.
From the following which uses OOP concept
(a) C
(b) C++
(c) Pascal
(d) Fortran
Answer:
(b) C++

Question 8.
______________ is the smallest individual unit
Answer:
Token

Question 9
Pick the odd one out
(a) float
(b) void
(c) break
(d) Alvis
Answer:
(d) Alvis, the others are keywords.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 10.
Reserved words for the compiler is ____________
(a) Literals
(b) Identifier
(c) Keywords
(d) None of these
Answer:
(c) Keywords

Question 11.
Pick an identifier from the following
(а) auto
(b) age
(c) float
(d) double
Answer:
(b) age

Question 12.
Pick the invalid identifier
(a) name
(b) Date of birth
(c) age
(d) joining time
Answer:
(b) Date of birth, because it contains space.

Question 13.
Pick the octal integer from the following
(a) 217
(b) 0 X 217
(c) 0217
(d) None of these
Answer:
(c) 0217, an octal integer precedes 0

Question 14.
Pick the hexadecimal integer from the following
(a) 217
(b) 0 × 217
(c) 0217
(d) None of these
Answer:
(b) 0 × 217, a hexadecimal integer precedes 0×

Question 15.
From the following pick a character constant
(a) ‘A’
(b) ‘ALL’
(c) ‘AIM’
(d) None of these
Answer:
(a) ‘A’, a character enclosed between single quote

Question 16.
Non graphic symbol can be represented by using ___________
Answer:
Escape Sequence

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 17.
Manish wants to write a program to produce a beep sound. Which escape sequence is used to get an alert (sound)
(a) \a
(b) \d
(c) Vs
(d) None of these
Answer:
(a) \a.

Question 18.
Ajo wants to print a matter in a new line. Which escape sequence is used for this?
(a) \a
(b) \n
(c) \s
(d) None of these
Answer:
(b) \n

Question 19.
To represent null character is used ______
(a) \n
(b) \0
(c) \f
(d) As
Answer:
(b) \0

Question 20.
State True/ False a string is automatically appended by a null character.
Answer:
True

Question 21.
From the following pick a string constant
(a) ‘a’
(b) “abc”
(c) ‘abc’
(d) None of these
Answer:
(b) “abc”, a character constant must be enclosed between double quotes.

Question 22.
C++ was developed by __________
(a) Bjarne Stroustrup
(b) James Gosling
(c) Pascal
(d) None of these
Answer:
(a) Bjarne stroustrup

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 23.
From the following which is not a character constant.
(а) ‘c’
(b) ‘e’
(c) ‘d’
(d) “c”
Answer:
(d) “c”, It is a string constant the others are character constant.

Question 24.
From the following which is a valid declaration.
(a) int 91;
(b) int x;
(c) int 9x;
(d) int “x”;
Answer:
(b) int x;

Question 25.
Symbols used to perform an operation is called ____________
(a) Operand
(b) Operator
(c) Variable
(d) None of these
Answer:
(b) Operator

Question 26.
Consider the following
C = A + B. Here A and B are called
(a) Operand
(b) Operator
(c) Variable
(d) None of these,
Answer:
(b) Operand

Question 27.
The execution of a program starts at ________ function
Answer:
main()

Question 28.
The execution of a program ends with ________ function
Answer:
main()

Question 29.
______ is used to write single line comment
(a) //
(b) /*
(c) */
(d) None of these
Answer:
(a) //

Question 30.
const k = 100 means
(a) const float k = 100
(b) const double k = 100
(c) const int k = 100
(d) const char k = 100
Answer:
(c) const int k = 100

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 31.
Each and every statement in C++ must be end with ________
(а) Semicolon
(b) Colon
(c) full stop
(d) None of these
Answer:
(a) Semicolon

Question 32.
From the following select the input operator
(а) >>
(b) <<
(c) >
(d) <
Answer:
(a) >>

Question 33.
From the following select the output operator
(a) >>
(b) <<
(c) >
(d) <
Answer:
(b) <<

Question 34.
From the following which is known as a string terminator.
(а) ‘\0’
(b) ‘\a’
(c) ‘As’
(d) ‘\t’
Answer:
(a) ‘\0’

Question 35.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated. From the following which file is a must to run the program
(a) sum.exe
(b) sum.obj
(c) sum.vbp
(d) sum.htm
Answer:
(a) sum.exe

Question 36.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated namely sum.obj and sum.exe. From this which file is not needed to run the program
Answer:
sum.obj is not needed and can be deleted.

Question 37.
From the following which is ignored by the compiler
(a) statement
(b) comments
(c) loops
(d) None of these
Answer:
(b) comments

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 38.
To write a C++ program, from the following which statement is a must
(a) sum()
(b) main()
(c) #include<iostream>
(d) #include<iomanip>
Answer:
(b) main(). A C++ program must contains at least one main() function

Question 39.
State True / False
Comment statements are ignored by the compiler
Answer:
True

Question 40.
More than one input/output operator in a single statement is called _______
Answer:
Cascading of I/O operator

Question 41.
Is 0X85B a valid integer constant in C++? If yes why?
Answer:
Yes. It is a hexadecimal number.

Plus One Introduction to C++ Programming Two Mark Questions and Answers

Question 1.
Mr. Dixon declared a variable as follows
int 9age. Is it a valid identifier. If not briefly explain. the rules for naming an identifier.
Answer:
It is not a valid identifier because it violates the rule
The rules for naming an identifier is as follows:

  1. It must be start with a letter(alphabet)
  2. Under score can be considered as a letter
  3. White spaces and special characters cannot be used.
  4. Key words cannot be considered as an identifier

Question 2.
How many bytes used to store ‘\a’.
Answer:
To store ‘\a’ one byte is used because it is an escape sequence. An escape sequence is treated as one character. To store one character one byte is used.

Question 3.
How many bytes used to store “\abc”.
Answer:
A string is automatically appended by a null character.

  • Here one byte for \a (escape sequence).
  • One byte for character b.
  • One byte for character c.
  • And one byte for null character.
  • So a total of 4 bytes needed to store this string.

Question 4.
How many bytes used to store “abc”.
Answer:
A string is automatically appended by a null character.

  • Here one byte for a.
  • One byte for character b.
  • One byte for character c.
  • And one byte for null character.
  • So a total of 4 bytes needed to store this string.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 5.
Consider the following code
int main()
{
cout<<“welcome to C++”;
}
After you compile this program there is an error called prototype error. Why it is happened? Explain.
Answer:
Here we used the output operator cout<<. It is used to display a message “welcome to C++” to use this operator the corresponding header file <iostream> must be included. We didn’t included the header file hence the error.

Question 6.
In C++ the size of the string “book” is 5 and that of “book\n” is 6. Check the validity of the above statement. Justify your answer.
Answer:
A string is automatically added by a null character(\0). The null character is treated as one character. So the size of string “book” is 5. Similarly, a null character (\0) is also added to “book\n”. \n and \0 is treated as single characters. Hence the size of the string “book\n” is 6.

Question 7.
Pick the odd man out. Justify
TOTSAL, TOT_SAL, totsal5, Tot5_sal, SALTOT, tot.sal
Answer:
tot.sal. Because it contains a special character dot(.). An identifier cannot contain a special character. So it is not an identifier. The remaining satisfies the rules of naming identifier. So they are valid identifiers.

Question 8.
Write a C++ statement to print the following sentence. Justify.
“\ is a special character”
answer:
cout<<“\\ is a special character”
\\ is treated as an escape sequence.

Question 9.
A student type a C++ program and saves it in his personal folder as Sample.cpp. After getting the output of the program, he checks the folder and finds three files namely Sample.cpp, Sample.obj and Sample.exe. Write the reasons for the generation of the two files in the folder.
Answer:
After the compilation of the program sample.cpp, the operating system creates two files if there is no error. The files are one object file (sample.obj) and one executable file(sample.exe). Now the source file(sample.cpp) and object file(sample.obj) are not needed and can be deleted. To run the program sample.exe is only needed.

Question 10.
Mention the purpose of tokens in C++. Write names of any four tokens in C++. (2)
Answer:
Token: It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens.

  1. Keywords
  2. Identifier
  3. Literals (Constants)
  4. Punctuators
  5. Operators

Question 11.
The following are some invalid identifiers. Specify its reason.

  1. Sum of digits
  2. 1 year
  3. First jan
  4. For

Answer:

  1. Sum of digits → space not allowed hence it is invalid
  2. 1 year → First character must be an alphabet hence it is invalid
  3. First.jan → special characters such as dot (.) not allowed hence it is invalid.
  4. For → It is valid. That is it is not the keyword for

Question 12.
Some of the literals in C++ are given below. How do they differ?(5, ‘5’, 5.0, “5”)
Answer:

  • 5 – integer literal
  • ‘5’ – Character literal
  • 5.0 – floating point literal
  • “5”- string literal

Question 13.
Identify the invalid literals from the following and write reason for each:

  1. 2E3.5
  2. “9”
  3. ‘hello’
  4. 55450 (2)

Answer:
1. 2E3.5 → The mantissa part (3.5) will not be a floating point number. Hence it is invalid

3. ‘hello’ → It is a string hence it must be enclosed in double quotes instead of single quotes. It is invalid.

Question 14.
Which one of the following is a user defined name?
(a) Keyword
(b) Identifier
(c) Escape sequences
(d) All of these
Answer:
(b) Identifier

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 15.
Identify whether the following are valid identifiers or not? If not give the reason.

  1. Break
  2. Simple.interest (2)

Answer:

  1. Break – It is valid( break is the keyword, not Break);
  2. Simple.interest – It is not valid, because dot(.) is used.

Question 16.
Identify the invalid literals from the following and write a reason for each:

  1. 2E3.5
  2. “9”
  3. ‘hello’
  4. 55450 (2)

Answer:
1. Invalid, because exponent part should not be a floating point number

2. valid

Plus One Introduction to C++ Programming Three Mark Questions and Answers

Question 1.
Rose wants to print as follows
\n is used for New Line. Write down the C++ statement for the same.
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<“\\n is used for New Line”;
}

Question 2.
Alvis wants to give some space using escape sequence as follows
Welcome to C++. Write down the C++ statement for the same
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<“Welcome to \t C++”;
}

Question 3.
We know that the value of pi = 3.14157, a constant (literal). What is a. constant? Explain it.
Answer:
A constant or a literal is a data item its value doe not change during execution.
1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
Eg. For decimal “100, 150, etc

  • For octal 0100, 0240, etc
  • For hexadecimal 0 × 100, 0x1 A, etc

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
eg: 3.14157,79.78, etc

3. Character literal:
A valid C++ character enclosed in single

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 4.
Write a program to print the message “TOBACCO CAUSES CANCER” on screen.
Answer:
#include<iostream>
using namespace std;
int main()
{
cout<<” TOBACCO CAUSES CANCER”;
}

Question 5.
You are supplied with a list of tokens in C++ program, Classify and Categorise them under proper headings.
Explain each category with its features. tot_mark, age, M5, break, (), int, _pay, ;, cin.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming 1

Question 6.
Write a program to print the message “SMOKING IS INJURIOUS TO HEALTH” on screen.
Answer:
#include<iostream>
using namespace std;
int mainO
{
cout<<” SMOKING IS INJURIOUS TO HEALTH”;
}

Plus One Introduction to C++ Programming Five Mark Questions and Answers

Question 1.
Consider the following code
The new line character is \n. The output of the following code does not contain the \n. Why it is happened? Explain.
Answer:
\n is a character constant and it is also known as escape sequence. This is used to represent the non graphic symbols such as carriage return key(enter key), tab key, backspace, space bar, etc. It consists of a backslash symbol and one more characters.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming 2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 5 Introduction to C++ Programming

Question 2.
You are about to study the fundamentals of C++ programming Language. Do a comparative study of the basics of the new language with that of a formal language like English or Malayalam to familiarize C++? Provide sufficient explanations for the compared items in C++ Language.
Answer:
1. Character set:
To study a language first we have to familiarize the character set. For example, to study English language first we have to study the alphabets. Similarly here the character set includes letters(A to Z & a to z), digits(0 to 9), special characters(+, -, ?, *, /, …..) white spaces(non printable), etc.

2. Token:
It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens

  • Keywords: These are reserved words for the compiler. We can’t use for any other purposes.
    Eg: float is used to declare variable to store numbers with decimal point. We can’t use this for any other purpose
  • Identifier: These are user defined words. Eg: variable name, function name, class name, object name, etc…
  • Literals (Constants): Its value does not change during execution
    eg: In maths % = 3.14157 and boiling point of water is 100.
  • Punctuators: In English or Malayalam language punctuation mark are used to increase the readability but here it is used to separate the tokens.
    eg:{,}, (,), ……..
  • Operators: These are symbols used to perform an operation(Arithmetic, relational, logical, etc…).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Students can Download Chapter 4 Chemical Bonding and Molecular Structure Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Plus One Chemistry Chemical Bonding and Molecular Structure One Mark Questions and Answers

Question 1.
The octet rule is not valid for
a) CO2
b)H2O
c) O2
d) CO
Answer:
d) CO

Question 2.
The stability of an ionic crystal depends principally on
a) High electron gain enthalpy of the anion forming species
b) The lattice enthalpy of the crystal
c) Low ionization enthalpy of the cation forming species
d) Low heat of sublimation of cation forming solid
Answer:
b) The lattice enthalpy of the crystal

Question 3.
Which of the following molecules has highest dipole moment?
a) H2S
b)CO2
c) CCl4
d) BF3
Answer:
a) H2S

Question 4.
The d-orbital involved in sp3d hybridization is .
Answer:
dz2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 5.
Which of the following is paramagnetic and has a bond order of ½?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 1
Answer:
H2+

Question 6.
Dipole moment µ electric charge ‘e’ and bond length ‘d’are related by the equation.
Answer:
M = e x d

Question 7.
In which of the following carbon atom is sp2 hybridised?
a) CO2
b) C2H6
c) C6H6
d) HCN
e) \(C{ H }_{ 3 }-C\equiv CH\)
Answer:
C6H6

Question 8.
AgF is ionic whereas Agcl is covalent. This can be explained by
Answer:
Faja’ens Rule

Question 9.
The shape of covalent molecule CIF3 is _________
Answer:
T – shaped

Question 10.
The C – O bond order in CO32- is
Answer:
1.33

Plus One Chemistry Chemical Bonding and Molecular Structure Two Mark Questions and Answers

Question 1.
The order of repulsion of electron pairs as written by student is given below:
lone pair-lone pair repulsion < lone pair-bond pair repulsion>bond pair-bond pair repulsion.
1. Can you see anything wrong in this?
If yes, correct it.
2. Name the theory behind this.
Answer:
1. Yes.
Repulsion decreases in the order: lone pair-lone pair repulsion>lone pair-bond pair repulsion> bond pair-bond pair repulsion,

2. VSEPR theory

Question 2.
During a small group discussion in the class room a student argued that in acetylene both the carbon atoms are in sp3 hybridised state.

  1. What is your opinion?
  2. What is the bond angle between carbon atoms in acetylene?

Answer:

  1. The student’s argument is wrong. In acetylene both the carbon atoms are triple bonded and are in sp hybridised state,
  2. 180°

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Classify the following compounds according to their hybridisation.
CH4, BF3, C2H4, BeF2, C2H2
Answer:
sp3 hybridisation – CH4
sp2 hybridisation – BF3, C2H4
sp hybridisation – BeF2, C2H2

Question 4.
A student arranged the halide ions in the increasing order of polarisability as: F < l < CI < Br
1. Is this the correct order? If not write it in correct order.
2. Justify.
Answer:
1. No.
Polarisability increases in the order: F < Cl <Br < l

2. Polarisability increases when the size of anion increases.

Question 5.
Give any two differences between sigma and pi bonds.
Answer:
Sigma bond (σ) is formed when two atomic orbitals under head-on overlapping. It is a strong bond. Pi(π) bond is formed when atomic orbitals undergo lateral (sidewise) overlapping. It is a weak bond.

Question 6.
Write the type of hybridisation of each carbon in the compound.
CH3-CH=CH-CN
Answer:
Carbon 1 → sp³
Carbon 2 → sp²
Carbon 3 → sp²
Carbon 4 → sp

Plus One Chemistry Chemical Bonding and Molecular Structure Three Mark Questions and Answers

Question 1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 2
1. What is meant by the above picture?
2. Which type of bond is present here?
3. Which type of overlapping leads to the formation of π bond?

Answer:

  1. s-s overlapping
  2. A strong sigma bond
  3. This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals.

Question 2.
‘In ethane there are 6 covalent bonds. Five are strong σ bonds and the remaining one is a weak π bond.’

  1. Do you agree with this?
  2. How is a bond different from π bond in the mode of formation?

Answer:

  1. Yes.
  2. Sigma bond is formed by the end to end overlapping of bonding orbitals along the internuclear axis, π bond is formed by the lateral or sidewise overlap of half filled atomic orbitals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Choose the correct molecules from the given clues: H2O, SF6, BF3
1. Clue -1 The central atom is in sp² hybridised state and the molecule has trigonal planar in shape. Clue-2 The bond angle is 120°
2. Clue-1 The number of electron pairs in this molecule is 6.
Clue -2 It has octahedral geometry.
3. Clue-1 The bond angle is reduced from 109° 28′ to 104.5°
Clue-2 It has a bent shape.
Answer:

  1. BF3
  2. SF6
  3. H2O

Question 4.
Give theoretical explanation for the following statements.
1. H2S is acidic while H2O is neutral.
2. Hydrogen chloride gas dissolves in water.
Answer:
1. S-H bond energy is less than that of O-H bond energy. So H+ can be easily generated from H2S.

2. When HCl is treated with H2O it undergoes hydrolysis as per the following reaction and dissolves.
HCl + H2O → H3O+ +Cl

Question 5.
The potential energy level diagram forthe formation of hydrogen molecule as drawn by a student is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 3
1. Resketch the graph with correct labelling.
2. How can you determine the radius of one hydrogen atom?
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 4
2. Bond length in hydrogen molecule is the intermolecular distance between two hydrogen atoms. So, half of the bond length is taken as the radius of hydrogen atom.

Question 6.
Ionisation enthalpy is one of the factors favoring the
formation of ionic bonds.
1. Will you agree with this statement?
2. Explain how?
3. Write anotherfactorfavouring the formation of ionic bonds.
Answer:
1. Yes.

2. In the formation of the ionic bond, a metal atom losses electrons to form cation. This process requires energy equal to the ionisation enthalpy. Lesser the ionisation enthalpy of the metal atom, easier will be the removal of electron from the atom to form cation and hence greater will be the tendency to form ionic bond.

3. Electron gain enthalpy of the element forming anion.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 7.
Complete the following table:

Sigma bondπ -bond
Formation………….………….
Strong/Weak………….………….
About rotation………….………….

Answer:

Sigma bondπ -bond
FormationSigma bond is formed by end to end (or axial) overlap of atomic orbitalsπ -bond is formed by the sidewise overlap of atomic orbitals
Strong/WeakStrongWeak
About rotationFree rotationFree rotation is not possible

Question 8.
a) The dipole moment of BF3 is zero even though the B – F bonds are polar. Justify.
b) Give the hybridisation involved in the following compounds

  1. NH3
  2. C2H4
  3. SF6
  4. PCl5

c) o-nitro phenol has a lower boiling point than its para isomer. Why?
Answer:
a) In BF3, dut to the symmetric trigonal planar geometry of the molecule, the B – F bond are oriented at an angle of 120°to one another. The three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 5
b)

  1. sp³
  2. sp²
  3. sp³d²
  4. sp³d

c) In o-nitrophenol, intramolecular hydrogen bonding is present and there is no association of molecules whereas in p-nitrophenol there is inter-molecular hydrogen bonding which causes association of molecules.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 6

Question 9.
1. How many a and n bonds are there in the following molecules i) ethane ii) acetylene?
2. BF3 and NH3 are tetra atomic molecules. But the shape of BF3 is different from that of NH3. Explain this using hybridisation.
Answer:
1. i) Ethane – 7σ bonds
ii) Acetylene-3σ bonds and 2 π bonds

2. In BF3 molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital. As a result, boron has three unpaired electrons. These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals oriented in a trigonal planar arrangement and overlap with 2 p orbitals of F to form three B – F bonds. Therefore, BF3 molecule has a planar geometry with FBF bond angle of 120°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 7
In NH3, the valence shell electronic configuration of N in ground state is 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\). These four orbitals undergo sp³ hybridisation to form four sp³ hybrid orbitals, three of them containing unpaired electrons and the fourth one containing lone pair. The three hybrid orbitals overlap with 1 s orbitals of hydrogen atoms to form three N – H sigma bonds. Since, the bp-lp repulsion is greater than the bp-bp repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. Thus, the geometry of NH3 molecule is trigonal pyramidal.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 8

Question 10.
Covalent bond is formed by the overlaping of atomic orbitals.
1. What is meant by orbital overlapping?
2. What are the 3 types of overlapping?
Answer:
1. Orbital overlapping is the partial interpenetration or merging of atomic orbitals. It results in the pairing of electrons. Greater the overlap the stronger is the bond formed between two atoms.

2. s-s overlapping
s-p overlapping
p-p overlapping

Question 11.
1. Which among the following will exist He2 or He2+? Explain.
2. H2S is a gas at ordinary condition, while H2O is liquid. Account for the above statement.
3. State the hybridisation in the following molecules,
i) PF6
ii) C2H6
Answer:
1. He2+
Helium molecule contains 4 electrons. Out of this 4 electrons, 2 are present in the bonding molecular orbital and the remaining 2 are present in the anti-bonding molecular orbital.
Bond order = ½ (Nb-Na)
= ½ (2-2) = 0
Hence, He2 cannot exist. The molecular orbital diagram is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 9
He2+ contains 3 electrons. Out of these 3 electrons, 2 are present in the σ1s level and the remaining one is present in the σ* 1s level.
Bond order = ½ (Nb-Na)
= ½ (2-1) = ½
Since the bond order is half the molecular ion exists but possesses low stability.

2. In H2S, there is no hydrogen bonding whereas in water, molecular association is possible due to intermolecular hydrogen bonding.

3. i) PF5 = sp³d hybridisation
ii) C2H6 = sp³ hybridisation

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 12.
Bond order is a term commonly used in MO theory.
1. How is it calculated?
2. How is it related to bond length and bond energy?
Answer:
1. Bond order = ½ (Nb-Na)
where Nb = No. of electrons occupying bonding orbitals and Na = No. of electrons occupying antibonding orbitals.

2. As the bond order increases, bond length decreases and bond energy increases, i.e., bond order is directly proportional to bond energy and inversely proportional to bond length.

Question 13.
1. Explain the hybridisation and geometry of ethyne.
2. What is the difference between bonding molecular orbital and antibonding molecular orbital?
3. How the magnetic nature of a molecule is related to its electronic structure?
Answer:
1. In the formation of ethyne (C2H2), both the carbon atoms undergo sp hybridisation having two unhybridised orbitals (2px and 2py). One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming o bonds. Each of the two unbybridised p orbitals of both the carbon atoms overlaps sidewise to form two K bonds between the carbon atoms. Thus, ethyne has a linear geometry with π bond angle of 180°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 10

2. The molecular orbital which has lower energy than the atomic orbital is called bonding molecular orbital and the molecular orbital which has greater energy than the atomic orbital is called anti bonding molecular orbital.

3. If all the molecular orbitals in a molecule are doubly occupied (i.e., paired), the substance is diamagnetic (repelled by magnetic field). It one or more molecular orbitals are singly occupied (i.e., unpaired) it is paramagnetic (attracted by magnetic field).

Question 14.
Molecular Orbital Theory (MOT) is an advanced theory
of chemical bonding.
1. Write the salient features of MOT.
2. What is meant by LCAO? Illustrate using hydrogen molecule.
3. What are the conditions for the combination of atomic orbitals?
Answer:
1.

  • The electrons in a molecule are present in the
    various molecular orbitals.
  • The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
  • iii) The electron in a molecular orbital is influenced by two or more nuclei depending upon the number of atoms in the molecule.
  • The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two
    molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
  • The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
  • The electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
  • The molecular orbitals are filled in accordance with the Aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule.

2. LCAO refers to the linear combination of atomic orbitals. It is an approximate method used to explain the formation of molecular obritals. Consider hydrogen molecule consisting of two atoms A and B. Each hydrogen atom has one electron in the 1s orbital. The atomic orbitals of these atoms can be represented by the wave functions ψA and ψB. Mathematically, the formation of molecular orbitals can take place by addition and by subtraction of wave functions of individual atomic orbitals.
ψMO = ψA ± ψB
Therefore, the two molecular orbitals σ and σ* are formed as:
σ* = ψA – ψB
The molecular orbital σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ* formed by the subtraction of atomic orbital is called antibonding molecular orbital. The energy level diagram is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 11

3.

  • The combining atomic orbitals must have the same or nearly same energy.
  • The combining atomic orbitals must have the same symmetry about the molecular axis.
  • The combining atomic orbitals must overlap to the maximum extent.

Question 15.
Consider a reaction PCl5(g) → PC3(g) + Cl2(g)
1. What is the change in hybridisation state of phosphorus?
2. Explain why does PCl5 decomposes easily?
Answer:
1. When PCl5 decomposes to PCl3, the hybridisation of P changes from sp³d to sp³.

2. In PCl5, the five sp³d orbitals of P overlap with the singly occupied p orbitals of Cl atoms to form five P-CI sigma bonds. Three P-Cl bonds which lie in one plane and make an angle of 120° with each other are called equatorial bonds. The remaining two P – Cl bonds, called axial bonds, one lie above and the other lie below the equatorial plane, make an angle of 90° with the plane. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, axial bonds are slightly longer and hence slightly weaker than the equatorial bonds. This makes PCl5 molecule more reactive and hence it decomposes easily.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 12

Question 16.
The electron dot structure (Lewis structure) of ammonia molecule is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 13
1. Write the number of bond pairs of electrons and lone pairs of electrons in ammonia molecule.
2. The structures of o-nitrophenol and p-nitrophenol are shown in the figure. The former is a steam volatile liquid whereas the latter is a solid. Justify your answer giving reason.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 14
Answer:
1. Ammonia molecule contains one lone pair of electrons and 3 bond pair of electrons,

2. In o-nitrophenol, there is intramolecular hydrogen bonding and there is no molecular association. But in p-nitro phenol intermolecular hydrogen bonding is present and hence molecular association is possible.

Plus One Chemistry Chemical Bonding and Molecular Structure Four Mark Questions and Answers

Question 1.
Hydrogen bonding is present in NH3 and H2O.
1. What is hydrogen bond?
2. What are different types of hydrogen bonds?
3. Explain the effect of hydrogen bonding.
Answer:
1. Hydrogen bond is defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O, N) of the same or another molecule.

2. Intermolecular hydrogen bond and Intramolecular hydrogen bond.

3. Compounds containing hydrogen bonds show higher melting and boiling points. Compounds whose molecules can form hydrogen bonds with water molecules are soluble in water.

Question 2.
Classify the following compounds according to their
shape.
BeF2, BeCl2, CH4, BF3, PCl5, SF6, SbCl5 NH4+, SiF4, AlCl3.
Answer:
Linear – BeF2, BeCl2
Trigonal planar-AlCl3, BF3
Tetrahedral – CH4, NH4+, SiF4
T rigonal bipyramidal – PCl5, SbCl5

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Benzene is an example of a compound exhibiting resonance.
1. What is meant by resonance?
2. Explain the resonance of ozone.
Answer:
1. When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more different structures, then the actual molecule is said to be a resonance hybrid of these canonical structures.

2. The resonance in ozone can be represented by the following structures:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 15
According to structures I and II, there is one single bond and one double bond in ozone molecule. But experiments show that both the oxygen- oxygen bonds are equal and the bond length 128 pm) is intermediate between single (148 pm) and double bond (121 pm) lengths. Hence it is assumed that ozone is a resonance hybrid (structure III) of structures I and II.

Question 4.
Match the following:

No. of electrons pairsShape of moleculeExamples
2Trigonal planarSF6
4LinearBeF2, BeCl2
3TetrahedralBF3, AlCl3
6OctahedralCH4, SiF4

Answer:

No. of electrons pairsShape of moleculeExamples
2LinearBeF2, BeCl2
4TetrahedralCH4, SiF4
3Trigonal planarBF3, AlCl3
6OctahedralSF6

Question 5.
In the formation of methane, carbon undergoes sp³ hybridisation.

  1. What do you mean by sp³ hybridisation?
  2. Give the % s-character and p-character of an sp³ hybrid orbital.
  3. What is the bond angle in methane?
  4. What is the geometry of methane molecule?

Answer:

  1. sp³ hybridisation involves mixing up of one – s and three-p orbitals of the valence shell of an atom to form four sp³ hybrid orbitals of equivalent energies and shape.
  2. Each sp³ hybrid orbital has 25% s-character and 75% p-character.
  3. The angle between the sp3 hybrid orbitals in methane is 109°28′.
  4. Tetrahedron.

Question 6.
Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 16
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 17

Question 7.
Fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 18
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 19

Question 8.
Dipole moment is used to predict the shape of
molecules.
1. Justify the statement based on the shapes of CO2 and H2O.
2. Which is having high dipole moment? NH3 or NF3? Why?
Answer:
1. Carbon dioxide is a linear molecule in which the two C=0 bonds are oriented in the opposite directions at an angle of 180°. Hence the two C=0 bond dipoles cancel each other and the resultant dipole moment of CO2 is zero. Thus CO2 is non¬polar molecule
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 20
On the other hand, water molecule has a bent structure in which two O-H bonds are oriented at an angle of 104.5°. Therefore, the bond dipoles of two O-H bonds do not cancel each other and the molecule will have a net dipole moment (1.85D).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 21

2. The dipole moment of NH3 is higher than that of NF3. In both cases, the central N atom has a lone pair whose orbital dipole points away from the N atom. In NH3 the orbital dipole due to the lone pair is in the same direction as the resultant bond dipole of the three N-H bonds. On the other hand, in the case of NF3, the resultant dipole of the three N-F bonds is in the opposite direction to the orbital dipole due to the lone pair. Thus, the orbital dipole due to the lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 22

Question 9.
The geometry of a covalent molecule is related to the hybridisation involved in the central atom. Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 23
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 24

Question 10
Depending upon the type of overlapping, covalent bonds are of two types.
a) Name them and give any two difference between them.
b) Find the total number of these two types of bonds ’ in propane and 2-butene.
Answer:
a) Sigma (σ) bond and pi (π) bond.
Sigma (σ) bond:
This type of covalent bond is formed by the end to end overlapping of half-filled atomic orbitals along the internuclear axis. The overlap is also known as head on overlap or axial overlap. The electrons constituting sigma bond are called sigma electrons.

Pi (π) bond:
This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals. The atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicularto the internuclear axis.
b) 10 σ bond in propane
11 σ bond and 1π bond in 2-butene

Plus One Chemistry Chemical Bonding and Molecular Structure NCERT Questions and Answers

Question 1.
Explain the formation of a chemical bond. (2)
Answer:
According to Kossel-Lewis approach, the formation of chemical bond between the two atoms takes place either by the transference of electrons or by mutual sharing of electrons. However, according to the modem view the formation of chemical bond between the two approaching atoms occurs only if there is a net decrease of energy because of attractive and repulsive forces.

Question 2.
Write the favourable conditions for the formation of ionic bond. (2)
Answer:
Ionic bond is formed by transference of electrons from one atom to another. The favourable conditions for its formation are:

  • Low ionisation enthalpy of element forming cation.
  • More negative value of electron gain enthalpy of element forming the anion and
  • High value of lattice enthalpy of the compound formed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that in ammonia. Discuss. (2)
Answer:
The difference in bond angles is due to the different numbers of lone pairs and bond pairs in the two species. In NH3, the N atom has two lone pairs and three bond pairs while in H2O, the O atom has two lone pairs and two bond pairs. The repulsive interactions of lone pairs and bond pairs in water are relatively more than those in NH3. Hence, bond angle around central atom in water is relatively smaller (104.5°) than that in NH3 molecule (107°).

Question 4.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? (2)
BF3 + NH3 → F3B.NH3
Answer:
During combination of species BH3 and NH3, N atom of NH3 is donor and B atom of BF3 is acceptor. The hybrid state of B in BF3 is sp² and that of N in NH3 is sp³. In the compound F3B+-NH3 both N and B atoms are surrounded by four bond pairs. Thus, the hyrid state of both is sp³. Hence during the reaction the hybrid state of B changes from sp² to sp³ but that of N remains the same.

Question 5.
Define hydrogen bond. Is it weaker or stronger than the van der Waals’ forces? (2)
Answer:
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Hydrogen bond is stronger than van der Waals’forces because it is a strong dipole-dipole interaction. The van der Waals’ forces, on the other hand, are weak dispersion forces.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Students can Download Chapter 2 Data Representation and Boolean Algebra Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Plus One Data Representation and Boolean Algebra One Mark Questions and Answers

Question 1.
___________ is a collection of unorganized fact.
Answer:
Data

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 2.
Data can be organized into useful ____________
Answer:
Information

Question 3.
___________ is used to help people to make decision.
Answer:
Information

Question 4.
Processing is a series of actions or operations that convert inputs into __________
Answer:
Output

Question 5.
The act of applying information in a particular context or situation is called ____________
Answer:
Knowledge

Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
(d) Adeline aged 17 years is in class 12. This is information. The others are data.

Question 8.
Raw facts and figures are known as _______
Answer:
data

Question 9.
Processed data is known as _______
Answer:
Information

Question 10.
Which of the following helps us to take decisions?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 11.
Manipulation of data to get information is known as ___________
Answer:
Data processing

Question 12.
Arrange the following in proper order
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:

  1. Data Capture
  2. Input
  3. Storage
  4. Process
  5. Output
  6. Distribution

Question 13.
Pick the odd one out and give reason:
(a) Calculation
(b) Storage
(c) Comparison
(d) Categorization
Answer:
(b) Storage. It is one of the data processing stage the others are various operations in the stage Process

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.

  1. (101)2, (111)2, (1001)2, ……….
  2. (1011)2, (1110)2, (10001)2, ………

Answer:

  1. 1011, 1101
  2. 10101, 10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _______________
Answer:
Base

Question 18.
The base of decimal number system is ________
Answer:
10

Question 19.
MSD is ________
Answer:
Most Significant Digit

Question 20.
LSD is _________
Answer:
Least Significant Digit

Question 21.
Consider the number 627. Its MSD is _________
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is __________
Answer:
7

Question 23.
The base of Binary number system is ___________
Answer:
2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series,
(101)2, (111)2, (1001)2
Answer:
1011, 1101

Question 26.
State True or False. In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is ________
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1, 2, 3, 4, 5, 6, 7, __
Answer:
10

Question 29.
The base of Hexadecimal number system is ________
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexadecimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 32.
Convert (1110.01011)2 to decimal.
Answer:
1110.01011 = 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 + 0 × 2 – 1 + 1 × 2 – 2 + 0 × 2 – 3 + 1 × 2 – 4 + 1 × 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)10

Question 33.
1 KB is bytes.
(a) 25
(b) 210
(c) 215
(d) 220
Answer:
(b) 210

Question 34.
The base of hexadecimal number system is ________.
Answer:
16

Question 35.
A computer has no _________
(a) Memory
(b) l/o device .
(c) CPU
(d) IQ
Answer:
(d) IQ

Question 36.
Pick the odd man out.
(AND, OR, NAND, NOT)
Answer:
NOT

Question 37.
Select the complement of X + YZ.
(a) \(\bar{x}+\bar{y}+\bar{z}\)
(b) \(\bar{x} .\bar{y}+\bar{z}\)
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)
(d) \(\bar{x}+\bar{y} \cdot \bar{z}\)
Answer:
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)

Question 38.
Select the expression for absorption law.
(a) a + a = a
(b) 1 + a = 1
(c) o . a = 0
(d) a + a . b = a
Answer:
(a) a + a . b = a

Question 39.
What is the characteristic of logical expression?
Answer:
Logical expressions yield either true or false values

Question 40.
Name the table used to define the results of Boolean operations.
Answer:
Truth Table

Question 41.
According to ________ law, \(\bar{x}+\bar{y}=\overline{x y}\) and \(\overline{x y}=\bar{x}+\bar{y}\)
Answer:
De Morgan’s law

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 42.
A NOR gate is ON only when all its inputs are
(a) ON
(b) Positive
(c) High
(d) OFF
Answer:
(d) OFF

Question 43.
The only function of a NOT gate is __________
Answer:
Invert an output signal

Question 44.
NOT gate is also known as _________
Answer:
Inverter

Question 45.
What is the relation between the following statements.
x + 0 = x and x . 1 = x
Answer:
One is the dual of the other expression.

Question 46.
The algebra used to solve problems in digital systems is called __________
Answer:
Boolean Algebra

Question 47.
Pick the one which is not a Basic Gate.
(AND, OR, XOR, NOT)
Answer:
XOR

Question 48.
Select the universal gates from the list. (NAND, NOR, NOT, XOR)
Answer:
NAND, NOR

Question 49.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 50.
Fill up the missing digit.
(41)8 = ( )16
Answer:

  • Step 1: Divide the number into one each and write down the 3 bits equivalent.
  • Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 1
Answer:
So the answer is 21.

Question 51.
Real numbers can be represented in memory by using __________
Answer:
Exponent and Mantissa

Question 52.
Consider the number 0.53421 x 10-8 Write down the mantissa and exponent.
Answer:
Mantissa: 0.53421
Exponent: -8

Question 53.
Characters can be represented in memory by using _________
Answer:
ASCII Code

Question 54.
ASCII Code of A’ is __________
Answer:
(100 0001)2 = 65

Question 55.
ASCII Code of’a’ is __________
Answer:
(110 0001)2 = 97

Question 56.
Define the term ‘bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 57.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 58.
ASCII stands for __________.
Answer:
American Standard Code for Information Interchange

Question 59.
List any two image file formats.
Answer:
BMP, GIF

Question 60.
Name the operator which performs logical multiplication.
Answer:
AND

Question 61.
Name a gate which is ON when all its inputs are OFF .
Answer:
NAND or NOR

Question 62.
Specify the laws applied in the following cases.

  1. a (b + c) = ab + ac
  2. (a + b) + c = a + (b + c)

Answer:

  1. Distribution law
  2. Associative law

Question 63.
Pick the correct Boolean expression from the following.
(a) \(A +\bar{A}=163.\)
(b) \(\text { A. } \bar{A}=1\)
(c) \(A \cdot \overline{A B}=A + B\)
(d) A + AB = A
Answer:
(a) & (d)

Question 64.
1’s complement of the binary number 110111 is _________
Answer:
Insert 2 zeroes in the left hand side to make the binary number in the 8 bit form 00110111
To find the 1’s complement, change all zeroes to one and all ones to zero. Hence the answer is 11001000

Plus One Data Representation and Boolean Algebra Two Mark Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get information as a result that must be stored in the computer for future references.

Question 2.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 3.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee. Here data is the ingredients – water, sugar, milk and coffee powder
Information is the final product i.e. Coffee Processing is the series of steps to convert the ingredients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 4.
ASCII is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because it is a 7 bit code so it can represent 27 = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 5.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact matvh

AB
(12)8(1110)2
F1625
(19)1610
(11)8(13)16
(17)8
9

Answer:

AB
1210
F(17)8
(19)1625
(11)89

Question 6.

  1. Name various number systems commonly used in computers.
  2. Include each of the following numbers into all possible number systems
    123 569 1101

Answer:

  1. The number system are binary, octal, decimal and hexa decimal
  2. All possible number systems are
    • 123 Octal, decimal and hexa decimal
    • 569 Decimal, hexa decimal
    • 1101 Binary, Octal, Decimal, Hexa decimal

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 7.
Fill up the missing digit. (Score 2)
If (220)a = (90)b then (451)a = (  )10
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 10. The values of b can be 10 or 16. L.H.S > R.H.S. a < b and a b also.
The possible values of a and b are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 2

Question 8.
Convert (106)10 = (  )2?
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 3

Question 9.
Convert (106)10 = (  )8?
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 4

Question 10.
(106)10 = (  )16?
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 5

Question 11.
Convert (55.625)10 = (  )2?
Answer:
First convert 55, for this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 6
Write down the remainders from bottom to top.
(55)10 = (110111 )2
Next convert 0.625, for this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 7
Write down the remainder from top to bottom. So the answer is
(55.625)10 = (110111.101)2

Question 12.
Convert (55.140625)10 = (  )8?
Answer:
First convert 55, for this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 8
Write down the remainders from bottom to top.
(55)10 = (67)8
Next convert 0.140625, for this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 9
Write down the remainders from top to bottom. So the answer is
(55.140625)10 = (67.11 )8

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 13.
Convert (55.515625)10 = (  )16
Answer:
First convert 55, for this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 10
Write down the remainders from bottom to top.
ie. (55)10 = (37)16
Next convert .515625
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 11
So the answer is
(55.515625)10 = (37.84)16

Question 14.
Convert (101.101)2 = ( )10?
Answer:
101.101 = 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2-1 + 0 × 2-2 + 1 × 2-3 = 4 + 0 + 1 + 1/2 + 0 + 1/8 = 5 + 0.5 + 0.125
(101.101)2 = (5.625)10

Question 15.
Convert (71.24)8 = (  )10?
Answer:
71.24 = 7 × 81 + 1 × 80 + 2 × 8-1 + 4 × 8=2
= 56 + 1 + 2/8 + 4/82
= 57 + 0.25 + 0.0625 (71.24)8
(71.24)8 = (57.3125)10

Question 16.
Convert (AB.88)16 = (  )10?
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 12
= 160 + 11 + 0.5 + 0.03125
(AB.88)16 = (171.53125)10

Question 17.
Convert (1011)2 = (  )8?
Answer:
Step I: First divide the number into groups of 3 bits starting from the right side and insert necessary zeroes in the left side.
0 0 1 | 0 1 1

Step II: Next write down the octal equivalent.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 13
So the answer is (1011)2 = (13)8.

Question 18.
Convert (110100)2 = (  )16
Answer:

  • Step I: First divide the number into groups of 4 bits starting from the right side and insert necessary zeroes in the left side.
  • Step II: Next write down the hexadecimal equivalent.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 14
So the answer is (110100)2 = (34)16.

Question 19.
(72)8 = ( )2?
Answer:
Write down the 3 bits equivalent of each digit.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 15
So the answer is (72)8 = (111010)2.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 20.
Convert (AO)16 = (  )2 ?
Answer:
Write down the 4 bits equivalent of each digit
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 16
So the answer is (AO)16 = (1010 0000)2.

Question 21.
Convert (67)8 = (  )16?
Answer:
Step I: First convert this number into binary equivalent for this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 17

Step II: Next convert this number into hexadecimal equivalent for this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 18
So the answer is (67)8 = (37)16

Question 22.
Convert (A1)16 = (  )8?
Answer:
Step I: First convert this number into binary equivalent. For this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 19

Step II: Next convert this number into octal equivalent. For this do the following.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 20
So the answer is (A1)16 = ( 241)8.

Question 23.
What is the use of the ASCII Code?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the keyboard is represented in memory by using ASCII Code.
eg: A’s ASCII Code is 65 (1000001), a’s ASCII Code is 97 (1100001)

Question 24.
Pick invalid numbers from the following.

  1. (10101)8
  2. (123)4
  3. (768)8
  4. (ABC)16

Answer:

  1. (10101)8 – Valid
  2. (123)4 – Valid
  3. (768)8 – Invalid. Octal number system does not contain the symbol 8
  4. (ABC)16 – Valid

Question 25.
Convert the decimal number 31 to binary.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 21
(31)10 = (11111)2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 26.
Find decimal equivalent of (10001 )2
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 22
= 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 16 + 0 + 0 + 0 + 1
= (17)10

Question 27.
If (X)8 =(101011 )2 then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 23
X = 53

Question 28.
Fill the blanks:
(a) (………..)2 = (AB)16
(b) (——D—–)16 = (1010 1000)2
(c) 0.2510 = (—–)2

Answer:
Write down the 4bit equivalent of each digit
(a) (………..)2 = (AB)16
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 24
= (10101011)2
(b) (——D—–)16 = (1010 1000)2
(A D 8)16 =(1010 1101 1000)2

(c) 0.2510 = (—–)2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 25
0.2510 = (0.01)2

Question 29.
Which is the MSB of representation of -80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 30.
Write 28.756 in Mantissa exponent form.
Answer:
28.756 = .28756 × 100
= .28756 × 102
= .28756 E + 2

Question 31.
Represent -60 in 1’s complement form.
Answer:
+60 = 00111100
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
-60 is in 1’s complement is 11000011

Question 32.
Define Unicode.
Answer:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 216 = 65536 characters(i.e, world’s all written language characters) can store by using this.

Question 33.
Substract 1111 from 10101 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here the subtrahend is 1111 and minuend is 5 bits. So insert a zero. So subtrahend is 01111. First take the 1’s complement of subtrahend and add 1 to it
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 26
Here is a carry. So ignore the carry and the result is +ve.
So the answer is 110

Question 34.
You purchased a soap worth Rs. (10010)2 and you gave Rs. (10100)2 and how much rupees will you get back in binary.
Answer:
Substract (10010)2 from (10100)2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 27
You will get rupees (10)2

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 35.
Draw the logic circuit diagram for the following Boolean expression.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 28

Question 36.
Simplify the expression using basic postulates and laws of Boolean algebra.

  1. \(\bar{x}+x \cdot \bar{y}\)
  2. \(x(y+y . z)+y(\bar{x}+x z)\)

Answer:

  1. \(\bar{x}+\bar{y}\)
  2. y

Question 37.
Show \(A(\bar{B}+C)\) using NOR gates only.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 29

Question 38.
The following statement Demorgan’s theorem of Boolean algebra. Identify and state ‘Break the line, change the sign’.
Answer:
Demorgan’s theorems,
Demorgan’s first theorem,
\(\overline{x + y}\) = \(\bar{x} . \bar{y}\)
Demorgan’s second theorem,
\(\overline{x – y}\) = \(\bar{x} + \bar{y}\)

Question 39.
Prove algebraically that
\(x \cdot y + x \cdot \bar{y} \cdot z\) = x . y + x . z
Answer:
\(x \cdot y + x \cdot \bar{y} \cdot z\) = \(x(y+\bar{y} . z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 40.
State which of the following statements are logical statements.
(a) AND is a logical operator
(b) ADD 3 to y
(c) Go to class
(d) Sun rises in the west.
(e) Why are you so late?
Answer:
(a) and (d) are logical statements because these statements have a truth value which may be true or false.

Question 41.
Express the integer number -39 in sign and magnitude represnetation.
Answer:
First find the binary equivalent of 39 for this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 30
In sign and magnitude representation -39 in 8 bit form is (10100111)2.

Question 42.

  1. Which logic gate does the Boolean expression \(\overline{\mathrm{AB}}\) represent?
  2. Some NAND gates are given. How can we construct AND gate, OR gate and NOT gate using them?

Answer:
1. NAND

2. AND gate
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 31

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 43.
Perform the following number conversions.

  1. (110111011.11011)2 = (….)8
  2. (128.25)10= (…..)8

Answer:
1. 110111011.11011
Step 1: Insert a zero in the right side of the above number and divide the number into groups of 3 bits as follows
110 111 011 . 110 110

Step 2: Write down the corresponding 3 bit binary equivalent of each group
6 7 3 .6 6
Hence the result is (673.66)8

2. It consists of 2 steps.
Step 1: First convert 128 into octal number for this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 32
Write down the remainders from bottom to top.
(128)10 = (200)8

Step 2: Then convert .25 into octal number for this do the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 33
(0.25)10 = (0.2)8.
Combine the above two will be the result.
(128.25)10= (200.2)8

Question 44.
Represent -38 in 2’s complement form.
Answer:
+38 = 00100110
First take the 1 ’s complement for this change all 1 to 0 and all 0 to 1
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 34
2’s complement of -38 is (11011010)8.

Plus One Data Representation and Boolean Algebra Three Mark Questions and Answers

Question 1.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.

In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to input data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 2.
Complete the series.

  1. 3248, 3278 ,3328, …., ….
  2. 5678, 5768, 605s, ……, …..

Answer:
1. (324)8 = 3 × 82 + 2 × 81 + 4 × 80
= 3 × 64 + 2 × 8 + 4 × 1
= 192 + 16 + 4 = (212)10

(327)8 = 3 × 82 + 2 × 81 + 7 × 80
= 192 + 16 + 7 = (215)10

(332)8= 3 × 82 + 3 × 81 +2 × 80
= 192 + 24 + 2 = (218)10
So the missing terms are (221)10, (224)10 we have to convert this into octal.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 35
So the missing terms are (335)8, (340)8

2. (567)8 = 5 x 82 + 6 x 81 + 7 x 80
= 5 x 64 + 6 x 8 + 7 x 1
= 320 + 48 + 7 = (375)10

(576)8 = 5 x 82 + 7 x 81 + 6 x 80
= 320 + 56 + 6 = (382)10

(605)8 = 6 x 82 + 0 x 81 + 5 x 80
= 6 + 64 + 0 + 5
= 384 + 0 + 5 = (389)10
So the missing terms are (396)10, (403)10 we have to convert this into octal.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 36
So the missing terms are (614)8, (623)8

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 3.
Fill up the missing digits.

  1. (4……)8 = (……110)2
  2. (…….7……)8 = (100…….110)2

Consider the following
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 37
Answer:
1. 4…….100 and 110……….6
So (46)8 = (100 110)2

2. 100…….4
7……111
110………6
So (476)8 = (100 111 110)2

Question 4.
Fill up the missing numbers.

  1. (A…….)16 = (……..1001)2
  2. (…….B…….)16 = (1000………1111)2

Consider the following:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 38
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 39
Answer:
1. A……..1010
1001………9
So (A9)16 = (1010 1001)2

2. B…….1011
1000………8
1111………F
So (8BF)16 = (1000 1011 1111)2

Question 5.
Complete the Series.

  1. 6ADD, 6ADF, 6AE1, ……., …….
  2. 14A9, 14AF, 14B5, …….., ……

Answer:
1. Consider the sequence
6ADD, 6ADF, 6AE1, ………….
Here the ‘numbers’ are
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11,——–
The difference between 6ADD & 6ADF is 2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 40
Similarly 6ADF & 6AE1 is 2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 41
So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5

2. Consider the sequence.
14A9, 14AF, 14B5, ———
The difference between 14A9 and 14AF is 6.
The normal sequence is
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 42
The difference between 14AF and 14B5 is also 6.

The normal sequence is
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 43

Similarly the next 6 terms in the sequence are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 44

Similarly the next 6 terms are
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 45
So the missing terms are 14BB and 14C1

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 6.
Find the octal numbers corresponding to the following numbers using shorthand method.

  1. (ADD)16
  2. (DEAD)16

Answer:
1. (ADD)16
Step 1: Write down the 4 bit binary equivalent of each digit
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 46

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 47
(ADD)(ADD)16 = (5335)(ADD)8

2. (DEAD)16
Step 1: Write down the 4 bit binary equivalent of each digit.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 48

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 49
(DEAD)16 = (157255)8

Question 7.
If (126)x = (56)y, then find x and y.
Answer:
L.H.S contains 2 & 6, so x ≠ 2
R.H.S contains 5 & 6, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also The possible values of x and y are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 50
Case I:
Let x = 8 then y = 10
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 51
It is grater than (56)10
so when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 52

Question 8.
If (102)x = (42)y then (154)x = (  )y.
Answer:
L.H.S contains 2, so x ≠ 2
R.H.S contains 5 & 4, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also
The possible values of x and y are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 53
case I:
let x = 8 and y = 10
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 54
So when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 55
So x = 8 and y = 16
then we have fo find the hexadecimal equivalent of (154)8 For this first convert this into binary thus again convert it into hexadecimal. First write down the 3 bit equivalent of 154.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 56
Then divide this number into groups of 4 bits starting from the right and write down the hexa decimal equivalent.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 57
so the result is (154)8 = (6C)16

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 9.
Fill up the missing digit.
If (121)a = (441)b then (121)b = (  )10
Answer:
L.H.S. contains 2, so a ≠ 2
R.H.S. contains 4, so b ≠ 2
L.H.S. < R.H.S. So a > b and a b also.
Hence the values of a can be 10 or 16.
The values of b can be 8 or 10.
The possible values of a and b are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 58
Case I:
Let a = 16 and b = 10
(121 )16 = (289)10, so b ≠ 10

Case II:
Let a = 16 and b = 8
(121)16 = (289)10
(441)8 = 4 × 82 + 4 × 81 + 1 × 80
= 256 +32 + 1
= (289)10.
So a = 16 and b = 8.
Then (121)8 = 1 × 82 + 2 × 81 + 1 × 80
= 64 + 16 + 1 = (81)10

Question 10.
Fill up the missing digit. (Score 3)
If (128)a = (450)b then (16)a = (  )10
Answer:
L.H.S. contains 2 & 8, so a 2 and a ≠ 8.
R.H.S. contains 4 and 5, so b ≠ 2.
L.H.S. < R.H.S. so a > b and a ≠ b also.
The possible values of a and b are given below.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 59
Case I:
a = 16 and b = 8
(128)16 = (296)10
(450)8 = (296)10 So a = 16 and b = 8.
Then (16)16 = 1 × 16 + 6 × 160 = (22)10

Question 11.
Fill up the missing digit.
(3A.6D)16 = (  )8
Answer:
Step I: Write down the 4 bits equivalent of each digit.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 60

Step II: Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010

Step III: Write the octal equivalent of each group. SO we will get. (72.332)8.
(3A.6D)16 = (72.332)8

Question 12.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:

  1. Sign Magnitude Representation (SMR)
  2. 1’s Complement Representation
  3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number +5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is – ve.
eg: If a computer has word size is 1 byte then
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 61
Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can represent 27 = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1 ‘s Complement Representation: To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1’s complement.
eg: To find the 1’s complement of 21 +21 = 00010101
To get the 1 ‘s complement change all 0 to 1 and all 1 to 0.
-21 = 11101010
1’s complement of 21 is 11101010

3. 2’s Complement Representation: To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of 21 +21 =00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 62
2’s complement of 21 is 1110 1011

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 13.
Write short notes about Unicode (3)
Answer:
It is like ASCII Code. By using ASCII, we can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
eg: Malayalam, Hindi, Sanskrit .

Question 14.
Match the following.

1. (106)10a. (171.53125)10
2. (71.24)8b. (6a)16
3. (AB.88)16c. (20)8
4. (10)16d. (10000000)2
5. (128)10e. (10)16
6. (16)10f. (57.3125)10

Answer:
1 – b, 2 – f, 3 – a, 4 – c, 5 – d, 6 – e

Question 15.
Find the largest number in the list.

  1. (1001)2
  2. (A)16
  3. (10)8
  4. (11)10

Answer:
Convert all numbers into decimal
1. (1001)2 = 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 8 + 0 + 0 + 1
= (9)10

2.) (A)16 = (10)10

3. (10)8 = 1 × 81+0 × 80
= (8)10
So the largest number is 4 – (11)10

Question 16.
Subtract 10101 from 1111 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here subtrahend is 10101 and minuend is 1111 First take the 1’s complement of subtrahend and add 1 to it.
1’s complement of 10101 is 01010 add 1 to it
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 63
Here is no carry. So the result, is -ve and take the 2’s complement of 11010 and put a -ve symbol. So 1’s complement of 11010 is 00101 add 1 to this
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 64
So the result is -00110

Question 17.
Mr. Geo purchased (10)2 kg sugar @Rs. (110 10)2 and (1010)2 kg Rice @Rs. (10100)2. So how much rupees he has to pay in decimal.
Answer:
Convert each into decimal number system multiply and sum it up.
(10)2 = (2)10

(11010)2 = 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 21
= 16 +8 + 0 +2 + 0
= (26)10

(1010)2 = 1 × 23 + 0 × 22 + 1 × 21 +0 × 20
= 8 + 0 + 2 + 0
= (10)2

(10100)2 = 1 × 24 + 0 × 23 + 1 × 22 +0 × 21 + 0 × 20
= 16 + 0 + 4 + 0 + 0
= (20)2
therfore 2 × 26 + 10 × 20
= 52 + 200
= 252
So Mr. Geo has to pay Rs. 252/-

Question 18.
Mr. Vimal purchased a pencil @ Rs. (101)2, a pen @ Rs. (1010)2 and a rubber @ Rs. (10)2. So how much rupees he has to pay in decimal.
Answer:
Add 101 + 1010 + 10
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 65
then convert (10001)2 into decimal
(10001)2 = 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 16 + 0 + 0 + 0 + 1
= (17)10
So Mr. Vimal has to pay Rs. 17/-

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 19.
Mr. Antony purchased 3 books worth Rs. a total of (1100100)2. Atlast he returned a book worth Rs. (11001)2. So how much amount he has to pay for the remaining two books in decimal number sys¬tem.
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 66
then convert (1001011 )2 into decimal
(1001011)2 = 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20
=64 + 0 + 0 + 8 + 0 + 2 + 1
= (75)10
So he has to pay Rs. 75/-

Question 20.
Mr. Leones brought two products from a super market a total of Rs. (11010010)2 and he got a dicount of Rs. (1111)2 So how much he has to pay for this products in decimal number system.
Answer:
Substract (1111)2 from (11010010)2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 67
then convert (11000011)2 into decimal
(11000011)2 = 1 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20
=128 + 64 + 0 + 0 + 0 + 0 + 2 + 1
= (195)10

Question 21.
A textile showroom sells shirts with a discount of Rs. (110010)2 on all barads. Mr. Raju wants to buy a shirt worth Rs. (11111000)2. So after discount how much amount he has to pay in decimal.
Answer:
Substract (110010)2 from (11 111 000)2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 68
then convert (11000110)2 into decimal
(11000110)2 = 1 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20
= 128 + 64 + 0 + 0 + 0 + 4 + 2 + 0
= (198)10

Question 22.
Mr. Lijo purchased a product worth Rs. (1110011)2 and he has to pay VAT @ Rs. (1100)2. Then calculate the total amount he has to pay in decimal.
Answer:
Add (1110011)2 and (1100)2
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 69
then convert (1111111)2 into decimal
(1111111)2 = 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= (127)10

Question 23.
By using truth table, prove the following laws of Boolean Algebra.

  1. Idempotent law
  2. Involution law

Answer:
1. A + A = A
A = A = A
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 70
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 71

2. (A1)1 = A
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 72

Question 24.
Consider the logical gate diagram.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 73

  1. Find the logical expression for the circuit given.
  2. Find the compliment of the logical expression.
  3. Draw the circuit diagram representing the compliment.

Answer:
1. \((x+\bar{y}) \cdot z\)

2. \((\bar{x} . y)+\bar{z}\)
3.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 74

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 25.
Draw the logic circuit diagram for the following Boolean expression.
\(A \cdot(\bar{B} + \bar{C})+\bar{A} \bar{B} \bar{C}\)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 75

Question 26.
Consider a bulb with three switches x, y and z. Write the Boolean expression representing the following states.

  1. All the switches x, y and z are ON
  2. x is ON and y is OFF or Z is OFF
  3. Exactly one switch is ON.

Answer:

  1. xy . z
  2. \(x \bar{y}+\bar{z}\)
  3. \(x . \bar{y} . \bar{z}+\bar{x} . y . \bar{z}+\bar{x} . \bar{y} . \bar{z}\)

Question 27.
Match the following.

AB
i. Idem potent lawa. x + (y + z)=(x + y)+z
ii. Involution lawb. x + xy = x
iii. Complementarity lawc. x + y = y + x
iv. Commutative lawd. xx- 0
v. Absorption lawe. x = x
vi. Associative lawf. x + x = x

Answer:
i – f, ii – e, iii – d, iv – c, v – b, vi – a

Question 28.
Explain the principle of duality.
Answer:
It states that, starting with a Boolean relation, another Boolean relation can be derived by

  1. Changing each OR sign (+) to a AND sign (.)
  2. Changing each AND sign (.) to an OR sign (+)
  3. Replacing each 0 by 1 and each 1 by 0.

The relation derived using the duality principle is called the dual of the original expression,
eg: x + 0 = x is the dual of x . 1 = x

Question 29.
Draw the circuit diagram for \(F=A \bar{B} C+\bar{C} B\) using NAND gate only.
Answer:
\(F=A \bar{B} C+\bar{C} B\)
= (A NAND (NOT B) NAND C) NAND ((NOT C) NAND B)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 76

Question 30.
Draw a logic diagram for the function f = YZ + XZ using NAND gates only.
Answer:
f = YZ + XZ
= (Y NAND Z) NAND (X NAND Z)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 77

Question 31.
How do you make various basic logic gates using NAND gates.
Answer:
1. AND operation using NAND gate,
A.B = (A NAND B) NAND (A NAND B)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 78

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 79

3. NOT operation using NAND gate,
NOT A = (A NAND A)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 80

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 32.
Which of the following Boolean expressions are correct? Write the correct forms of the incorrect ones.

  1. A + A1 = 1
  2. A + 0 = A
  3. A . 1 = A
  4. A . A1 = 1
  5. A + A . B = A
  6. A . (A + B) = A
  7. A + 1=1
  8. \((\overline{\mathrm{A} . \mathrm{B}})=\overline{\mathrm{A}} . \overline{\mathrm{B}}\)
  9. A + A1B = A + B
  10. A + A = A
  11. A + B . C = (A+B) . (B+C)

Answer:

  1. Correct
  2. Correct
  3. Correct
  4. Wrong, A . A1 = 0
  5. Correct
  6. Correct
  7. Correct
  8. Wrong \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
  9. Correct
  10. Correct
  11. Wrong, A + B . C = (A + B) . (A + C)

Question 33.
Prove algebraically that (x + y)’ . (x’ + y’) = x’ . y’
Answer:
LHS = (x + y)’ . (x’ + y’)
= (x’ . y’) . (x’ . y’)
= x’ . y’ . x’ + x’ . y’ . y’
= x’ . y’ + x’ . y’
= x ‘. y’ = RHS
Hence proved.

Question 34.
Give the complement of the following Boolean Expression.

  1. (A + B) . (C + D)
  2. (P + Q) + (Q + R) . (R + P)
  3. (B + D’) . (A + C’)

Answer:
1. ((A+B) . (C+D))1 = (A+B)’ + (C+D)’
= A’ . B’ + C’ . D’

2. ((P+Q) + (Q+R) . (R.P))’ = (P+Q) ‘. ((Q+R) . (R+P))’
= P’ . Q’ . (Q+R)’ + (R+P)’
= P’ . Q’ . (Q’ . R’ + R’ . P’),

3. ((B+D’).(A+C’))’ = (B+D’)’0 + (A+C’)’
= B’ . D” + A’ . C”
= B’ . D + A’ . C

Question 35.
State and prove the idempotent law using truth table. Idempotent law
Answer;
Idempotent law states that

  1. A + A = Aand
  2. A . A = A Proof

1. A + A = A
Truth table is as follows:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 81
ie. A + A = A as it is true for both values of A. Hence proved.

2. A . A = A
Truth table is as follows:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 83
ie. A . A = A itself. It is true for both values of A. Hence proved.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 36.
State the Absorption laws of Boolean algebra with the help of truth tables.
Answer:
Absorption law states that
A + A . B = A and A . (A + B) = A
Proof:
The Truth table of the expression A + A . B=A is as follows.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 89
Here both columns A and A + A . B are identical. Hence proved.
For A . (A + B) = A, the truth table is as follows:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 90
Both columns A & A . (A + B) are identical. Hence proved

Question 37.
State Demorgen’s laws. Prove anyone with truth table method.
Answer:
Demorgan’s first theorem states that (A + B)’ = A’ . B’
ie. the complement of sum of two variables equals product of their complements,

The second theorem states that (A . B)’ = A’ + B’
ie. The complement of the product of two variables equals the sum of the complement of that variables.
Proof:
Truth table of first one is as follows:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 86
From the truth table the columns of both (A + B)’ and A’ . B’ are identical. Hence proved.

Question 38.
Fill in the blanks:

  1. (0.625)10 = (……….)2
  2. (380)10 = (……..)16
  3. (437)8 = (………)2

Answer:

  1. (0.101)2
  2. (17C)16
  3. (100 011 111)2

Question 39.
What do you mean by universal gates? Which gates are called Universal gates? Draw their symbols.

OR

Construct a logical circuit for the Boolean expression \(\bar{a} \cdot b+a \cdot \bar{b}\). Also write the truth table.
Answer:
Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.
NAND gate:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 87

NOR gate:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra - 88

Truth table:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 91

Logical circute:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 92

Question 40.
Computers uses a fixed number of bits to respresent data which could be a number, a character, image, sound, video etc. Explain the various methods used to represent characters in memory.
Answer:
Representation of characters.
1. ASCII(American Standard Code for Information Interchange):
It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by the U.S. government. Each character in the keyboard has a unique number.
eg: ASCII code of ‘a’ is 97.

When you press ‘a’ in the keyboard , a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 27 = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2. EBCDIC(Extended Binary Coded Decimal Interchange Code):
It is an 8 bit code introduced by IBM(International Business Machine). 28 = 256 characters can be represented by using this.

3. ISCII(Indian Standard Code for Information Interchange):
It uses 8 bits to represent data and introduced by standardization committee and adopted by Bureau of Indian Standards(BIS).

4. Unicode:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 216 = 65536 characters (i.e, world’s all written language characters) can store by using this.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 41.
Draw the logic circuit for the function
\(f(a, b, c)=a . b . c+\bar{a} . b+a . \bar{b}+a . b . \bar{c}\)

OR

Prove algebrically.
\(x . y+x . \bar{y} . z=x . y .+x . z\)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 93

OR

\(x \cdot y+x \cdot \bar{y} \cdot z=x \cdot(y+\bar{y} \cdot z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 42.
Following are the numbers in various number systems. Two of the numbers are same. Identify them:

  1. (310)8
  2. (1010010)2
  3. (C8)16
  4. (201)10

OR

Consider the following Boolean expression:
(B’ + A)’ = B . A’
Identify the law behind the above expression and prove it using algebriac method.
Answer:
1. (310)8 = 3 * 82 + 1 * 81 + 0 * 80
= 192 + 8 + 0
= (200)10

2. (1010010)2 = 1 × 26+ 0 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 0 × 20
= 64 + 0 + 16 + 0 + 0 + 2 + 0
= (82)10

3. (C8)16 = C × 16 + 8 × 160
= 12 × 16 + 8 × 1
= 192 + 8
= (200)10
Here (a) (310)8 and (C8)16 are same

OR

This is De Morgan’s law (B’ + A’) = (B’)’ . A’
= B . A’
Hence it is proved

Question 43.
Find the decimal equivalent of hexadecimal number (2D)16. Represent this decimal number in 2’s complement form using 8 bit word length.
Answer:
Convert (2D)16 to binary number for this write down the 4 bit binary equivalent of each number
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 94
(2D)16 = (00101101 )2
First find the 1’s complement of (00101101 )2 and add 1 to it
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 95
Hence 2’s complement is (11010011)2

Question 44.
Answer any one question from 15(a) and 15(b).
1. Draw the logic circuit for the Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

2. Using algebraic method prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:
1.
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 96

OR

2. L.H.S. = \(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y Z+Y\)
= \(\bar{y} \cdot(\bar{z}+z)+y \cdot(z+1)\)
= \(\bar{y}. 1+\bar{y} \cdot 1=y \cdot y=1\)

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 45.
With the help of a neat circuit diagram, prove that NAND gate is a universal gate.
Answer:
1. AND operation using NAND gate,
A . B = (A NAND B) NAND (A NAND B)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 97

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 98

3. NOT operation using NAND gate,
NOT A = (A NAND A)
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 99

Question 46.
Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

OR

Using algebraic method, prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:
Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra 100

OR

= Y . Z + Y . Z + Y . Z + Y
= Y . (Z + Z) + Y . (Z + 1)
= Y . 1 + Y. 1
= Y + Y
= 1
Hence the result.

Plus One Data Representation and Boolean Algebra Five Mark Questions and Answers

Question 1.
Explain the components of Data processing.
Answer:
Data processing consists of the techniques of sorting, relating, interpreting and computing items of data in orderto convert meaningful information. The components of data processing are given below.

  1. Capturing data: In this step acquire or collect data from the user to input into the computer.
  2. Input: It is the next step. In this step appropriate data is extracted and feed into the computer.
  3. Storage: The data entered into the computer must be stored before starting the processing.
  4. Processing/Manipulating data: It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
  5. Output of information: In this stage we will get the results as information after processing the data.
  6. Distribution of information: In this phase the information(result) will be given to the concerned persons/computers.

Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Question 2.
Define computer. What are the characteristics?
Answer:
A computer is an electronic device used to perform operations at very high speed and accuracy.
Following are the characteristics of the computer.

  1. Speed: It can perform operations at a high speed.
  2. Accuracy: It produces result at a high degree of accuracy.
  3. Diligence: Unlike human beings, a computer is free from monotony, tiredness, lack of concentration etc. We know that it is an electronic ma chine. Hence it can work four hours without making any errors.
  4. Versatility: It is capable of performing many tasks. It is useful in many fields.
  5. Power of Remembering: A computer consists of huge amount of memory. So it can store and recall any amount of information. Unlike human beings, it can store huge amount of data and can be retrieved when needed.

Disadvantages of computer:

  1. No. IQ: It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
  2. No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers)

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

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Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Plus One Accountancy Applications of Computers in Accounting One Mark Questions and Answers

Question 1.
The physical components of computer is called as …………..
(a) Software
(b) Hardware
(c) Liveware
Answer:
(b) Hardware

Question 2.
Set of programs which governs the operation of a computer system is termed …………..
(a) System software
(b) Software
(c) Application of window
Answer:
(b) Software

Question 3.
A centrally controlled integrated collection of data is called ……………
(a) DBMS
(b) Information
(c) Database
Answer:
(c) Database

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 4.
Tally is a
(a) Utility software
(b) Application software
(c) Operating system
(d) Connecting software
Answer:
(b) Application software

Question 5.
…………… is a software system that manages the creation of use of database.
(a) Database
(b) DBMS
(c) Management system
Answer:
(b) DBMS

Question 6.
Which one of the following is an output device of a computer?
(a) Mouse
(b) Keyboard
(c) Monitor
(d) Barcode reader
Answer:
(c) Monitor

Question 7.

…………. is the storehouse of a computer.
Answer:
Memory

Question 8.
…………… are set of program designed to carry out operations for a specified application.
Answer:
Application software.

Question 9.
The output obtained from VDU (Visual Display Unit) is termed ………..
Answer:
Hard copy.

Question 10.
The part.of the computer which controls the various operations of a computer is called ………..
Answer:
Control unit.

Question 11.
………….. is temporary memory and anything stored in it will remain there a long as the system is on.
Answer:
RAM (Random Access Memory)

Question 12.
Modern computerised accounting systems are based on the concept of ………..
Answer:
Database.

Question 13.
A sequence of actions taken to transform the data into decision-useful information is called ………..
Answer:
Data processing

Question 14.
The joystick is a …….. device of a computer.
Answer:
Input.

Question 15.
VDU is also called ……….
Answer:
Monitor.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 16.
Complete the series using the hint given.
Hint: System analyst → Human beings → liveware
a. Windows → Operating system → ?
Answer:
Software.

Plus One Accountancy Applications of Computers in Accounting Two Mark Questions and Answers

Question 1.
Match the following.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting img1
Answer:
1-b
2-e
3-d
4-c
5-a

Question 2.
What is a computer?
Answer:
Sp A computer is an electronic device that accepts data and instruction as input, stores them, process the data according to the instructions and communicate the results as output.

Question 3.
Hardware includes different devices. Name any four devices.
Answer:
Keyboard, Mouse, Monitor, Processor.

Question 4.
Redraw the given block diagram of a computer correctly:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting img2
Answer:
Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting img3

Question 5.
List out any four features of a computer.
Answer:

  1. Highspeed
  2. Large volume of data can be stored.
  3. Accuracy is very high.
  4. Computers are multipurpose information machine ie. versatility.

Question 6.
List out any four limitations of a computer.
Answer:

  1. Computers lacks common sense.
  2. Lack of decision-making skills
  3. Computers have no intelligence.
  4. Computers cannot make judgments based on feelings.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 7.
What is Accounting Information System?
Answer:
Accounting Information System (AIS) is a collection of resources (people and equipment), designed to transform financial and other data into information. Such information is organised in a manner that correct decisions can be based on it.

Question 8.
What are the basic requirements of a computerised accounting system?
Answer:
Every computerised accounting system has two basic requirements.

  1. Accounting Framework: It consists of a set of principles, coding and grouping structure of accounting.
  2. Operating procedure: It is a well defined operating procedure blended suitably with the operating environment of the organisation.

Question 9.
State the various essential features of an accounting report.
Answer:
The accounting report must have the following features it.

  1. Relevance
  2. Timelines
  3. Accuracy
  4. Completeness
  5. Summarisation

Question 10.
Give examples of the relationship between a Human Resource Information System and MIS.
Answer:
There is a relationship between the Human Resource Information System and Management Information System, the following are the example of it.

  1. Hiring employees as per the requirement.
  2. Evaluating the performance of the workers.
  3. Enrolling employees in benefit.

Question 11.
Give examples of two types of Operating System.
Answer:

  • DOS – Disk Operating System
  • Windows – Windows Operating System

Plus One Accountancy Applications of Computers in Accounting Three Mark Questions and Answers

Question 1.
Explain the term ‘Liveware’.
Answer:
People interacting with computers are called Live-ware of the computer system. It consists of the following three groups.
1. System analysts:
System analysts are the people who design data. processing systems.

2. Programmers:
Programmers are the people who write programs for processing data.

3. Operators:
Operators are the people who participate in operating the computer.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 2.
What is the Transaction processing system? Name three components of a Transaction processing system.
Answer:
Transaction processing systems (TPS) are among the earliest computerised systems catering to the requirements of large business enterprises. The purpose of a TPS is to record, process, validate and store transactions that occur in the various functional areas of business for subsequent retrieval and usage.
TPS system has three components:

  • Input-Processing-Output
  • ATM facility.
  • Telephone Account and Airline Seat Reservation System are examples of TPS.

Question 3.
Discuss the different types of accounting packages. The accounting packages are classified into the following categories.
Answer:
1. Ready to use accounting software:
It is relatively easier to learn and people adaptability is very high. It is suited to small/ conventional organisations. The level of secrecy is relatively low. This software offers little scope of linking to other information systems.

2. Customised Accounting software:
Helps to meet the special requirement of the user. It is suited to large and medium organisations and can be linked to the other information system.

3. Tailored:
The accounting software is generally tailored in large business organisations with multi-users and geographically scattered locations. This software requires specialised training for users. The level of secrecy is relatively high and they offer high flexibility in terms of number of users.

Question 4.
“Computers are the servants or masters of human beings.” Elucidate.
Answer:
A computer system have certain special features or advantages which in comparison to human beings become its capabilities.
The advantages of computers are as follows:

  1. High speed
  2. Accuracy
  3. Storage of huge data
  4. Versatility
  5. Deligence

Even though computers possess the above-mentioned features, it suffers from the following limitations:

  1. Computers lack common sense.
  2. Lack of IQ
  3. Lack of decision making skill
  4. No feeling

Question 5.
Find out the odd one and state reasons.

  1. Mouse, Monitor, Programmers, Processor.
  2. DACEASY, FORTRAN, ALU, LINUX
  3. Monitor, Barcode reader, Printer, Plotter

Answer:

  1. Programmers, others are hardware components.
  2. ALU, others are software
  3. Barcode reader, others are output devices.

Plus One Accountancy Applications of Computers in Accounting Four Mark Questions and Answers

Question 1.
Find the odd one and state reason.

  1. Keyboard, Mouse, Light pen, Printer
  2. System Analysts, Language Processors, System software, Utility Programmes.
  3. RAM, Floppy disk, Compact disk, Hard disk
  4. COBOL, C++, DOS, BASIC

Answer:

  1. A printer is an output unit, all others are input units.
  2. System Analyst is a human ware.
  3. RAM is the Internal memory unit, all others are. external memory unit.
  4. DOS is an operating system, all others are computer languages.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 2.
Classify the following into input unit and output unit devices.
Keyboard, Mouse, VDU (Visual Display Unit), Printer, Magnetic tape, Magnetic disk, Light pen, Optical scanner, Plotter, Speech synthesiser, MICR, OCR, Barcode reader, Smart card reader, Speaker, LCD projector.
Answer:

Input devicesOutput devices
KeyboardVDU
MousePrinter
Magnetic tape MagneticPlotter
disk Light Pen OpticalSpeech Synthesiser
scanner MICR OCRSpeaker
Barcode reader SmartLCD projector
card reader

Question 3.
What are the generic consideration before sourcing accounting software?
Answer:
The following factors are considered before sourcing accounting software:

  1. Flexibility
  2. Cost of installation and maintenance
  3. Size of organisation
  4. Ease of adaptation and training needs
  5. Utilities / MIS reports
  6. Expected level of secrecy (Software and data)
  7. Exporting/importing data facility
  8. Vendors reputation and capability

Question 4.
Classify the following components as Hardware, soft-ware and liveware.

  1. Programmers
  2. Keyboard
  3. Windows or Linux
  4. COBOL or C++
  5. Mouse
  6. Assembler or Compiler
  7. Operators
  8. Virus/Antivirus/ Scanners
  9. Monitor
  10. Processor
  11. System Analysts
  12. MS-Excel or MS Office

Answer:
a. Hardware:

  • Keyboard
  • Mouse
  • Monitor
  • Processor

b. Software:

  • Windows or Linux (Operating System)
  • COBOL or C++ (Computer Language)
  • Assembler or Compilers (Language processor)
  • Virus/Antivirus and Scanner (Utility programs)
  • MS Excel or MS Office

c. Liveware:

  • System analyst
  • Programmers
  • Operators

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 5.
Complete the following diagrams showing the functional relationship of the various components of computers.
Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting img4
Answer:
a. Input devices:

  1. Keyboard
  2. Mouse
  3. Light pen

b. Output devices:

  1. Monitor
  2. Printers
  3. Plotters

c. CPU:

  1. Memory unit
  2. ALU
  3. Control unit

d. Secondary storage devices:

  1. Floppy disk
  2. Hard disk
  3. Optical disk

Plus One Accountancy Applications of Computers in Accounting Five Mark Questions and Answers

Question 1.
Computerised Accounting is different from Manual accounting. Explain.
Answer:
Computerised accounting is different from manual accounting, the following are the main difference between these two:

Computerised AccountingManual Accounting
1. In computerised accounting data can be easily processed and statements can be prepared with high speed and accuracy.1. In manual accounting financial statements cannot be prepared with such speed and accuracy.
2. Mass data can be stored in very small space and brought back very easily.2. Data are stored in large number of books and retrieval of data is a very tedious job.
3. Coding is essential in computerised accounting.3. Coding is not essential.
4. Closing entries are not necessary.4. Closing entries are necessary.
5. The possibility of errors are less in computerised accounting.5. The possibility of errors are more.

Plus One Accountancy Applications of Computers in Accounting Six Mark Questions and Answers

Question 1.
What are the elements of a computer system?
Answer:
A computer system is a combination of six elements. They are as follows:
1. Hardware:
The physical components of a computer system is termed as Hardware. Eg: Mouse, Keyboard, Monitor, Processor, etc.

2. Software:
Set of programs that govern the operations of a computer system is termed as soft-ware. There are six types of software as follows.

  1. Application software
  2. Operating system
  3. Utility programs
  4. Language processors
  5. System software
  6. Connectivity software

3. People:
People interacting with computers are also called the “live-wave” of the computer system. It consists of the following three groups.

  1. System analysis
  2. Programmers
  3. Operators

4. Procedures:
The procedure means a series of operations in a certain order or manner to achieve desired results. There are three types of procedures which constitute part of computer system

  1. Hardware oriented
  2. Software oriented
  3. Internal procedure

5. Data:
These are facts and may consist of numbers, text, etc. These are gathered and entered into a computer system.

6. Connectivity:
Tie manner in which a particular computer system is connected to others says through telephone lines, microwave transmission, satellite, etc. is the element of connectivity.

Plus One Accountancy Chapter Wise Questions and Answers Chapter 10 Applications of Computers in Accounting

Question 2.
Define computerised accounting. List out various advantages and limitations of computerised accounting system.
Answer:
A computerised accounting system is an accounting information system that processes financial transactions and events to produce reports as per user requirements.
a. Advantages:

  1. Speed
  2. Accuracy
  3. Reliability
  4. Efficiency
  5. Storage and Retrieval
  6. Automated document production
  7. Quality reports
  8. Real-time user interface

b. Limitations:

  1. Huge training costs
  2. Staff opposition
  3. System failure
  4. Breaches of security
  5. Inability to check unanticipated errors