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SSLC Chemistry Chapter 5 Compounds of Non-Metals Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 5 Compounds of Non-Metals Solutions

Text Book Page No: 79

Sslc Chemistry Chapter 5 Kerala Syllabus  Question 1.
Take a little ammonium chloride (NH₄Cl) in a watch glass and add a little calcium hydroxide (Ca(OH)₂) to it. Stir well. Can you sense any smell?
Answer:
Irritating smell is experienced.

Compounds Of Non Metals Class 10 Kerala Syllabus Question 2.
Show wet blue and red litmus papers over the watch glass one by one. Which litmus paper shows a color change ?
Answer:
Red litmus turns to blue

5 nonmetals Question 3.
Is the gas acidic or basic?
Answer:
It is basic

Text Book Page No: 80

Compounds Of Nonmetals Kerala Syllabus 10th Standard Question 4.
Why ammonia gas is passed over quick lime (CaO)?
Answer:
To remove the water content in the ammonia gas.

Sslc Chemistry Chapter 5 Notes Kerala Syllabus Question 5.
What may be the reason for collecting ammonia in this manner ?
Answer:
Ammonia is lighter than air.

Chemistry Class 10 Chapter 5 Kerala Syllabus  Question 6.
What is your inference about the density of ammonia from this ?
Answer:
The density of ammonia is lighter than air

Sslc Chemistry Chapter 5 Questions And Answers Question 7.
Arrange the apparatus as shown in figure (Figure 5.2).

What do you observe?
Answer:
Water in the beaker lifts up in the flask which is filled with ammonia.

Chemistry Chapter 5 Class 10 Kerala Syllabus Question 8.
What inference can be made about the solubility of ammonia in water? Why does water rush into the flask?
Answer:
Ammonia dissolves abundantly in water. The pressure in the flask decreases with more ammonia dissolves in a water. So more quantity of water lifts up in the flask.

Compounds Of Nonmetals Class 10 Kerala Syllabus  Question 9.
Why does water entering the flask change its color?
Answer:
The ammonium hydroxide formed by dissolving ammonia in water is an alkali.

Compounds Of Non Metals Class 10 Notes Kerala Syllabus Question 10.
Which property of ammonia is responsible for this change in color?
Answer:
The property of base/alkali

Kerala Syllabus 10th Standard Chemistry Chapter 5 Question 11.
Complete the chemical equation given below and find the product obtained when ammonia is dissolved in water.
Answer:
NH₃ + H₂O → NH₂OH

Sslc Chemistry Chapter 5 Notes Pdf Kerala Syllabus Question 12.
Tick ✓ which is applicable to ammonia

Answer:

Compounds Of Non Metals Kerala Syllabus 10th Standard Question 13.
When an Ammonia tanker leaks, water is sprayed to reduce its intensity. What is the reason for this?
Answer:
Ammonia dissolves abundantly in water. Ammonia gas is released when the tanker leaks, it dissolves in water and can minimize the spreading of gas into atmosphere.

Compounds Of Non-Metals Class 10 Kerala Syllabus Ques. 14.
Some uses of ammonia are given below.
Answer:

• For the manufacture of chemical fertilizers like ammonium sulfate, ammonium phosphate, urea, etc.
• As a refrigerant in ice plants.
• To clean tiles and window panes.
• Ammonia is also used as a refrigerant gas, for purification of water sup¬plies, and in the manufacture of plastics, explosives, textiles, pesticides, dyes and other chemicals.
• It is found in many household and industrial-strength cleaning solutions.

Text Book Page No: 82

Sslc Chemistry Chapter 5 Pdf Kerala Syllabus Question 15.
Take some ammonium chloride (NH₄Cl) in a boiling tube and heat it. Don’t you sense a peculiar smell?
Answer:
Yes. Pungent smell

Kerala Syllabus 10th Chemistry Chapter 5 Kerala Syllabus Question 16.
Which is the gas evolved here?
Answer:
Ammonia.

Hsslive Guru Chemistry 10 Kerala Syllabus Question 17.
Show a wet red litmus paper on the mouth of the boiling tube. What change can you observe?
Answer:
Turns blue

10th Class Chemistry Chapter 5 Kerala Syllabus Question 18.
Keep the litmus paper for some more time at the mouth of the boiling tube and then observe its color change. What is the change occurred?
Answer:
The wet litmus paper has changed again to red color due to the presence of the hydrogen chloride (HCl) gas. When ammonium chloride (NH₄Cl) is heated. Lighter NH₃ comes out first then the denser HCl comes out.

Class 10 Chemistry Kerala Syllabus Question 19.
Write the chemical equation of this reaction.
Answer:
NH₄Cl(S) → NH₃(g) + HCl (g)

Text Book Page No: 83

Question 20.
A glass rod dipped in cone. HCl is, shown into the jar which is filled with ammonia. What do you observe?
Answer:
White smoke is formed.

Question 21.
Complete the equation and find out the product?
Answer:
NH₃ + HCl → NH₄C1
NH₄Cl is the product. The formation of NH₄Cl is the reason for the white smoke.

Question 22.
What happens to the white powder on heating?
Answer:
It disappears

Text Book Page No: 84

Question 23.
Examine the chemical equations given below and write the forward and backward reactions in each.
1. N₂ (g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
2. 2SO₂ (g)+ O₂ (g) \(\rightleftharpoons\) 2SO₃(g)
3. H₂(g) + I₂ (g) \(\rightleftharpoons\) 2HI (g)
Answer:
1. N₂ (g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
Forward reaction:
N₂ (g) + 3H₂(g) → 2NH₃(g) 1
Backward reaction:
2NH₃(g) → N₂ (g) + 3H₂(g)

2. 2SO₂ (g)+ O₂ (g) \(\rightleftharpoons\) 2SO₃(g)
Forward reaction:
2SO₂ (g)+ O₂ (g) → 2SO₃(g)
Backward reaction:
2SO₃(g) → 2SO₂ (g)+ O₂ (g)

3. H₂(g) + I₂ (g) \(\rightleftharpoons\) 2HI (g)
Forward reaction:
H₂(g) + I₂ (g) → 2HI (g)
Backward reaction:
2HI (g) → H₂(g) + I₂(g)

Text Book Page No: 85

Question 24.
What happens to the rates of forward and backward reactions as time progresses?
Answer:
As the time passes, rate of forward reaction decreases and rate of backward reaction increases.

Question 25.
Identify the point at which the rates of both forward and backward reactions become equal?
Answer:
A

Text Book Page No: 86

Question 26.
The rate of which reaction increases when the concentration of nitrogen is increased? Forward reaction/ Backward reaction.
Answer:
Forward reaction

Question 27.
What happens if the concentration of ammonia is increased?
Answer:
Rate of backward reaction increases

Question 28.
What will be the effect of removing ammonia continuously from the system?
Answer:
Rate of forward reaction increases.

Question 29.
Complete the table writing the effect of change in concentration in the system at equilibrium.

Answer:

Question 30.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
In this equation what is the total number of moles of the reactant molecules?
Answer:
Total number of moles of the reactant = 4

Text Book Page No: 87

Question 31.
What about the products?
Answer:
2

Question 32.
Forward reaction : 4 mole reactant molecule → 2 mole product molecules ( volume decreases)
Backward reaction : …. (a) …..mole product molecules → ….(b) ….. mole reactant molecules (volume ……(c) …….)
Answer:
a. 2
b. 4
c. increases

Question 33.
In the manufacture of ammonia, the reaction in which direction results in the decrease in the number of molecules?
Answer:
Forward reaction.

Question 34.
What happens when the pressure of system is decreased?
Answer:
Rate of forward reaction increases.

Question 35.
What if the pressure of the system is decreased?
Answer:
Rate of backward reaction increases.

Question 36.
In the manufacture of ammonia, why is a pressure of 150-300 atm used?
Answer:
When pressure is increased, rate of forward reaction increases. As a result, there will be an increase in the yield of product.

Question 37.
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
What is the total number of moles of reactants?
Answer:
2

Question 38.
What about the products?
Answer:
2

Text Book Page No: 88

Question 39.
N₂ (g) + 3H₂ (g) – 2NH₃(g) + Heat
Which is the endothermic reaction in this ?
Forward reaction / Backward reaction
Answer:
Backward reaction.

Text Book Page No: 89

Question 40.
Sulphuric acid is formed also by the direct dissolution of sulfur trioxide in water. Still, sulfur trioxide is not directly dissolved in water. Why?
Answer:
The dissolution of sulfur trioxide in water is an exothermic process. It may turn sulphuric acid initially formed into fine fog-like particles (smog) which will hinder further dissolution.

Text Book Page No: 90

Question 41.
Complete the flow chart.

Answer:
a. SO₃
b. H₂O

Question 42.
Take 5mL water in a test tube and slowly add concentrated sulphuric acid to it. Touch the bottom of the test tube. What do you feel? Is the reaction exothermic or endothermic?
Answer:
Feels hot. It is a exothermic reaction.

Text Book Page No: 91

Question 43.
What are the constituent elements of sugar?
Answer:
Carbon, Hydrogen, and Oxygen

Question 44.
Which is the black substance in the product formed?
Answer:
Carbon

Question 45.
What is the ratio of hydrogen and oxygen in sugar?
Answer:
2: 1

Question 46.
Which is the substance that absorbed hydrogen and oxygen from sugar in the ratio as in water?
Answer:
Conc. H₂SO₄

Question 47.
Complete the table by involving the activities given below.

Answer:

Question 48.
Why is concentrated sulphuric acid not used as a drying agent in the preparation of ammonia?
Answer:
Ammonia is a base. Ammonia reacts with sulphuric acid and the salt ammonium sulfate is produced.
2NH₃ + H₂SO₄ → (NH₂)2SO₄

Text Book Page No: 92

Question 49.
Add concentrated sulphuric acid to a test tube containing a small quantity of carbon. Heat it. What do you observe?
Answer:
Gases are formed. CO₂ and SO₂ are the gases.

Question 50.
Analyze the chemical equation and find the reason for your observation.

a. What is the oxidation state of elemental carbon?
Answer:
Zero

b. What about the carbon in carbon dioxide?
Answer:
The oxidation state of carbon in carbon dioxide is +4.

c. Was carbon oxidized or reduced in this reaction?
Answer:
Carbon is oxidized.

d. Which is the oxidizing agent?
Answer:
Sulphuric acid is the oxidizing agent.

Question 51.
See the reaction between concentrated sulphuric acid and copper.

a. Is copper oxidized or reduced in this case?
Answer:
Copper is oxidized.

b. Which is the oxidizing agent in this reaction? Which is the reducing agent?
Answer:
Copper is the reducing agent.
Sulphuric acid is the oxidizing agent.

Text Book Page No: 93

Question 52.
Analyze the given chemical equation.
Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl

a. Which substance is soluble in water among the products?
Answer:
NaCl

b. Which substance is the white precipitate?
Answer:
Barium sulphate (BaSO₄).

c. Does the white precipitate dis-solve when dilute hydrochloric acid is added to it?
Answer:
No. It is not soluble in dilute HCl

Question 53.
Write down the observation in the table given below, when lmL Barium chloride solution is added to the solutions given in the table.

Answer:

Let Us Assess

Question 1.
In which of the following reversible reactions does change in pressure not influence equilibrium? What is the reason?
i. H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI
ii. N₂ + 3H₂(g) \(\rightleftharpoons\) 2NH₃
Answer:
i. H₂(g) + I₂(g) \(\rightleftharpoons\) 2 Hg(g)
In this reaction, the number of moles of the reactants and the number of moles of the products are same.

Question 2.
What is the use of applying high pressure during the formation of ammonia from nitrogen and hydrogen?
Answer:
In this reaction, the number of moles of the products is less than the number of moles of the reactants. So if the pressure is increased, rate of forward reaction is increased. Thereby the yield of the product will be more.

Question 3.

a. Identify the reactants and products.
b. Products are frequently removed from the system? Explain the reason.
Answer:
a. Reactants : C(s), H₂O(g)
Products: CO(g), H₂(g)
b. When the products are frequently re moved from the system, rate of forward reaction increases. Hence more products will be formed.

Question 4.
2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g) + heat
In this reaction how do the following changes influence the amount of the product?
a. Decrease in temperature
b. Increase in pressure
c. Increase in concentration of oxygen
Answer:
a. When the temperature is decreased, rate of forward reaction increases. Hence more products is formed.
b. When pressure is increased, rate of forward reaction increases. Hence more product is formed.
c. When the concentration of oxygen is increased the rate of forward reaction increases. Hence more product is formed.

Question 5.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + heat
a. What change in pressure is required for the maximum yield of the product?
b. What is the change in concentration required for increasing the rate of the forward reaction?
Answer:
a. On increasing the pressure, yield of the product will be maximum.
b. Increase the concentration of reactants nitrogen or hydrogen. Removal of product ammonia from the system.

Question 6.
The chemical equation of one of the different stages of manufacturing sulphuric acid by contact process is given below. Find out the influence of the following factors in the reaction given below.
2SO₂ (g) + O₂ (g) \(\rightleftharpoons\) 2SO₃(g) + heat
1. Increase the amount of oxygen
2. Pressure is increased
3. Catalyst vanadium pentoxide (V₂O₅)is added
4. SO₃ is removed
Answer:
1. Rate of forward reaction increases.
2. Rate of forward reaction increases
3. Attains the equilibrium quickly.
4. Rate of forward reaction increases

Question 7.
Calcium oxide (CaO) is used as drying agent in the preparation of Ammonia in laboratory. Can concentrated H₂SO₄ be used as drying agent instead of CaO? Justify your answer.
Answer:
Sulphuric acid can not be used instead of CaO. it is because sulphuric acid reacts with ammonia and the salt ammonium sulfate is formed.

Question 8.
Which property of sulphuric acid is shown in the following situations.
a. During the preparation of chlorine, the gas is passed through concentrated H₂SO₄
b. Wooden cupboards appeared to be burnt when concentrated sulphuric acid happened to fall on it.
Answer:
a. It’s character as a drying agent
b. It’s character as a dehydrating agent.

Extended Activities

Question 1.
The graph for the reaction N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)+ heat is given below.

a. Identify and write the reactions C and D
b. What happens to the position of point A in the graph when a catalyst is used? Redraw the graph.
Answer:
a Reaction C – Forward reaction,
Reaction D – Backward reaction,
b. Equilibrium is attained quickly. Point A shift to the left side.

Question 2.
It is often said that the production of sulphuric acid is a bench mark of the industrial development of a country. Prepare a note based on the various uses of sulphuric acid.
Answer:
Sulphuric acid is an important industrial chemical which is used in the manufacturing processes of many goods over a wide range of applications. Sulfuric acid used in pulp and paper industry for chlorine dioxide generation, tall oil splitting and pH-adjustments.Sulfuric acid is a strongly acidic, oily liquid that may be clear to cloudy in appearance. Concentrated sulfuric acid acts as both an oxidizing and dehydrating agent.

Sulfuric acid was once known as oil of vitriol. Here are some of the growing number of end-users and applications using sulfuric acid is Agricultural chemicals Aluminum Sulfate, Batteries, Cellophane, Detergents, Explosives, Fertilizers, Gasoline, Herbicides, Iron and steel pickling, Jet Fuel, Kerosene, Leather, Lubricating Oils, Medicinal processes, Oil additives, Paper, Rayon and rubber, Sugar, Synthetic fibers, Veterinary drugs, Water softener regeneration, Water treatment, Yellow pigments.

Question 3.
Fill half of a beaker of capacity 50 mL with sugar. Add concentrated sulphuric acid so that the sugar is immersed in it. Observe the changes. What are the products formed? Which property of sulphuric acid is revealed here?
Answer:
The sulphuric acid removes water from the sugar in a highly exothermic reaction, releasing heat, steam, and sulfur oxide fumes. Aside from the sulfurous odor, the reaction smells a lot like caramel. The white sugar turns into a black carbonized tube that pushes itself out of the beaker. Dehydrating properties of H₂SO₄ are shown in the experiment. Concentrated sulphuric acid has the ability to absorb chemically combined water, or hydrogen and oxygen from substances in the ratio corresponding to that of water. This process is known as dehydration. Concentrated sulphuric acid is a strong dehydrating agent.

Compounds of Non-Metals Orukkam Questions and Answers

Question 1.
a. Find out the relation between Temperature and rate of a reaction?
Select the materials from the list above
Sodium the Sulphate, Test tube, dil HCl, Cu, Mg ribbon, Beaker, Water, Spirit lamp, boding Tube
b. Prepare a write up for finding the relation between temperature and rate of a reaction?
c. Why rate of a reaction increase when temperature increases?
Answer:
a. Essential Materials:- Sodium Thiosu- phate, Hydrochloric Acid, Water, Boiling Tube, Spirit Lamp. Prepare a solution of diluted sodium Thiosulphate in a beaker. Take equal amount of this solution in two Boiling Tubes. Heat one of the boiling tubes for some time. Add equal amount of diluted HCl to both the boiling tubes.
b. The Essential material/Materials Required Mg Ribbon, Cone HCl, dil HCl
Write up
Take Magnesium Ribbons of equal mass in two test tubes. Add dil HC1 to one test tube and dilute HC1 to the other in equal volumes.
c. Energy as well as the speed of molecules increases when reactants are heated. That is as temperature increases the number of molecules with three hold energy increases. As a result, the number of effective collisions increases and these rate of reaction also increases.

Question 2.
Some chemical reactions are given below.
1. Zn + 2HCI → ZnCl₂ + H₂
2. 2Mg + O₂ → 2MgO
3. NH₄Cl \(\rightleftharpoons\) NH₃ +HCl
a. What are the peculiarities of first two reactions
b. Conduct an experiment for viewing the dissociation and association taking place in the third equation.
c. In the three reactions, reactants turned into product and products are converted into reactants, is it true? d What type of reactions are they all represent?
e. Write down the characteristics of the re-action.
Answer:
a. In these reactions, the reactants change to products but the products cannot be changed back to reactants.
b. Heat some Ammonium chloride in a boiling tube.
c. No
d. First two Reactions are irreversible reactions. Third Reaction is a reversible Reaction.
e. Characteristics of Irreversible Reaction:- Reactants become products but products cannot be changed back to Reactants.
Characteristics of Reversible Reaction:- Reaction takes place in both directions.

Question 3.
Fe(NO₃)₃ + 3KCNS → Fe(CNS) + 3KNO₃ This balanced chemical equation is wrote on the black board when the teacher is going to conduct an experiment on chemical equilibrium.
a. In the above reaction, which chemical has red color.
b. Fe(NO₃)₃, KCNS combined together, put it on the test tube stand, is there any col our change? Is the color is diminishing while it kept in test tube stand?
c. Convert the solution reared into four be a kers dilute each with equal amount of water.
In the first beaker add Fe (NO₃)₃, in the second one KCNS, in the third KNO₄ like wise. Compare the color change with the fourth beaker. Find out the reason.
d. Point out the characteristics of equilibrium based on the experiment done.
e. In minute level chemical equilibrium is Kinetic energy why?
f. How and when a reversible reaction attains chemical equilibrium.
g. In the graph given below, when the reactant and product attain the level A? What are the characteristics of the point A?

Answer:
a. Fe (CNS), (Ferric Thiocyanate)
b. Yes, Red color increases
c.

d. At equilibrium both the reactants and products coexist.

• The rates of forward and backward reactions become equal at equilibrium.
• Chemical equilibrium is dynamic at the molecular level.
• The system can attain equilibrium even when the concentration of the reactants and products are not equal. Once equilibrium has been attained, there will be no change in the concentration of reactants as well as that of products.
• Chemical equilibrium is attained in closed system.

e. Even in equilibrium state reactants, molecules form products and products form reactants. Hence in minute level chemical equilibrium is kinetic energy.
f. The concentration of reactants and products do not change a reversible reaction attain chemical equilibrium,
g. A is the point at which the rates of both forward and backward reactions become equal.
At equilibrium both the reactants and products coexist.

Question 4.
N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃
H₂+I₂ \(\rightleftharpoons\) 2Hl
N₂O₄(g) \(\rightleftharpoons\) 2NO₃
2SO₂ + O₂ \(\rightleftharpoons\) 2SO₃
Write down detail how amount of the prod¬ucts increases in the above reactions (based on Le chandelier principle)
Hints: reference must be given on each of the following, concentration, pressure, temperature, catalyst
Answer:

• Increase the concentration of reactants or anyone reactants.
• Increase the temperature – Increases the rate of endothermic reactions.
• Decrease the temperature- Increase the rate of exothermic reactions. Increase the pressure- Reaction takes place faster in the direction where the molecules are less.
• Decrease the pressure- Reaction takes place faster in the direction where molecules are more.
• Catalysts are added at the beginning of a chemical reaction.

Compounds of Non-Metals SCERT Questions and Answers

Question 5.
The graph showing the progress of the reaction N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃ is given

a. Identify the reactions represented by B and C?
b. What is the significance of the state A?
c. Is there any change in the concentration, as time passes after attaining the stage A? Explain.
Answer:
a. B – Forward reaction, C – Backward reaction
b. Equilibrium
c. No At equilibrium rate of both forward and backward reactions are equal.

Question 6.
Coldwater is taken in one test tube an hot water in another one. Mg ribbon with same size is dropped in each of the test tubes,
a. In which test tube, hydrogen is formed with greater speed?
b. Which factor influences the rate of reaction? Explain the reason.
Answer:
a. Test Tube which contains hot water.
b. Temperature, When temperature increases the number of molecules with thresh-old energy increases. As a result number of effective collisions increases.

Question 7.
The chemical equation of the industrial preparation of ammonia is given below.
N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃ + Heat
Suggest the methods to get more NH₃.
Answer:
Increase the Pressure and Temperature.

Question 8.
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g) + Heat
The following circumstances influence the reaction
a. Increase the concentration of H₂.
b. Increase the pressure
c. Increase the temperature.
Answer:
a. Increases the rate of forward reaction
b. No effect
c. Increases the rate of backward reaction

Question 9.
The formation of SO₃ in the industrial preparation of sulphuric acid is given below.
2SO₂ + O₂ \(\rightleftharpoons\) 2SO₃+ Heat
a. Explain the effect of concentration of 02 to get maximum yield of SO₃? State rea son.
b. Identify the law related to it. State it.
Answer:
a. Increase the concentration of oxygen, which increases the rate of forward reaction,
b. When the concentration, pressure or tern perature of a system at equilibrium is changed, the system will readjust itself so as to nullify the effect of that change and attain a new state of equilibrium. This is Le Chatelier’s principle.

Question 10.
The chemical equation of a stage in the industrial preparation of sulphuric acid is given below.
2SO₂ + O₂ \(\rightleftharpoons\) 2SO₃+ Heat
a. Which is the catalyst used in this reaction?
b. What is the influence of the catalyst in equilibrium?
Answer:
a. Vanadium pentoxide / V₂O₅
b. Catalyst increases the rate of both the forward and backward reactions to the same extent.

Question 11.
The chemical equation of the industrial preparation of ammonia is given below.
N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃+ Heat
a. Temperature is to be decreased to get maximum yield of ammonia, according to the Le Chatelier principle. Why?
b. What is the reason for taking an optimum temperature in this reaction?
Answer:
a. When the temperature decreases exothermic reaction increases ie. forward reaction increases.
b. At low-temperature rate of forward and backward reaction is slow.

Question 12.
Analyse the following equations and answer the questions

a. Which of these reactions are affected by change in pressure? What are the reasons?
b. How the increase in pressure influence the reaction which you have already identified ?
Answer:
a. iii. N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃
In first reaction NH₄Cl is solid compound In second reaction number of reactants and product molecules are equal.
b. When pressure increases, rate of forward reaction increases.

Question 13.
Catalysts are substances which influence the rate of chemical reactions.
Explain how the catalysts influence the rate of reversible reaction?
Answer:
The catalyst increases the rate of both the forward and the backward reactions to the same extent. As a result, the system reaches equilibrium at a faster rate.

Question 14.
Some features of a reversible reaction are given below.
1. Product formation increases when the temperature is increased.
2. There is no effect when the pressure is increased.
Explain the reason for above inferences.
Answer:
1. Forward reaction is an exothermic.
2. Number of reactant and product molecules

Question 15.
A + B + Heat \(\rightleftharpoons\) 2C + D
This reversible reaction is in equilibrium. What happens to the amount of products under the following conditions?
a. C is removed from the system
b. B is added in excess
c. Temperature is increased
d. A suitable catalyst is added.
Answer:
a. Increase the amount of product
b. Increase the amount of product
c. Decrease the amount of product,
d. Increase the amount of product

Compounds of Non-Metals Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 16.
Select the correct statements which are related to the influence of catalyst in a reversible reaction.
a. Forward reaction takes place when a cata lyst is used in a reversible reaction, h, Attains equilibrium faster,
c. Catalyst does not help to form more pro-duct
d The catalyst increases the rates of both the forward and the backward reactions to the same extent.
e. Increases the speed of backward reaction.
f. Does not helps to produce more product
Answer:
Statements b, c, d are correct.

Question 17.
2X(g) + heat \(\rightleftharpoons\) Y(g) + Z(g) Explain the influence of pressure and temperature in this reaction.
Answer:
Pressure — Pressure has no effect here. Because the number of molecules in the reactants and products side are equal.
Temperature — Here the forward reaction is endothermic. So increasing temperature increases the rate of forward reaction. Decreasing temperature decreases the rate of forward reaction.

Short Answer Type Questions (Score 2)

Question 18.
Observe the equation of the chemical reactions given below.
i. NaOH (aq) + HCl (aq) → 2NaCl (aq) + H₂O(l)
ii. N₂(g) + O₂(g) \(\rightleftharpoons\) 2NO(g)
iii. Zn(s) + ZHCl → ZnCl₂(ql) + H₂(g)
iv. 2SO₂(g) + O₂(g) \(\rightleftharpoons\) 2SO₃(g)
a In which chemical reaction pressure can influence its speed ?
b. In this reaction what changes will be made in pressure for increasing forward reaction ? Why ?
Answer:
Reaction (iv)
2SO₃(g) + O₃(g) \(\rightleftharpoons\) 2SO₃(g)
b. In this reaction, the number of moles of the products is less than the number of moles of the reactants. So if the pressure is increased, rate of forward reaction is increased. Thereby the yield of the product will be more.

Question 19.
A graph given below deals with the reversible reaction.

a What does A, B, C indicate?
b. What inference can be drawn about the concentration of the reactants and products at the point D and E?
Answer:
A – Forward reaction
B – Backward reaction
C – Equilibrium
b. There is no change in the concentration, of reactants and products at the points D and E. Because at the point ‘C’ the process attains equilibrium.

Question 20.
Some chemicals are given below. Sodium chloride, Ammonium hydroxide, Nitric acid, cone. Sulphuric, Sodium hydroxide.
a. Which are the substances needed to produce hydrogen chloride?
b. Suggest a method to identify hydrogen chloride gas.
Answer:
a. Sodium chloride
b. Show a glass rod dipped in ammonia, just above the jar in which hydrogen chloride gas is taken. A white smoke of NH₃₄Cl is produced.

Question 21.
When ammonia was leaked two solutions were arised.
i) Spray water
ii) Spray HCl
Which method you adopt? Justify your answer.
Answer:
Spray water. Ammonia dissolves abundantly in water. So we can prevent the easy spreading of ammonia. When HCl is used the white smoke of NH₄Cl spread in the atmosphere.

Short Answer Type Questions (Score 3)

Question 22.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃ (g) + heat
How do the following factors influence the forward reaction?
a. One of the products is removed
b. Increase in pressure
c. More N₂ is added.
Answer:
a. Speed of forward reaction increases
b. Speed of forward reaction increases
c. Speed of forward reaction increases

Question 23.
A graph given below deals with the reversible reaction.

a. What happened to the forward and backward reactions as time passes?
b. In which minute does the system attain equilibrium?
c. What change occurs to the equilibrium, when a catalyst is used?
Answer:
a. As time passes the speed of forward reaction decreases and backward reaction increases.
b. 25th-minute
c. Attain equilibrium fast ie. before 25th minute.

Question 24.
Ammonia is an industrially useful compound of nitrogen.
a. Name the industrial production of ammonia
b. Write the equation of the reaction,
c. Write any two uses of ammonia
Answer:
a. Haber process
b. N₂ + 3H₂ → 2NH₃
c. i. For the manufacture of the fertilizers
ii. As a coolant in ice plants.

Question 25.
Ammonia is a pungent-smelling gas.
a. How does ammonia convert into
i. liquor ammonia
ii. liquid ammonia
b. Write the color, smell, solubility in water and density of ammonia.
Answer:
a. i. Prepare a concentrated solution of ammonia by dissolving it in water. It is liquor ammonia.
ii. The gaseous ammonia can be converted into liquid by applying pressure. It is called liquid ammonia.
b. Ammonia has no color and irritating smell. It dissolves abundantly in water. It has density less than air.

Question 26.
Sulphuric acid is industrially manufactured by using contact process. Write the equation of this process.
Answer:

Question 27.
Write equations for the following chemicals to prepare from sulphuric acid.
a. Hydrogen chloride
b. Oleum
c. Sodium sulfate
Answer:
a. NaCl + H₂SO₄ → NaHSO₄ + HCl
b. H₂SO₄ + SO₃ → H₂S₂O₇
c. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Question 28.
Observe the following chemical equation
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
a Which one has oxidation state as 0?
b. What is its oxidation state after the chemical reaction?
c. Is the change oxidation or reduction?
d. Which chemical nature of sulphuric acid is found here?
Answer:
a. Cu
b. 2+
c. Oxidation
d. Nature of oxidation

Question 29.
Litmus has colored in the solution.
A. When BaCl₂ solution was added to this.
B. Solution, a white precipitate insoluble in HCl is produced.
1. Write the chemical formula of A.
2. Write the chemical name and chemical formula of the white precipitate produced when it was added with BaCl2.
c. Complete the equation
C + 2(A) →…….. + …… +……..
Answer:
a. H₂SO₄
b. Barium sulphate (BaSO₄)
c. C + 2H₂SO₄ → CO₂ + 2H₂O + SO₂

Question 30.
Analyze the table given below and complete it.

Answer:
a. White precipitate is formed
b. Ba₂CO₃
c. Clear solution
d. White precipitate is formed
e. BaSO₄
f. No change
g. No substance is formed

Question 31.
Sulphuric acid is a drying agent as well as a dehydrating agent.
a. Find out the difference between these two activities.
b. Write one example each for these nature of sulphuric acid.
Answer:
a. Dehydration is the process of absorbing hydrogen and oxygen from substances in the ratio of water.
Drying activity is the process of absorbing water content present in a substance.
b. Sugar bums to black when concentrated sulphuric acid is added in it. It is an example for dehydration.
Concentrated H₂SO₄ is used to remove water particles from HCl gas during its manufacture – drying agent.

Question 32.
Some incomplete equations are given below:
i. MgSO₄ + BaCl₂ → …….. + MgCl₂
ii. K₂CO₃ + BaCl₂ → ……… + 2KC₁
iii. KCl + AgNO₃ → …….. + KNO₃
a. Complete the equations.
b. Find out the white precipitate in each of the reactions.
Answer:
a. i. MgSO₄ + BaCl₂ → BaSO₄ + MgCl₂
ii. K₂CO₃ + BaCl₂ → BaCO₃ + 2KCl
iii. KCl + AgNO₃ → AgCl + KNO₃
b. i. BaSO₄
ii. BaCO₃
iii. AgCl

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SSLC Maths Chapter 10 Polynomials and Algebra Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 10 Polynomials and Algebra Notes

Textbook Page No. 237

Free Degree and Leading Coefficient Calculator – Find the leading coefficient of a polynomial function step-by-step.

Polynomials Class 10 Kerala Syllabus  Questions 1.
Write the second degree polynomials. given below as the product of two first degree polynomials. Find also the solutions of the equation p(x) = 0 in each.
i. p(x) = x² – 7x+12
ii. p(x) = x² + 7x + 12
iii.p (x) = x² – 8x +12
iv. p(x) = x² + 13x +12
v. p (x) = x² + 12x – 13
vi. p (x) = x² – 12x – 13
Answer:
i. p (x) = x² – 7x + 12
a + b = –7, ab = 12
a = –3, b = –4
x² – 7x + 12 = (x – 3) (x – 4)
x² – 7x + 12 = 0
(x – 3) (x – 4) = 0
x – 3 = 0, x – 4 = 0
x = 3, x = 4
ii. p(x) = x² + 7x + 12
a + b = 7, ab= 12 a = 3, b = 4
x² + 7x + 12 = (x + 3) (x + 4)
x² + 7x + 12 = 0
(x + 3)(x + 4) = 0
x = 3, x = 4
iii. p(x) = x^{2 –} 8x + 12
a + b = 8, ab= 12
a = 6, b = –2
x² – 8x + 12 = (x – 6) (x – 2)
(x – 6) (x – 2) = 0
x = 6, x = 2
iv. p(x) =x² + 13x + 12
a+b = 13, ab = 12
a =12, b=1
(x² +13x + 12) = (x + 12) (x + 1)
x²+13x + 12 = 0
(x+ 12) (x+ 1) = 0
x+ 12 = 0, x+ 1 = 0
x = -12 ,x = –1
v. p(x) = x² + 12x – 13
a = –13, b = 1
x² + 12x – 13 = (x + 13)(x – 1)
x + 13 = 0, x – 1 =0
x = –13, x= 1 .
vi. p(x) = x² – 12x – 13
x² – 12x – 13 = (x – a) (x – b)
= x² – (a + b) x + ab
a + b= 12 ab = –13
(a – b)² =(a + b)² – 4ab
= (12)² – 4x – 13 = 196
a – b = 14
a + b= 12
a= 13; b = –1
x² – 12x – 13 = (x – 13)(x + 1)
x – 13 = 0, x + 1 =0
x= 13 ,x = –1

Textbook Page No. 240

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Sslc Maths Chapter 10 Kerala Syllabus Questions 1.
In each pair of polynomials given below, find the number to be subtracted from the first to get a polynomial for which the second is as factor. Find also the second factor of the polynomial got on subtracting the number.
i. x² – 3x + 5, x – 4
ii. x² – 3x + 5, x + 4
iii. x² + 5x – 7, x – 1
iv. x² – 4x – 3, x – 1
Answer:
p(x) = x² – 3x + 5
If x – 4 is a factor, p(4) = 0
p(4) = (4)² – 3 x 4 + 5 = 9
For x – 4 to become a factor of p (4) must be equal to zero.
For p (4) = 0 here we have to subtract 9 from p(x).
That is, add -9 to p(x) for (x – 4) become a factor.
∴ p(x) = x² – 3x + 5 – 9 = x² – 3x – 4
x² – 3x – 4 = (x – a) (x – b)
= x² – (a + b) x + ab
a + b = 3
ab = –4
(a – b)² = (a + b)² – 4 ab
= (3)² – 4x – 4 = 25
a – b = 5
a + b = 3
a = 4; b = –1
x² – 3x – 4 = (x – 4)(x + 1)
Second factor is (x +1)
ii. p(x) = x² – 3x + 5
If x + 4 is a factor, p(–4) = 0
p(–4) = (–4)² – 3x – 4 + 5 = 33
For x + 4 to become a factor of p (–4)
must be equal to zero.
For p (–4) = 0 here we have to subtract 33 from p(x).
That is, add –33 to p(x) for (x + 4) become a factor.
∴p(x) = x² – 3x + 5 – 33 = x² – 3x – 28
x² – 3x – 28 = (x – a) (x – b)
= x² – (a+b) x + ab
a + b = 3
ab = -28
(a – b)² = (a + b)² – 4ab
= (3)² – 4x – 28 = 121
a – b= 11
a + b = 3
a = 7;
b = -4 x² – 3x – 28 = (x – 7)(x + 4)
Second factor is (x – 7)
iii. p(x) = x² + 5x – 7
If x – 1 is a factor, p(1) = 0
p(1) = (1)² +5 x 1 – 7 = –1
For x – 1 to become a factor of p (1) must be equal to zero.
For p (1) = 0 here we have to subtract –1 from p(x).
That is, add 1 to p(x) for (x – 1) become a factor.
p(x) = x² + 5x – 7 + 1 = x² + 5x – 6
x² + 5x – 6 = (x – a) (x – b)
a + b = 5
ab = –6
a = –6; b= 1
x² + 5x – 6 = (x + 6)(x – 1)
Second factor is (x + 6)
iv. p(x) = x² – 4x – 3
If x – 1 is a factor, p(1) = 0
p(1) = (1)² – 4 x 1 – 3 = –6
For x – 1 to become a factor of p (1) must
be equal to zero.
For p (1) = 0 here we have to subtract -6 from p(x).
That is, add 6 to p(x) for (x – 1) become a factor.
∴ p(x) = x² – 4 x – 3 + 6 = x² – 4x +3
x² – 4x + 3 = (x – a) (x – b)
a + b = –4 ab = 3
a =1; b = 3
x² – 4x +3 = (x – 1)(x – 3)
Second factor is (x – 3)

Still, using a computation tool like a factor complex polynomials calculator, you can streamline things for you.

Polynomials Class 10 State Syllabus Questions 2.
In the polynomial x² + kx + 6, what number must be taken ask to get a polynomial for which x –1 is a factor? Find also the other factor of that polynomial.
Answer:
p (x) = x² + kx + 6
If (x – 1) is a factor of p(x)
then p(1) = 0
p(1) = 12 + k x 1 + 6 = 7 + k
7 + k = 0
k= –7
∴ p(x) = x² – 7x + 6
a + b = 7
ab = 6
a= 1, b = 6
factors are (x – 1 )(x – 6)
Second factor is (x – 6)

Polynomials Class 10 Kerala Syllabus Questions 3.
In the polynomial kx² + 2x – 5, what number must be taken ask to get a polynomial for which x –1 is a factor?
Answer:
p(x) = kx² + 2x – 5
If (x – 1) is a factor, then p(1) = 0
p(1) = k(l)² + 2 x 1 – 5
= k + 2 – 5 = k – 3
k – 3 = 0 k = 3

Textbook Page No. 242

Sslc Polynomials Questions And Answers Kerala Syllabus Question 1.
Write the second degree polynomials given below as die product of two first degree polynomials:
i. x² – 20x + 91
ii. x² – 20x + 51
iii. x² + 5x – 84
iv. 4x² – 16x +15
v. x² – x – 1
Answer:
p(x) = x² – 20x + 91
We must solve the equation p(x) = 0
x² – 20x + 91 = 0

Sslc Maths Polynomials Solutions Kerala Syllabus Question 2.
Prove that none of the polynomials below can be factored into a product of first degree polynomials:
i. x² + x + l
ii. x² – x + l
iii. x² + 2x + 2
iv. x² + 4x + 5
Answer:
i. x² + x+1
We must solve the equation p(x) = 0
x² + x +1 =0

no solutions
p(x) doesn’t have any first degree factors.
ii. x² – x + 1
We must solve the equation p(x) = 0
x² – x + 1 = 0

no solutions
p(x) doesn’t have any first degree factors
iii. x² + 2x + 2
We must solve the equation p(x) = 0
x² + 2x + 2 = 0

no solutions
p(x) doesn’t have any first degree factors.
iv. x² + 4x + 5
We must solve the equation p(x) = 0
x² + 4x + 5 = 0

no solutions
p(x) doesn’t have any first degree factors.

Sslc Maths Chapter 10 Solutions Kerala Syllabus Question 3.
In the polynomial p(x) = x² + 4x + k, up to what number can we take ask, so that p(x) can be factorized as a product of two first degree polynomials?
Answer:
p(x) = x² + 4x + k
p(x) = 0
x² + 4x + k = 0

p(x) can be factorized as a product of two first degree polynomials, the \(\sqrt{16-4 k}>0\) So, k can take value up to 4.

Polynomials Orukkam Questions & Answers

Worksheet 1

Polynomials Class 10 Hsslive Kerala Syllabus Question 1.
Write the product (x – 1) x (x + 1)
Find the product of (x – 1),(x + 1),(x + 2) If the poduct is p(x)find p(1), (-1), p(-2) Write the solution of the equation p(x) = 0.
Answer:
(x – 1)x (x + 1) = x² – 1
(x – 1) (x + 1) (x + 2)=(x² – 1) (x + 2)
= x³ – x + 2x² – 2 = x³ + 2x² – x – 2 = 0
p(x) = x³ + 2x² – x – 2 = 0
p(1) =1 + 2 – 1 –  2 = 0
p(–1) = – 1 + 2 + 1 – 2 = 0
p(–2) = – 8 + 8 + 2 – 2 = 0
1, – 1, – 2 are the solutions of the equation p(x) = 0.

Hss Live Guru 10th Maths Kerala Syllabus Question 2.
Expand (x – a)(x – b). If x² – 7x + 12 = (x – a)(x – b) then find a+b.
Also find ab Calculate the values of a, b. Write the factors of (x² – 7x +12 ).
Find the solutions of (x² – 7x + 12).
Answer:
(x – a) (x – b) = x² – bx – ax + ab
= x² – x (a + b) + ab
= x² – 7x + 12 = (x – a)(x – b)
a + b = 7
ab = 12
(a + b)² – 4ab = 7² – 4 x 12 = 49 – 48 = 1 = a – b
a + b = 7
a – b = 1, 2a = 8 a = \(\frac { 8 }{ 2 }\) =4
b = 7 – 4 = 3 a = 4 and b = 3.
Solution of x² – 7x + 12 is x² – 7x + 12 = (x – 4) (x – 3)
Solutions = 4, 3

Maths Questions And Answers For Class 10 Kerala Syllabus Question 3.
If p(x) = x3 – 6×2 + 11x – 1 then find p(1), p(2), p(3). Find p(x) – p(1), p(x) – p(2), p(x) – p(3), p(x) – p(1). Write the solutions of p(x) – p(1) = 0
Polynomials Class 10 Worksheet with Answer:
p(x) = x³ – 6x² + 11x – 1
p(1) = 1 – 6 + 11 – 1 = 5
p(2) = 8 – 24 + 22 – 1 = 5
p(3) = 27 – 54 + 33 – 1 = 5
p(x) – p(1) = x³ – 6x² + 11x – 6
p(x) – p(2) = x³ – 6x² + 11x – 6
p(x) – p(3) = x³ – 6x² + 11x – 6
p(x) – p(1) = x³ – 6x² + 11x – 6 = 0
If x = 1, 2, 3 then p(x) – p(1) = 0.
Factors of the equations are (x – 1), (x – 2), (x – 3).
Solutions of the equations are 1, 2, 3.

Hss Live Maths 10th Kerala Syllabus Question 4.
When p(x)is divided by (ax + b), the quotient is q(x)and the remainder is c. p(x) = (ax + b) x q(x) + c
When does the value of p(x) equal to c \(p\left(\frac{-b}{a}\right)=\left(a \times \frac{-b}{a}+b\right) \times q\left(\frac{-b}{a}\right)+c\)
What is the remainder when p(x)is divided by ax + b. When does(ax + b)becomethe factor of p(x).
Answer:
The remainder obtained when p(x) is divided by (ax + b) = \(p\left(\frac{-b}{a}\right)\)
When \(p\left(\frac{-b}{a}\right)\) = 0, then (ax + b) will be a factor of p(x).

Worksheet 2

Hsslive Maths 10th Kerala Syllabus Question 5.
Write the following as the product of first degree polynomials
1. x² + 7x + 12
2. x² + 3x + 2
3. x² – 9x – 22
4. 2x² + 5x – 3
Answer:
1. x² + 7x + 12 = (x + 4)(x + 3)
2. x² + 3x + 2 = (x + 2)(x + 1)
3. x² – 9x – 22 =(x – 11)(x + 2)
4.2x² + 5x – 3 =\(\left(x-\frac{1}{2}\right)\)(x+3)=(2x-l)(x+3)

Hss Live Maths 10 Kerala Syllabus Question 6.
Write a polynomial p(x) in which p(1) = 0, P(-2) = 0, p(2) = 0.
Answer:
If p( 1) = 0, then x – 1 is a factor.
If p(–2) = 0, then x + 2 is a factor.
If p(2) = 0, then x – 2 is a factor.
p(x) = (x – 1) (x + 2) (x – 2) = x³ – x² – 4x + 4

Class 10 Kerala Syllabus Maths Solutions Question 7.
Write a second degree polynomial p(x) in which \(p(\sqrt{2}+1)=p(\sqrt{2}-1)=0\)
Answer:

Hsslive 10th Maths Kerala Syllabus Question 8.
Prove that x² + 2x +2 cannot be written as the product of first degree polynomials
Answer:
b2 – 4ac = 22 – 4 x 1 x 2 = –4 < 0
∴ x² + 2x + 1 cannot be written as the product of first degree polynomials

Worksheet 3

Polynomials Class 10 In Malayalam Kerala Syllabus Question 9.
Find the remainder and quotient obtained by dividing x³ – 5x² + 7x + 3by (x + 2).
Answer:
Let quotient = x² + ax + b and remainder be c, then
x³ – 5x² + 7x + 3 =(x + 2)(x² + ax + b) + c +
= x³ – 5x² + 7x + 3 = x³ + ax² + bx + 2x² + 2ax + 2b + c
= x³ +( a + 2) x² + (b + 2a)x + 2b + c From this equation,
a + 2 = –5,
b + 2a = 7, 2b + c = 3
i.e., a = –7
b= 21
c = -39
quotient = x² – 7x + 21,
remainder = –39

Maths Polynomials Class 10 State Syllabus  Question 10.
Given x – 1 is a factor of x² + ax + b. Prove that (a + b = –1)
Answer:
Let (x – 1) be a factor, then p(1)=0
p(1) = 12 + a x 1 + b = 0
i.e., a + b = –1

Hss Guru Maths 10 Kerala Syllabus Question 11.
p(x) = (4x² – 1)(x + 2)Write p(x) as the product of first degree factors.Write p(x) in the form of a trird degree polynomial What is the remainder obtained by dhpding 4x³ + 6x² – x + 2by (x + 2) .What is the re¬mainder obtained by dividing 4x³ + 6x² – x + 1 by (2x – 1).
Answer:

Polynomials SCERT Questions & Answers

Question 12.
Write the second-degree polynomial p(x)= x² + x – 6 as the product of first-degree polynomials. Find also the solution of the equation p (x)=0 [Score: 4, Time: 7 minute]
Answer:

Question 13.
For what values of x, the polynomial 2x² – 7x – 15 is equal to zero? Write this polynomial as the product of two first degree polynomials. [Score: 4, Time: 7 minute]
Answer:

Question 14.
Write the polynomial p(x) = x² + 4x + 1 as the product of two first degree polynomials. Find the solution of the equation p (x) = 0. [Score: 4, Time: 5 minute]
Answer:
x² + 4x + 1 = (x – a) (x – b) = x² – (a + b) x + ab(1)
a + b = –4, ab = 1, a – b = 2√3
a = –2 + √3, b = –2 – √3 (1)
x² + 4x + 1 =(x + 2+ √3 ) (x + 2 – √3 ) (1)
x² + 4x + 1 = 0 =>(x + 2 + √3 )(x + 2 –√3 ) = 0 (1)
x = –2 – √3 ,or x = –2 + √3

Question 15.
In the polynomial p (x) = x²+ ax + b p (3 + √2 )= 0, p (3 – √2 ) = 0, write this polynomial after finding a and b. [ Score: 4, Time: 5 minute]
Answer:
p (x) = x² + ax + b
p(3 + √2) = 0, (x – 3 – √2) is a factor (1)
p(3 – √2 ) = 0, ( x – 3 + √2) is a factor (1)
p(x) = x² + ax + b = (x – 3 – √2) (x – 3 + √2)
= (x – 3)² – ( √2)² (1)
x² + ax + b = x² – 6x + 7 (1)

Question 16.
What number should be added to the polynomial p(x) = x² + x – 1, so that (x – 2)is a factor of the new polynomial. [Score: 4, Time: 6 minute]
Answer:
p (x) = x² + x – 1, remainder p(2) (1)
p (2) = (2)² + 2 – 1 = 5 (1)
For x – 2 to become a factor of p (2) must be equal to zero.
For p (2) = 0 here we have to substract 5 ffomp(x). (1)
That is, add –5 to p(x) for (x – 2) become a factor. (1)

Question 17.
What is the smallest natural number k, for which the polynomial 2x² + kx + 6 can be written as a product of two first degree polynomials? Write down the polynomial using k and express it as the product of two first degree polynomials [Score: 4, Time: 8 minutes]
Answer:

Question 18.
Method to check whether(x – a), and (x + a)are factors of a polynomial P(x).
Check whether (x + 2) and (x – 5) are factors of the polynomial p(x) = x² + 7x + 10 [Score: 4, Time: 6 minute]
Answer:
When a polynomial p(x) is divided by (x – a), if p(a) = 0 then (x – a) is a factor of p(x). When a polynomial p(x) is divided by (x + a), if p(–a) = 0 then (x + a) is a factor of p(x).
p(x) = x² + 7x + 10
p(–2) = 4 – 14 + 10 = 0 (1)
∴ x + 2 is a factor (1)
Remainder p(5) = (5)² + 7(5) +10 (1)
= 25 + 35 + 10 ≠ 0 (1)
∴ x – 5 is not a factor

Question 19.
When dividing x² + ax + b by (x – 2) and (x – 3) the remainder is zero. What are the numbers a and b. [Score: 3, Time: 5 minute]
Answer:
p(x) = x² + ax + b = (x – 3)(x – 2)
= x² – 5x + 6
a = –5, b = 6 (3)

Polynomials Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 20.
Check whether x – 1 is a factor of 3x³ – 2x² – 3x + 2.
Answer:
P(1)=3 x 13 – 2 x 12 – 3 x 1 + 2 = 0
Therefore x – 1 is a factor.

Question 21.
If (x – 1) is to be a factor of p(x) = a²x² – 4ax + 4a – 1. What should be the value of ‘a’ ?
Answer:
p(1) = 0
a2 – 4a + 4a – 1 = 0
a2 – 1 =0
a = +1 or –1

Question 22.

Answer:

Question 23.
Prove that (x – 1) is a factor of x¹³ – 1
Answer:
p(x) = x¹³ – 1
p(1)= 1¹³ – 1 = 1 – 1=0
p(1) = 0
(x – 1) is a factor of p(x)

Question 24.
Write the solution of polynomials p(x) = x² – 7x + 12.
Answer:
x² – 7x + 12 = (x – 4)(x – 3)
p(x) = 0 (x – 4) ( x – 3) = 0
x = 4, x = 3

Question 25.
The quotient is x2 – 5x + 6 when the polynomial p(x) is divided by (x – 1), and the remainder is 7. Then
a. p(x) = (………. ) ( ……… ) + 7. Complete it
b. Find P(2).
Answer:
a. p(x) = (x² – 5x + 6) (x – 1) + 7
b. p(2) = 4 – 10 + 6 + 7 = 7

Short Answer Type Questions (Score 3)

Question 26.
a. Find the remainder when
x³ – 4x³ + 12x – 45 is divided by (x – 2)?
b. Find the value of k if the remainder is zero on dividing 2x³ + 4x² – 10x + k by (x – 1)
Answer:
a. p(x) = x³ – 4x² – 12x – 45
remainder = p(2) = 23 – 4(2)² + 12 x 2 – 45 = 32 – 61 = –29
b. p(x) = 2x³ + 4x² – 10x + k; p(l) = 0
=> 2 x 13 + 4 x 12 – 10 x 1 + k = 0
2 + 4 – 10 + k = 0 – 4 + k = 0; k = 4

Question 27.
Factorise 3x² + 5x + 2 completely.
Answer:

Question 28.
Which first-degree polynomial is added to the polynomial 5x³ + 3x² to get x² – 1 as a factor.
Answer:
p(x) = 5x³ +3x² + ax + b
∴ x² – 1 isafoctorthen p(1), p(–1) will be zero.
p(1) = 5 x 1 + 3 x 1 + a x 1 + b = 0
= 5x – 1 + 3 x 1 + ax – 1 + b = 0
a + b = –8 (1)
= 5x – 1 + 3 x 1 + ax – 1 + b = 0
–a + b = 2 (2)
Find the solutions of the equation
b = –3, a = –5 added polynomial = –5x – 3

Long Answer Type Questions (Score 4)

Question 29.
a. Find the value of k if remainder when 5x³ + 4x – 11x + k is divided by (x – 1) is 0.
b. When x³ – 2x² + kx + 7 is divided by (x – 4) remainder is 11. Find k.
Answer:

Question 30.
a Show that the polynomial x + x + 1 has no first degree factors,
b. What is the remainder when the polynomial (x – 1) (x – 2) (x – 3) is divided by (x – 1)?
c. When (x – 1) (x – 2) (x – 3) + 2x + k is divided by (x – 1), the remainder is 10.
Then find out the remainder when it is divided by (x – 2).
Answer:

Long Answer Type Questions (Score 5)

Question 31.
Write 2x² + 5x + 3 as a product of two first degree polynomials.
Answer:

Question 32.
If(x + 1)and(x – 1) are factors of x3 + 2x² + px +q, find p and q.
Answer:
.

Polynomials Memory Map

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SSLC Maths Chapter 6 Coordinates Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 6 Solutions Malayalam Medium

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You can Download Refraction of Light Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 5 Refraction of Light Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 5 Refraction of Light Solutions

Textbook Page No. 103

Sslc Physics Chapter 5 Kerala Syllabus Question 1.
Do we observe the objects underwater at their original position?
Answer:
No.

Sslc Physics Chapter 5 Notes Kerala Syllabus Question 2.
Observe the path of light. Do you observe anything out of the ordinary? Can you de¬pict it in the science diary?
Answer:
The light rays below the water undergo deviation at the surface of water.

Textbook Page No. 104

Refraction Of Light Class 10 Kerala Syllabus Question 3.
Which are the media involved here?
Answer:
Air & Water.

Sslc Physics Refraction Of Light Kerala Syllabus Question 4.
What happens to the path of the light?
Answer:
Path of light undergoes deviation.

Sslc Physics Chapter 5 Refraction Of Light  Question 5.
Where does the deviation of the ray take place?
Answer:
At surface of water.

Physics Class 10 Chapter 5 Kerala Syllabus Question 6.
Why does the ray of light undergo a deviation here?
Answer:
The difference of speed of light rays in different media.

Physics Chapter 5 Class 10 Kerala Syllabus Question 7.
Does light pass through all the media at the same speed?
Answer:
No

Class 10 Physics Chapter 5 Kerala Syllabus Question 8.
As the optical density of a medium increases, the speed of light through it decreases.
What if the optical density decreases?
Answer:
Speed of light increases.

Class 10 Physics Chapter 5 Notes Kerala Syllabus Question 9.
Can the media given in the table be arranged in the increasing order of their optical densities?
Air<……….. <……… <……….
Answer:
Air < Water < Glass < Diamond

Textbook Page No. 105

10th Class Physics Chapter 5 Notes Kerala Syllabus Question 10.
Which is the incident ray on the surface of separation CD?
Answer:
QR

Refraction Of Light Class 10 State Board Kerala Syllabus Question 11.
The angle between the incident ray and the normal is called the angle of incidence. If so, can you explain what is angle of refraction??
Answer:|
∠r Between the refracted ray and the normal

Class 10 Physics Chapter 5 Kerala Syllabus Question 12.
Is the angle of refraction greater or lower than the angle of incidence when it goes from air to glass?
Answer:
Lower

Sslc Physics Chapter 5 Questions And Answers Kerala Syllabus Question 13.
What about from glass to air?
Answer:
Greater

Chapter 5 Physics Class 10 Kerala Syllabus Question 14.
Which is of greater optical density – air or glass?
Answer:
Glass

Physics Chapter 5 Class 10 Notes Kerala Syllabus Question 15.
While going from air to glass, the refracted ray deviates towards the normal/deviates away from the normal.
Answer:
Deviates towards the normal. While entering from air to glass (from a medium of lower optical density to that of a greater one) the refracted ray deviates away from the normal.

Textbook Page No. 106

Chapter 5 Physics Class 10 Notes Kerala Syllabus Question 16.
What happens while it goes from glass to air?
Answer:
Deviates away from the normal. Normal. While entering from glass to air (from a medium of greater optical density to that of a lower one) the refracted ray deviates away from the normal.

Physics Class 10 Refraction Of Light Solution Kerala Syllabus Question 17.
Are the angle of incidence, angle of refraction and the normal at the point of incidence on the same plane?
Answer:
Yes. The angle of incidence, angle of refraction and the normal at the point of incidence are in the same plane.

10th Class Physics Chapter 5 Kerala Syllabus Question 18.
Does refraction take place for a ray while entering a glass slab normal to it?
Answer:
No

Sslc Physics Chapter 5 Solutions Kerala Syllabus Question 19.
Examine using a ray of light from a laser torch. Ray diagrams of a light ray passing through different media are depicted. Find out the appropriate figures by observing these figures and also based on the concepts you have developed from the textbook.


Answer:

Textbook Page No. 107

Sslc Physics Chapter 5 Notes Pdf Kerala Syllabus Question 20.
Its an experiment, can you find out the path of light through a triangular prism using a laser torch? Record it in the science diary.

Answer:

10th Class Physics Refraction Of Light Kerala Syllabus Question 21.
What change occurs in the angle of refrac¬tion with the increase in the angle of inci¬dence when a ray enters a medium?
Answer:
Increases

Physics 10th Class Chapter 5 Kerala Syllabus Question 22.
Can you analyze the table and find out the change? What other conclusions can you arrive at from the table?
Answer:
When light passes through different pairs of media, the angle of refraction increases with the angle of incidence. The ratio of the sine of the angle of incidence to the sine of the angle of refraction \(\left(\frac{\sin i}{\sin r}\right)\) will be a constant. This constant is known as refractive index.

Textbook Page No. 108

Reflection Of Light Class 10 Kerala Syllabus  Question 23.
What specialty is observed in the value of ratio of sine of the angle of incidence to the sine of the angle of refraction, \(\left(\frac{\sin i}{\sin r}\right)\)?
Answer:
Will be a constant.

Textbook Page No. 109

Physics Class 10 Chapter 5 Notes Kerala Syllabus Question 24.
What will be refractive index n₁₂? Refractive index, n₁₂ =
Answer:

Textbook Page No. 110

Question 25.
Complete the table 5.6 (a)

Answer:

Question 26.
If the speed of light in water is 2.25 x 10⁸ m/s
a. Calculate the speed of light in vacuum
b. Calculate the speed of light in glass.
Answer:
a. Speed of light in vacuum = Speed of light in water × Refractive index

Question 27.
Place a pencil in an inclined position in a glass trough and fill three fourth of the trough with water. Depict in your science diary, the changes after adding water. What shall be the reason for the change?
Don’t you see that the position of the portion of the pencil underwater has changed? What may be the reason? Discuss.

Answer:
The light rays from pencil undergo deviation at the surface of water.

Textbook Page No. 111

Question 28.
Does the ray of light coming after reflection from the pencil undergo a deviation? What is the reason?
Answer:
Due to refraction

Question 29.
Now, will you be able to explain why a frog was obtained through the arrow was aimed at the fish?
Answer:
Light rays from fish and frog undergo deviation at surface of water. So they don’t see on exact place.

Question 30.
What is observed here?
Answer:
The coin seems to be lifted up.

Question 31.
What is the reason for this observation?
Answer:
As light from the coin comes from denser to rarer medium if bends away from normal so the position of the coin seems to be lifted up due to refraction.

Question 32.
Draw a thick line on a paper using a pen. Place a glass over it and observe as suggested below.
a Look from one side as shown in Fig. 5.8 (a)

b. Look from one side as shown in Fig. 5.8(b)

c. Look from one side as shown in Fig. 5.8 (c)

Answer:
a. The line seems to be rising on glass slab.
b. The line seems to be the surface of glass slab.
c. The line seems to be below the glass slab.

Textbook Page No. 112

Question 33.
Can you easily pick up the coin?
Answer:
We can’t easily pick up the coin.

Question 34.
What is the reason for the failure?
Answer:
Because coin’s actual place is not there. The light rays from the coin comes from denser to rarer medium it bends away from the normal, So the position of coin seems to be lifted up due to refraction.

Question 35.
Find out more examples of refraction from daily life.
Answer:
The stars in the sky appear glittering

• We can see the sun when it falls below the horizon.
• The bottom of water appears raised.
• We cannot locate the position of fish in the water

Question 36.
A person who looks at an aquarium as shown in Fig. 5.10 can see the base on the surface of water. What is the reason?
Answer:
The light rays from the bottom of the aquar¬ium reflect without undergoing refraction (Due to total internal reflection)

Textbook Page No. 113

Question 37.
What will be the angle of refraction when the refracted ray passes along the surface of water?
Answer:
Angle of refraction to becomes 90°.

Question 38.
Allow light to fall at an angle of incidence greater than the critical angle. What do you observe?
Answer:
The ray is reflected back to the same medium without undergoing refraction.

Textbook Page No. 114

Question 39.
Which are the figures that show total internal reflection?
Answer:

Question 40.
What is the critical angle of glass?
Answer:
42°

Question 41.
Will total internal reflection take place when light passing through water is inci¬dent on the surface of separation with air at an angle of incidence of 45°? Why?
Answer:
No. The critical angle of water 48.6°. You have already realized that total internal re-flection takes place only if the angle of incidence is greater than the critical angle. The total internal reflection takes place when in the incident ray is more than 48.6°

Question 42.
Find out the practical applications of total internal reflection in our day to day life.
1. Medical field – Endoscope
2. In the field of communication – Optical fiber cables.
Answer:
1. In the field of treatment- In the endoscope
2. In the field of telecommunication, optical fiber cables are used here.
3. Prism is used in the binoculars
4. Prism is used in the periscopes instead of mirrors.

Textbook Page 117

Question 43.
Complete the table 5.7

Answer:

Textbook Page 119

Question 44.
Where will the rays coining parallel to the principal axis converge?
Answer:
At principal focus

Question 45.
Where is the image formed?
Answer:
At principal focus

Question 46.
Write down the characteristics of the image.
Answer:
Position of the image: Between F and 2F
Nature of the image: Real, Inverted
Size of the image: Diminished

Textbook Page 120

Question 47.
Object at 2F

Answer:
Position of the image: At 2F on the other side Characteristic of the image: Inverted Size of the image: Same size of the object.

Question 48.
Object between F and 2F

Answer:
Position of the image: Beyond 2F
Characteristic of the image: Inverted
Size of the image: Bigger than the object

Question 49.
Object at F

Answer:
Position of the image: At the infinity
Characteristic of the image: Inverted
Size of the image: Big

Textbook Page 121

Question 50.
Object between F and Lens

Answer:
Position of the image: On the same side of the object
Characteristic of the image: direct image
Size of the image: Big in size

Question 51.
Have you observed images formed by a concave lens? What is the nature of the image?

Answer:
Images is smaller, virtual and direct

Textbook Page 122

Question 52.
Record the measurement shown in the figure as per the Cartesian System.

a. Distance of the object from the lens (u) = ………….
b. Distance of the image from the lens (v) = ………….
c. Height of object (OB) = …………..
d. Height of image (IM) = …………..
Answer:
a. u = -25cm
b. v = + 100cm
c. OB = +1cm
d. IM = -4 cm

Question 53.
Complete the Table 5.8

Answer:

Average f = 19.99 cm

Textbook Page 123

Question 54.
Is there any relation between the height of an object and the height of its image?
Can it be related to the ratio between the distance to the object and that to the image?
Answer:
The height of the object height of the image depends on the ratio of the distance to the object and distance to the image. When the object is kept at different positions, the height of the image can be varied.

Question 55.
When an object is placed at different positions in front of a lens, isn’t there a change in the height of the image?
Answer:
Yes

Question 56.
Calculate the magnification of the image formed by convex lens in Fig. 5.33
Answer:
Height of the image = (h_i) = -4 cm
Height of the object = (h₀) = +1 cm
Magnification = \(\frac{h_{1}}{h_{0}}=\frac{-4}{+1}=-4\)
or in another way
m = \(\frac { v }{ 4 }\)
v = +100, u = -25
m = \(\frac { 100 }{ -25 }\) = -4

Textbook Page 125

Question 57.
Find out the uses of lenses in our day to day life and record them in the science dairy.
Answer:

• In telescope
• In eye lenses
• In-camera
• In the microscope
• In the magnifying a lenses

Question 58.
What has the doctor indicated in the prescription?
Answer:
The doctor indicated about the type and power of lens.

Question 59.
Calculate the power of a lens of focal length + 25 cm.
Answer:

Question 60.
You can guess what the +2D in the prescription stands for. What is its focal length? What kind of lens is it?
Answer:
p = + 2D
It is a convex lense
f = \(\frac { 1 }{ p }\) \(\frac { 1 }{ 2 }\) = 50cm

Question 61.
You might have seen the twinkling of stars at night. But planets do not twinkle. Why?
Answer:
Light coming from distant stars passes through different layers of air. Each layer differs from the other in their optical densities. Hence light undergoes successive refraction. Since stars, at a greater distance, they appear like a point source. The rays of light appear to come from different points on reaching the eye after refraction. This is the reason for the twinkling of stars.

Let Us Assess

Question 1.
Refractive indices of different materials are given. Find out the medium through which light passes with maximum speed.
Answer:
Water

Question 2.
The nature of images formed by two lenses are given.
i. An erect and magnified virtual image.
ii. An erect and diminished virtual image
a. What type of lens is used in each case?
b. By using which type of lens will we get an image having the same size as the object? What is the position of the object?
Answer:
a. i. Convex lense
ii. Concave lense
b. Convex lense. Object will be at 2F.

Question 3.

a. MN represents a lens. What type of lens is this?
b. What are the characteristics of the image?
c. Copy the ray diagrams in the science diary and complete it
Answer:
a. Convex
b. Magnified, real, inverted

Question 4.
What do you mean by power of a lens? What is the SI unit of the power of a lens? Calculate the power of a concave lens of focal length 25 cm.
Answer:
Power is a term related to the focal length of a lens. Power of a lens is the reciprocal of focal length expressed in meters. Power
p = 1/f
Unit of power is dioptre. It is represented by

Question 5.
Observe the figure. Light falling on two different media are shown.

a. Which medium has greater optical density? Why?
b. Which medium has greater refractive index?
Answer:
a. Medium 1. Angle of incidence is same, but angle of refraction is different. When a ray of light passes from a medium of lower density to greater density the refracted ray deviates towards the normal. Medium 1 has lowest angle of refraction compared to medium 2.
b. Medium 1

Question 6.
An object of height 3 cm is placed in front of a convex lens of focal length 20 cm at a distance of 30 cm.
a. What is the distance to the image formed?
b. What is the nature of the image?
c. What is the height of the image?
Answer:

b. Magnified, inverted, real

Question 7.
In the table, the absolute refractive indices of certain transparent media are given.

a. Find out from the table the medium of highest and lowest optical densities,
b. If the speed of light in air is 3 x 108 m/s, what will be the speed of light through kerosene?
c. W ill a ray of light deviates towards the normal or away from the normal when it enters from air to diamond obliquely?
d. The refractive index of diamond is 2.42. What do you mean by this? Calculate the speed of light through diamond.
Answer:
a. Highest – Diamond
Lowest – air

c. Towards the normal
d. The speed of light in air (Vaccum) is 2.42 times greater than the speed of light in diamond or other hands. Speed of light in diamond is 2.42 times lesser than speed of light in air or vacuum.
Speed of light through diamond
\(=\frac{3 \times 10^{8}}{2.42}=1.25 \times 10^{8} \mathrm{m} / \mathrm{s}\)

Extended Activities

Question 1.
Half portion of a convex lens is wrapped with a black paper. Can this lens give a complete real image of an object? Explain.

Answer:
Yes. This lens gives a complete real image of an object. However, the intensity of image may be less.

Question 2.
The refractive indices of different media are given. Analyze the table and answer the following.
In which medium will light have the highest speed?
Answer:
Water

Question 3.
Glycerine, water and sunflower oil are taken in two beakers. A glass rod is dipped in one and a pyrex glass rod is dipped in the other. Do they appear in the same way? In which media are they visible? Justify.

Answer:
Objects are visible because they reflect some of light that shines on them. Refractive index of glycerine, sunflower oil, and pyrex glass are same. Glass rod clearly visible in them. But pyrex glass is not clearly visible because light ray are passed without deviation through medium having same refractive index.

Question 4.

Take a clean mineral water bottle and fill it with water. Put a hole on one side of the bottle. Allow water to flow out while passing a laser ray through it What do you observe? Why?
Answer:
As water throws out of the opening observe the case beam remains trapped in the water stream because of total inemal reflection. As light tries to pass from the more dense water to the less dense air. it bends. In narrow column of water, the light wave continues bouncing off the boundaries of the stream of water but cannot pass through into the air.

Refraction of Light Exam Oriented Questions & Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The power of a convex lens of focal length 2 m is …………..
Answer:
\(P=\frac{1}{f}=\frac{1}{2}=0.5 \text { dioptre }\)

Question 2.
What change occurs for power of a lens when it’s focal length increases?
Answer:
Power decreases

Question 3.
An object is placed at 2F in the case of a convex lens what is its magnification? (greater than 1, less than 1, zero, one)
Answer:
one

Question 4.
Figure shows a ray of light obliquely travels from air to glass. Find out the correct one.

Answer:
Figure iii
It bent towards the normal

Question 5.
Find out the relationship in the first pair and fill in the second pair appropriately.
Focal length – meter
Power of the lens – …………..
Answer:
Dioptre

Very Short Answer Type Questions (Score 2)

Question 6.
The image is direct and is above principal axis an object is kept at a distance of 30 cm away from a lens. An image of same size is formed. What is the focal length of the lens?
Answer:
Image of size is formed when the object is kept at 2F
∴ 2f = 30 cm
∴ f = 15 cm
∴ \(\frac { 30 }{ 2 }\) = 15cm

Question 7.
Write two situations when angle of refraction is 90°
Answer:
i. The angle of incidence should be equal to critical angle.
ii. Rays of light should travel from the medium of greater optical density to lesser optical density

Question 8.
Represent the image formation by a concave lens. Why the focus of this lens is called virtual?

Answer:
The rays of light pass through a concave lens do not really meet. They appear to meet. So the focus is called virtual.

Question 9.
What are the situation where total internal reflection take place?
Answer:
i. Rays of light travel from medium of greater optical density to that of lesser one.
ii. Angle of incidence should be greater than critical angle.

Question 10.
An object ‘OB’ is placed in front of lens is given below.

a. Name the lens?
b. Complete the figure, then find out the position of the image.
c. Identify the two nature of the image.
Answer:
a. Convex lens

Images formed in between F and 2F.
b. Real and inverted images formed.

Question 11.
Complete the figure then find out the principal focus of the concave lens.

Answer:

Question 12.
Terms related to lens are given below? Complete the questions using these terms?
(Pole, Focal length, Centre of curvature, Principal axis)
a. The midpoint of a lens is called …………..
b. The distance between pole to focus is called ……………
c. The center of imaginary spear of two sides of the lens is called ……………
d. Imaginary line passing through the center of curvature of the lens is called …………..
Answer:
a. Pole
b. Focal length
c. Centre of curvature
d. Principal axis

Question 13.
Corrected the wrong statement from the following.
a. Optical density of two mediums are different, it is reason for refraction.
b. Velocity of light is very high in optically denser medium.
c. Optical density of glasses lower than that of water
d. Velocity of light in vacuum is 3 × 108 m/s
Answer:
Wrong statements – b, c
b. Velocity of light is comparatively low in optically denser medium.
c. Optical density of glasses higher than that of Water.

Question 14.
Calculate the power of the lens of focal length 10 cm?
Answer:
f = 10 cm = 0.1 m
\(P=\frac{1}{f}=\frac{1}{0.1}=+10 \text { dioptre }\)

Question 15.
‘IM’ is an image of the object ‘OB’. Find out the given values related to new cartesian sign conventions.

a. Distance between lens to object (u) ……………..
b. The distance between lens to image (v) …………
c. Height of the object (OB) ………….
d. Height of the image (IM) …………
Answer:
a. u = -40 cm
b. v = +24 cm
c. OB = +2 cm
d. IM = -1cm

Question 16.
The statements associated with a lens are given by the new cartesian symbol. Write the correct one?
a. All distance are measured from the focus.
b. All distances measured along the direction of incident light is positive.
c. Lightray is assumed to travel from right to left.
d. X-axis is taken as the origin.
Answer:
b. All distances measured along the direction of incident light is positive
d. X-axis is taken as the origin

Short Answer Type Questions (Score 3)

Question 17.
Observe the figure. Light rays enter from medium 1 to medium 2

a. Which medium has greater optical density?. How do you reach in such a conclusion?
b. How do optical density and velocity of light related?
c. What is the speed of light in vacuum?
Answer:
a. medium 1
When the light rays enter from medium 1 to 2, they more away from the normal. The speed of light is also greater in medium 2 then medium.
b. When optical density increases the speed of light decreases.
c. 3 × 10⁸ m/s

Question 18.
A virtual image of an object is formed at 30 cm away from a lens when an object is kept at a distance 20 cm away from it. What is the focal length of the lens? Find the magnification of the image?
Answer:
u = -20 cm, v = -30 cm
(as the image is virtual, it is formed at the same side)

Question 19.
It is written by the doctor for the eye lens as -2.50.
a. What the defect of eye of the patient?
b. What is the focal length of the lens?
c. What type of lens is it?
Answer:
a. near sightedness
b. p = -2.5 D
\(\mathrm{f}=\frac{1}{\mathrm{p}}=\frac{1}{2.5}=\frac{-10}{25}\)
= -0.4 m = -40 cm
The focal length of the lens = – 40 cm
c. Concave lens

20. Observe the figure given below.

a. Why the ray PQ reflected back as in the above figure?
b. Name this phenomenon?
c. What happens to the ray of light when angle of incident is 30°.
Answer:
a. Angle of incidence is greater than that of critical angle.
b. Total internal reflection
c. Refraction takes place when it travels from water to air.

Question 21.
Nature of some images are given below. Find out the real and virtual images and then write it in separate column.
a. Inverted image
b. Do not obtained on a screen
c. Obtained on a screen
d. If the light rays really intersect to each other.
e. Erect image
f. The image distance cannot directly
Answer:
Real images: a, c, d
Virtual images: b, e, f

Question 22.
A burning candle is placed in front of a convex lens to obtain an image on the screen.
Find out the positions of the object in each of the conditions given below?
a. We get the size of the image is equal to the size of the object.
b. The size of the image is smaller than that of the object.
c. We get large real image than the object.
Answer:
a. 2F or C
b. Beyond 2F
c. In between 2F and F.

Question 23.
A person uses a spectacle of power of lens -1.25D.
a. What type of lens is this?
b. What is meant by power of a lens?
c. Find the focal length of the lens?
Answer:
a. Power of the lens is negative. So it is a concave lens.
b. Power of a lens is the reciprocal of its focal length.

Question 24.
Observe the figure carefully OB is an object placed in front of the convex lens.

a. Complete the ray diagram then find out the position of the image.
b. An object is placed at 2F, in which position image is formed?
c. Where we have to place an object to get a virtual image from a convex lens?
Answer:

a. Beyond 2F
b. Image is formed at 2F
c. An object is placed in between optic center and focus

Short Answer Type Questions (Score 4)

Question 25.
Combine suitable

Answer:
1 – (c)
2- (a)
3 – (e)
4 – (d)

Question 26.
Light obliquely travels from glass to air, critical angle of Glass is 42°.
a. What is mean by critical Angle?
b. Angle of incidence is 42° find its refracted angle?
c. In which phenomenon we observe the incident angle is 40°? Explain the phenomenon?
d. What happens to the ray of light when the angle of incidence is 45°? Explain the phenomenon?
Answer:
a. When a Ray of light passes from a medium of greater optical density to that of lower optical density, the angle of incidence at which the angle of refraction becomes 90° is the critical angle.
b. 90°
c. Reflection takes place when a Ray of light entering obliquely from one transfer in the medium to another its path undergoes a deviation at the surface of separation. This is called refraction of light.
d. Total internal reflection. When a Ray of light passes from a medium of higher optical density to a medium of lower optical density at an angle of incidence greater than the critical angle, the ray reflected back to the same medium without undergoing refraction this phenomenon is known as total internal reflection.

Students can Download Physics Chapter 6 Vision and the World of Colours Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 6 Vision and the World of Colours Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 6 Solutions Malayalam Medium

You can Download Vision and the World of Colours Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 6 Vision and the World of Colours Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 6 Vision and the World of Colours Solutions

Text Book Page No. 133

Sslc Physics Chapter 6 Kerala Syllabus Question 1.
Place a lighted candle at a distance 13 cm from the lens. Do you get a clear image of the candle on the screen?
Answer:
No

Sslc Physics Chapter 6 Notes Kerala Syllabus Question 2.
Place lenses of focal length 10 cm, 15cm and 20cm on the lens holder without changing the distance between the lens and the screen. On using which lens is the image clear?
Answer:
Lens of focal length 10 cm

Text Book Page No. 134

Vision And The World Of Colours Kerala Syllabus Question 3.
Given below are the ray diagrams of image formation in the eye.
i. Where is the image formed in each case?
ii. In which case is the image formed on the retina itself ?

Answer:
(i) a) In front of retina
b) Behind the retina
c) On retina
ii.

World of Vision Question 4. Will an image be formed on the retina when the object is at the near point

Will it be possible to see a clear image?
Answer:
No. It will not be possible to see a clear image.

Textbook Page No. 135

Physics Chapter 6 Class 10 Kerala Syllabus Question 5.
Will an image be formed on the retina when the object is at a far distance

Will a clear image be formed?
Answer:
Yes. A clear image will be formed.

Sslc Physics Chapter Wise Questions And Answers Question 6.
What shall be the reasons behind this defect?
1. Can you find out a reason based on the size of the eyeball?
Size larger/ smaller?
Answer:
Size larger

2. What if it is related to the focal length (or power)?
Power is high/ low
Answer:
Power low

Physics Class 10 Chapter 6 Kerala Syllabus Question 7.
What is the remedy for long – sight-edness?
Answer:
This can be rectified by using a convex lens of suitable power.

Text Book Page No. 136

Class 10 Physics Chapter 6 Kerala Syllabus Question 8.
Where is the image formed in these cases? Write down the answer analysing.

1. When the object at O is away from the eye, where will be the image formed? Can the object be seen clearly?
Answer:
In front of retina. Objects cann’t be seen clearly.

2. Can the object be seen clearly when it is at O₁?
Answer:
Yes

3. Why is it not possible to see objects placed at long distances?
Answer:
dmages are formed in front of retina.

4. What is its remedy?
Answer:
This can be overcome by using concave lens of suitable power.

Physics Chapter 6 Question 9. When a person suffering from problem in vision met a doctor, he wrote in his prescription the following figures. +1.5 D, -2D

1. What has the doctor indicated in the prescription?
Answer:
+1.5 D, -2D indicates the power of lense they required. The power of a lens is defined as the ability of the lens to converge or diverge a beam of light falling on it.

2. Which are the types of lenses prescribed here?
Answer:
The power of a convex lens is positive while the power of a concave lens is neg-ative.
+1.5D represents convex lense.
-2D represents concave lense.

Sslc Physics Chapter 6 Notes Pdf Kerala Syllabus Question 10.
For a healthy vision, what is the dis-tance to the near point?
Answer:
25cm

Text Book Page No. 137

Kerala Syllabus 10 Physics Chapter 6 Question 11.
Pass sunlight through a prism and allow it to fall on a screen. What are the colours seen on the screen?

Answer:
Violet, Indigo, Blue, Green, Yellow, Orange, Red.

Text Book Page No. 138

Hss Live Guru 10th Physics Kerala Syllabus Question 12.
The beam of white light from a torch is allowed to fall obliquely on the prism.

1. Which are the colours formed on the screen?
Answer:
Violet, Indigo, Blue, Green, Yellow, Orange, Red.

2. Aren’t these colours the same as the component colours obtained from the sunlight?
Answer:
Yes

Hsslive Guru 10th Physics Kerala Syllabus Question 13.

1. Which colour deviates the most due to dispersion?
Answer:
Violet

2. Which colour deviates at least?
Answer:
Red.

3. What may be the reason behind this difference in deviation?
Answer:
Differences in wavelength

Physics 10th Class Notes Kerala Syllabus Question 14.

1. Which colour has the shortest wavelength?
Answer:
Violet.

2. Which one has the longest?
Answer:
Red

Question 3.
When light passes through the prism, as the wavelength increases, how does the deviation change? Will it increase or decrease?
Answer:
The deviation of colours is minimum for the colours with more wavelength when the composite light passes through the prism. When the wavelength of the colour decreases, the deviation increases.

Text Book Page No. 139

Question 15.
When is the rainbow formed?
Answer:
In the morning and in the evening.

Question 16.
Where will be the sun when the rainbow is seen in the east?
Answer:
In the opposite direction (west)

Question 17.
Where will be the sun when the rainbow is seen in the west?
Answer:
East

Question 18.
How many times does a ray of light undergo refraction when it passes through a water droplet?
Answer:
The sunlight undergoes two times refraction in the water drop.

Question 19.
What about internal reflection?
Answer:
One time

Question 20.
What is the colour seen at the upper edge of the rainbow?
Answer:
Red

Question 21.
What is the colour seen at the lower edge?
Answer:
Violet

Text Book Page No. 140

Question 22.
What happened to the light when it passed through the first prism?
Answer:
The white light separates into its component colours when it passes through the first prism

Question 23.
What happened when it passed through the second one?
Answer:
The colours formed by the first prism recombine to form white light when it passes through the second prism.

Text Book Page No. 141

Make a Newton’s disc by painting the constituent colours of white col-our of white light in the same order and proportion.

Question 24.
In which colour does the disc appear when rotated fast?
Answer:
Very near to the white colour

Question 25.
Give Reason.
Answer:
Light from the seven colours of the colour disc falls continuously on the retina and due to the persistence of vision, disc appears as white.

Question 26.
The disc appeared white due to persistence of vision. Find out more examples of persistence of vision and write them down.
Answer:

• A torch rotated rapidly appears as an illuminated circle.
• Sparkler’s trail effect.
• Colour – top
• Thaumatrope.
• Kaleidoscopic colour-top.
• Rubber pencil trick.
• LED POV displays
• Revolving wheels,
• Rotation of fan leafs

Question 27.
During sunset, you might have no- I ticed that the western horizon becomes reddish. Why is it so?
Answer:
During sunset, the sunlight travels I maximum distance through the atmosphere to reach in our eyes. So red also I undergo scattering. So the horizon appears red.

Text Book Page No. 142

Question 28.
Is the reflection of light here regular or irregular?
Answer:
Irregular

Question 29.
Is the distribution of sunlight to all regions made possible by this type of scattering. Discuss.
Answer:
Yes, scattering is the partial and irregular reflection of light. The colours in sunlight undergo scattering when they fall on the particles in the atmosphere. Sunlight reaches in the rooms and under the trees by this method.

Text Book Page No. 143

Question 30.
When hydrochloric acid was added to the solution, which colour did spread first in the solution?
Answer:
Blue colour spreads first in the solution when acid was added.

Question 31.
Write down the colour changes observed on the screen in the order it occurred.
Answer:
Violet → indigo
blue → green → yellow
orange → red

Question 32.
What was the final colour observed on the screen?
Answer:
Red colour is seen at the end.

Question 33.
Which component colour in white light does undergo maximum scattering?
Answer:
Blue has the maximum scattering

Question 34.
During sunset, the horizon appears to have red colour. What may be the reason?
Answer:
During sunrise and sunset, light reaching us from the horizon has to travel long distances through the atmosphere. During this long journey, colours of shorter wavelength would be almost fully lost due to scattering. Then, the red light which undergoes only less amount of scattering decides the colour of the horizon. That is why the sun appears red during sunset and sunrise.

Text Book Page No. 144

Question 35.
Which are the occasions when sun-light has to travel greater distance through the atmosphere before reaching the eyes of ait observer on the earth?
Answer:
Morning and evening.

Question 36.
As sunlight passes through the atmosphere, which colour in it undergoes maximum scattering? Which colour undergoes minimum scattering?
Answer:
Colour in it undergoes maximum scattering: Violet
Colour in it undergoes minimum scattering: Red

Question 37.
When light reaches the observer after travelling long distances through the atmosphere, which colour reaches the eye? What is the reason?
Answer:
Red, it has highest wavelength and least scattering.

Question 38.
The western horizon remains reddish for some more time even after sunset. Why?
Answer:
During sunset, the sunlight travels maximum distance through the atmosphere to reach in our eyes. So red also undergoes scattering. So the horizon appears red.

Question 39.
Can you now guess why red colour has been given to the tail lamps of vehicles and signal lights? Discuss and note it down in your science diary.
Answer:
The primary reason why the colour red is used for danger signals is that red light is scattered the least by air molecules. So red light is able to travel the longest distance through fog, rain, and the like.

Text Book Pace No. 145

Question 40.
Write down what else can be ‘done to minimize the light pollution.
Answer:

• Start with the light switch
• Check with your power company to see if you’re paying for outdoor lighting.
• Consider replacing outdoor lights with intelligently designed, low- glare fixtures.
• Place motion sensors on essential outdoor lamps.
• Replace conventional high-energy bulbs with efficient outdoor CFLs and LED floodlights.
• Reduce the use decorative lighting.
• Use of covered bulbs that light facing downwards.
• The use of automatic systems to turn off street light at certain times.
• Have all information arid facts about light pollution.
• Use of glare-free bulbs, installing low hanging bulbs, having the lights facing downwards, and covering the bulbs to reduce bright skies at night.
• Share with family and friends.

Let Us Assess

Question 1.
How is the condition of the ciliary muscles while watching a distant object? How does this influence the focal length of the eye lens?
Answer:
While looking at far objects the ciliary muscles are relaxed and the curvature of the lens decreases. The focal length of the lens increases.

Question 2.
A child sitting at the backbench of a classroom is unable to see the letters on the board clearly. What is the defect of the eye of the child? How can it be remedied? Draw its ray diagram.
Answer:
Myopia or Nearsightedness. This can be rectified by using a concave lens of suitable power.

Question 3.
A person is not able to see objects beyond 1.3 m. What remedy can you suggest for this defect?
Answer:
Myopia or Nearsightedness. This can be rectified by using a concave lens of suitable power.

Question 4.
In what colour does the sky appear for an astronaut?
Answer:
Black

Question 5.
Red light is used as signal lamps to indicate danger. Explain.
Answer:
The primary reason why the colour red is used for danger signals is that red light is scattered the least by air molecules. So red light is able to travel the longest distance through fog, rain, and the a like.

Question 6.
What is the reason for using yellow light as fog lamps?
Answer:
Yellow lights are strictly used for fog situations, and for construction, so you can tell what is what on the road. Yellow has high wavelength, so yellow light is scattered the least by air molecules.

Question 7.
Which is the phenomenon behind dispersion of light?
a. Reflection
b. Refraction
c. Tyndal Effect
d. Scattering
Answer:
b. Refraction

Question 8.
During dispersion, different colours deviate differently. Explain why.
Answer:
The rate of refraction for different col-ours during dispersion is different due to the difference in wavelength of the colours. The rate of refraction decreases with increase in wavelength. Dispersion. is the phenomenon of splitting of light into its component colours.

Question 9.
The telescope called ‘Chandra X-ray Observatory’ is placed in the outer space. What is the advantage of placing it there? Explain with reference to the scattering of light in the atmosphere.
Answer:
Light can reach the telescope which is kept in space without scattering as there are no gases and tiny particles in space. So we can observe the heavenly bodies clearly there. The pictures may not be clear on earth due to the scattering of light.

Extended Activities

Question 1.
White light is allowed to fall on the bright side of a compact disc (CD). The reflected light is allowed to fall on a white wall. Observe the colours available in the spectrum and write them down in your science diary.
Answer:
Violet, indigo, blue, green, yellow, Orange, Red.

Vision and the World of Colours Orukkam Questions and Answers

Question 1.
Observe the figure

a. Which are the colours formed on the sc-reen?
b. Explain the phenomenon that causes the formation of array of colours
c. Which are the colours denoted by ‘A’ and ‘B’?
d. Which colour deviates the most?
e. Which colour deviates at least?
f. Different colours undergo different deviation. Why?
Answer:
a. Violet, Indigo, Blue, Green, Yellow, Orange, Red
b. Dispersion. The phenomenon of splitting, of a composite light into its constituent colours is known as dispersion of light.
c. A – Violet B – Red
d. Violet
e. Red
f. Different colours undergo different deviation due to difference in wavelength

Question 2.
Newton’s colour disc is made by painting the constituent colours of white light in the same order and proportion.
a. In which colour does the disc appear when rotated fast?
b. The reason behind such an appearance is a phenomenon related to our eyes. What is it explain?
c. If the disc is rotated slowly, what will be the observation?
d. Why does a torch rotated appear as an illuminated circle?
Answer:
a. White colour
b. Persistence of vision. When a person views an object, its image remains in the retina of the eye for a time interval of 0.0625 s
c. If the disc is rotated slowly it is easy to identify the different colours.
d. Due to the Persistence of vision.

Evaluation Activities

Question 3.
The figure shows a ray of sunlight falls obliquely on a water drop.

a. Complete the figure
b. How many times does the ray of light undergo refraction?
c. A natural phenomenon is there that connected with this figure. Name it?
Answer:

b. Twice
c. Rainbow

Question 4.
Match columns. A, B and C suitably.

Answer:
a.Magenta
b. White light
c. Cyan

Question 5.
The telescope ‘Chandra’ is placed in the space. What is the advantage of placing it there?
Answer:
Light can reach the telescope which is kept in space without scattering as there are no gases and tiny particles in space. So we can observe the heavenly bodies clearly there. The pictures may not be clear on. earth due to the scattering of light.

Vision and the World of Colours SCERT Questions and Answers

Question 6.
The figure shows a ray of light falling obliquely on a droplet of water for the formation of rainbow.

a. Copy the figure and complete the unfinished part.
b. What is the phenomenon that took place inside the droplet?
c. What is the colour of ‘X’?
d. What is the reason behind the red app-earing at the outer edge of a rainbow?
Answer:

b. Internal reflection
c. Violet
d. Sunlight undergoes refraction and internal reflection while passing through drops of water. All the drops that appear in the same colour are seen in the same arc. Thus red having highter wavelength is seen in the outer edge at higher angle.

Question 7.
Analyse the following statements and find out the reason behind them.
a. Stars can be seen even in day time while viewed from the moon
b. Raindrops falling down during rain appear like a glass rod.
c. A rainbow is seen in the shape of a circle when viewed from a height.
d. The sky of cities mostly appears in grey colour.
Answer:
a. There is no scattering for the light around the moon since there is no atmosphere around it. Hence the sky of moon appears dark.
b. Raindrops come down faster during rain. The distance travelled by a drop in 1/16 of a second appears like a glass rod due to persistence of vision.
c. When viewed from a height the observer can see points at 42.70 upwards, downwards as well as sideways. Hence rainbow is seen as a circle.
d. In cities there will be large particles in the atmosphere. Hence all colours of light scatter equally.

Question 8.
Two teams A and Bare taking part in a volley ball competition held on a ground illuminated by a sodium vapour lamp (yellow light). Team A is wearing white jersey and blue shorts while Team B is wearing yellow jersey and black shorts.
a. Can you distinguish between the teams based on the colour of their dress?
b. Justify your answer.
Answer:
a. Cannot be distinguished
b. In yellow light the white jersey and yell-ow jersey appear in yellow colour. This is because both white and yellow surfaces can reflect yellow light. The black shorts appears black itself and blue shorts appears dark since both of them absorb yellow light.

Question 9.
A disc painted with different colours is shown

a. Which colour is X if the disc appears white on rotating fast?
b. Why did the disc appear white on being rotated fast?
c. In what colour will the disc appear if green light alone is made to fall on the disc when it is being rotated fast?
Answer:
a. Cyan
b. Due to persistence of vision
c. In green colour
On rotating the disc faster, the disc app-ears white due to persistence of vision. The cyan appears as green since it reflects green light.

Question 10.
Colour filters are materials that will allow only certain colours of light to pass.
a. What is the colour of the filter that can transmit blue and red colours of light?
b. Which are the radiations that will be absorbed completely by an infrared filter?
c. Write down:
(i) Two specialities of infrared rays
(ii) Two uses of infrared rays
Answer:
a. Magenta
b. Visible light, ultra violet radiations
c. (i) Higher wavelength than that of visible light; becomes the reason for the heat . of the sunlight.
(ii) Used to take photograph of distant objects, remote sensing, secret signalling, for controlling robots.

Question 11.
The figure shows sunlight falling on a white screen after being passed through yellow and cyan filters.

a. Supply the omitted portion and complete the diagram
b. Which colours is‘X’?
c. (i) What is the complementary colour of this colour?
(ii) What do you mean by complementary colours?
Answer:
a.

b. Green
c. (i) Magenta
(ii) If a primary colour and a secondary colour can given white light on mixi-ng, then they are complementary colours.

Question 12.
Match the columns A, B and C suitably

Answer:

Question 13.
A spectrum is obtained by passing white light from a torch through a glass prism.
a. Write down the steps of this experiment.
b. Draw a figure that represents the experiment.
Answer:
a. Fix a black paper in from of a torch. Put a small hole in the middle of the paper. Arrange a screen on the other side. Make the beam of light fall obliquely on the prism.

Question 14.
The telescope ‘Chandra’ is installed on the outer space. Identify the correct statements related to it.
a. There is no scattering of light in the outer space
b. There is a greater scattering in the outer space
c. The vision is more accurate and clear
d. Presence of dust in the outer space helps in better vision and clarity
Answer:
Correct statements: a and c

Question 15.
Which of the following does not belong to the group?
a. red, yellow, blue, green
b. visible light, sound, X-rays, radio waves
Answer:
a. Yellow – it is a secondary colour
The rest are primary colours
b. Sound- it is a mechanical wave The rest are electromagnetic waves

Question 16.
Find out the relation between the first pair and complete the other accordingly:
a. Black: absorbs all colours of light white:
Answer:
a. Reflects all colours of light.

Vision and the World of Colours Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 17.
Find the odd one in the group and write the reason. [Red, Green, Blue, Yellow]
Answer:
Yellow. Others are primary colours

Question 18.
Using the relation from the first pair, complete the other.
Tyndal effect – scattering
Dispersion – ………………
Answer:
Refraction

Question 19.
Which of the following causes skin cancer,
a. Ultraviolet
b. Infrared
c. Radio
d. Microwave
Answer:
Ultraviolet

Question 20.
What is the colour seen at the inner edge of a rainbow?
Answer:
Violet

Question 21.
Fill in the blanks.
Green + ………. = Yellow
Answer:
Red –

Question 22.
What is the change in the deviation of those rays having lesser wavelength compared to those rays having higher wavelength? (more, less, no change)
Answer:
less

Question 23.
Name the phenomenon responsible for dispersion of light.
Answer:
Refraction

Question 24.
In the given figure which is correct.

Answer:

Question 25.
Distance to the near point of a healthy person is? (10 cm, 50 cm, 100 cm, 25 cm)
Answer:
25cm

Short Answer Type Questions (Score 2)

Question 26.
Arrange the following colours in the in-creasing “order of their wavelengths.
Violet, Red, Green, Blue
Answer:
Violet, Blue, Green, Red

Question 27.

Yellow light from a torch falls on a green paper and then falls on a white screen.
a. Which colour is seen on the screen?
b. Write the reason in two or three sentences.
Answer:
a. Green colour
b. When yellow light falls on a green paper it absorbs yellow and reflects green. So . the green colour is seen on the screen.

Question 28.
Myopia and Hypermetropia are the eye defect of human beings, identify the given statement then separate the reason for Myopia and Hypermetropia.
a. Image is formed behind the retina
b. Images formed in front of the retina
c. Power of the eye lens decreases
d. Power of the eye lens increases
e. Suitable power of concave lens is used to solve this problem
f. Suitable power of convex lens is used to solve this problem
Answer:
Myopia -b, d, e
Hypermetropia -a, c, f

Question 29.
Why concave lens always create virtual and erect image of the object?
Answer:
In this case, refracted Ray do not actually intersect to each other. It appears to intersect the images formed the same side of the lens.

Short Answer Type Questions (Score 3)

Question 30.

a. Complete the ray diagrams in the figures A and B when the rays of light passes through the prisms and reach out of them with the changes occurred to them.
b. If there are differences between the figures explain the reason
Answer:

b. There is only one colour in laser. Sunlight is the composite light of seven colours and so undergoes refraction

Question 31.
Explain the reason for the following.
a. At the time of sunset, western horizon is seen red.
b. Sky in the moon ha» a dark colour
Answer:
a. At the time of sunset, Light rays have to travel very long distances through air. Hence Light rays are scattered. Light rays like blue with shorter wavelength get lost due to scattering but rays with longer wavelength remain because it gets less scattering.
b. In moon there is no atmosphere and therefore scattering of light does not take place.

Question 32.
The figure shows a ray of light falling obliquely on a drop of water in atmosphere

a. Copy the diagram and complete it show-ing the internal reflection and refractions.
b. How does the sunlight appear as rainbow in water droplets?
Answer:

b. All the water droplets of the same colour appear to be in a same arc of a circle. The rays of light incident on the water droplets must be parallel to the line of vision. Each colour ray emerging from the water drop makes a definite angle from 40.8° to 42.7°. Red makes the higher angle of 42.7° and violet ma-kes a lower angle of 40.8°. Hence red colour is seen at the outer edge and violet colour at the inner edge. The other colours are seen in between depending on their wavelengths.

Question 33.
The image formation of a defected eye is given below

a. In which position images formed on a normal eye?
b. What is this eye defect?
c. How to solve this defect? Draw the diagram.
Answer:
a. On the retina.
b. Near-sightedness(Myopia).
c. Suitable power of concave lenses is used to solve this problem.

Short Answer Type Questions (Score 4)

Question 34.
Find the appropriate terms from the box for the statements given below.
Red, Scattering, Refraction, Dispersion
a. Phenomenon responsible for Rainbow.
b. Phenomenon responsible for Tyndal effect.
c. Colour seen at the outer edge of the rainbow.
d. Phenomenon responsible for Dispersion of Light.
Answer:
a. Dispersion
b. Scattering
c. Red
d. Refraction

Question 35.
Rearrange the table given below correctly.

Answer:

Question 36.
Analyse the picture and answer the questions.

a. What is the colour of the light falling on the white screen?
b. What colour is obtained on the screen if blue filter is used instead of green filter?
Answer:
a. Green
b. When blue filter paper is used, no light will fall on the screen. The yellow, green and red colours coming out of the yellow filter will not pass through blue filter paper.

Question 37.
When Newton’s colour disc rotates fast it appears white.
a. Which phenomenon is responsible for this?
b. Define this phenomenon.
c. Write another example related to this phenomenon.
Answer:
a. Persistence of Vision.
b. When an object is viewed by a person, the image remains in the retina for a time interval of 1/16 second. This phenomenon is called persistence of vision.
c. Raindrops look like glass rods during rain.

Students can Download Physics Chapter 4 Reflection of Light Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 4 Reflection of Light Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 4 Solutions Malayalam Medium

Students can Download Chemistry Chapter 7 Chemical Reactions of Organic Compounds Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 7 Chemical Reactions of Organic Compounds Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 7 Solutions Malayalam Medium

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SSLC Maths Chapter 11 Statistics and Algebra Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 11 Statistics and Algebra Notes

Textbook Page No. 245

Statistics Class 10 Kerala Syllabus Questions 1.
The distance covered by an athlete in long jump practice are
6.10, 6.20, 6.18, 6.20, 6.25, 6.21, 6.15, 6.10
in meters. Find the mean and median. Why is it that there is not much difference between these?
Answer:
Mean = \(\begin{array}{l}{=\frac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}} \\ {=\frac{49.39}{8}=6.17}\end{array}\)
If distances are written in ascending order
6.10, 6.15, 6.18, 6.20, 6.21, 6.25
Median = \(\frac{6.18+6.20}{2}=6.19 \mathrm{m}\)
Mean and median are gives the average dis¬tance covered by a person. Hence they will not have much difference..

Polynomial in Descending Order Calculator is a free online tool that determines the descending order of a polynomial in just a few taps.

Sslc Maths Statistics Kerala Syllabus Questions 2.
The table below gives the rainfall during one week of September 2015 in various districts of Kerala.

Calculate the mean and median rainfall in Kerala during this week. Why is the mean less than median?
Answer:
Mean = Total amount of rain / No. of districts
= \(\frac { 700 }{ 14 }\) = 50
In ascending order:
13.5, 20.5, 30.5, 33.5, 45.5, 50.3, 53.4, 56.3, 56.4, 56.9, 56.9, 66.7, 70.6, 89
Median \(=\frac{53.4+56.3}{2}=54.85\)
The mean is less than median because the number contains are far small and large numbers than mean.

Use this number sequence calculator to easily calculate the n-th term of an arithmetic, geometric or fibonacci sequence.

Sslc Statistics Solutions Kerala Syllabus Questions 3.
Prove that for a set of numbers arithmetic sequence, the mean and median are equal.
Answer:
Let a, a+d, a+3d, a+4d are the numbers of an arithmetic sequence, then median = \(\frac{a+a+4 d}{2}=\frac{2 a+4 d}{2}=a+2 d\)
Median will be the term which is at center = a + 2d
∴ A set of numbers in arithmetic sequence, the mean and median are equal.

Textbook Page No. 248

Sslc Maths Statistics Notes Kerala Syllabus  Questions 1.
35 households in a neighborhood are sorted according to their monthly income in the table below.

Calculate the median income.
Answer:

In the table monthly income up to 18th place be 6000 rupees.
That is 18^th place family also includes in the middle of total number of families.
∴ Median of the income = Rs. 6000

Sslc Maths Chapter 11 Statistics Kerala Syllabus Questions 2.
The table below shows the workers in a factory sorted according to their daily wages.

Calculate the median daily wage.
Answer:

Total number of workers = 30
Half= 15
So median is the wage of 15th worker.
The daily wages between the place 11 and 18 = 700 rupees
Median of daily wages = 700 rupees

Sslc Maths Chapter 11 Kerala Syllabus Questions 3.
The table below gives the number of babies born in a hospital during a week, sorted according to their birth weight.

Calculate the median birth-weight
Answer:

Total number of babies = 70
Half =35
So median is the weight of 35^th baby.
The weight of 35 111 child be in between 29 and 40 placed child, its weight will be 3 kg.
∴ Median of the weight = 3 kg.

Textbook Page No. 254

Sslc Statistics Notes Kerala Syllabus Questions 1.
The table shows some households sorted according to their usage of electricity:

Calculate the median usage of electricity.
Answer:

Half of the number of houses = 20
We have to find the electricity usage of the 20th house. According to this, we can divide

Statistics Class 10 State Syllabus Question 2.
Answer:

Total number of children = 34
It is an even number, so we will take the half of sum of weight of the children, those are in the 17 and 18 positions. According to this child between 13 and 22 have weight between 50.5 and 55.5. Our required children (between and 17 and 18) are in these positions. Divide 5 years from 50.5 to 55.5 into 10 equal, parts. Let consider each part have one child.

Weight of each part = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\).
Hence weight of a child in 13th place is in middle of 50.5 and 51. That is 50.75. Further, each student’s weight is increased by 5.
Hence weight of the man in the 17th position

Median \(=\frac{52.75+53.25}{2}=53\)

Statistics 10th Standard State Syllabus Question 3.
The workers of a company are arranged as given below. Calculate median

Answer:

Total number of workers = 30
Half = 15
So median is the wage of 15th worker.
The daily wage between the place 10 and 18 = 600 rupees
Median of daily wage = 600 rupees

Statistics SCERT Questions & Answers

Scert Class 10 Maths Statistics Kerala Syllabus Question 5.
10 households in a neighborhood are sorted according to their monthly income are given below
16500, 21700, 18600, 21050, 19500, 17000, 21000, 18000, 22000, 75000
a. What is the mean income of these 10 families?
b. How many families have monthly incomes less than the mean income? Prove that in such situation this average is suitable or not?
Answer:
a. Mean = \(\frac { sum }{ number }\) = \(\frac { 248000 }{ 10 }\) = 24800
b. 9 families have monthly income less than the mean income. So in this situation this is not a suitable average.

Sslc Maths Statistics Questions And Answers Kerala Syllabus Question 6.
Number of members in 10 families, collected by mathematics club survey are given. Calculate mean; median and explain. Which is the suitable average?.
4, 2, 3, 5, 4, 3, 2, 20, 4, 3
Answer:
a. Mean = 5
b. Median = 3.5
Suitable average median = 3.5

Sslc Statistics Questions Kerala Syllabus Question 7.
Weekly Wages of 9 persons Working in a factory are given. Find the median 2100; 3500, 2100, 2500, 2800, 4900, 2300, 2200, 3300
Answer:
Write the number in order.
2100, 2100, 2200, 2300, 2500, 2800, 3300, 3300, 3500 (1)

Median = 2500

Statistics Class 10 State Board Kerala Syllabus Question 8.
The table shows the workers doing different jobs in a factory according to their daily wages.

Calculate median of daily wages.

Answer:

The worker in the 12th position to 20th position has daily wage 270 ie, Median

Sslc Maths Chapter 11 Solutions Kerala Syllabus Question 9.
The table below shows the 60 children in a class sorted according to their heights

Find the median height?
Answer:

Sslc Maths Chapter Statistics Kerala Syllabus  Question 10.
Answer:
a. 100
b. 25
c.

Mean = \(\frac { 17840 }{ 100 }\) = 178.4

Hss Live Guru 10th Maths Kerala Syllabus Question 11.
The mean of the frequency table given below is 50. Then find out the values of a and b.

Answer:

mean = 50
∴ \(\frac{3480+30 a+70 b}{120}=50\)
30a + 70b = 6000 – 3480
30a + 70b = 2520 (1)
17 + 32 + 19 + a + b = 120
68 + a + b = 120,
a + b = 52 (2)
(2) x 30 => 30a + 30b = 1560
30a + 70b = 2520,
30a + 30b = 1560,
40b = 960,
b = \(\frac { 960 }{ 40 }\) = 24,
a + 24 = 52
a = 52 – 24 = 28,
∴ a = 28, b = 24

Long Answer Type Questions (Score 5)

10th Standard Maths Statistics Kerala Syllabus Question 21.
The table below shows groups of children in a class according to their heights:

a. If the children are lined up according to their heights, the median is the height of the child in which position?
b. According to the table, the height of this child is between what limits?
c. What are the assumptions used to compute the median?
d. What is the median height according to these assumptions?
Answer:

a. Height of the 21st child is the median height.
b. Height of the 21st child is between 145 cm and 150 cm.
c. Methods to find the median are.
1. Divide 5cm in between 145 cm and 150 cm into 10 equal sections.
2. Consider that the height of each sub-group is exactly on the midpoint of the subgroup.
Height of the 14th child is in between 145 cm and 145 \(\frac { 5 }{ 10 }\) cm.
i.e., 145 \(\frac { 5 }{ 20 }\) cm.
Similarly, the height of the 15th student is in between 145 \(\frac { 5 }{ 10 }\) cm and

Hence height of each child can be increased by \(\frac { 5 }{ 10 }\) cm.
There are 7 children to reach the 21st child from 14th child.
i.e., 14 th term is 145 \(\frac { 5 }{ 20 }\) and common difference is \(\frac { 5 }{ 10 }\).
Mean is the 21 st term of the arithmetic sequence.

Use our simple online Limit Of Sequence Calculator to find the Limit with step-by-step explanation.

Statistics Menton Map

Arithmetic mean is the sum divided by the number of terms.

Mean = \(\frac { sum of terms }{ Number of terms }\)

When the numbers are arranged in a ascending order, then the middle term is the median.
i.e., half of the total frequency will give the median.