Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

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Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration

Periodic Table and Electronic Configuration Text Book Questions and Answers

Kerala Syllabus 10th Standard Chemistry Chapter 1 Text Book Page No: 7

→ What is the basis of classification of elements in the periodic table?
Answer:
Atomic Number

Sslc Chemistry Chapter 1 Questions And Answers Text Book Page No: 8

→ Atomic number of sodium is 11 Electronic configuration – 2,8,1
GroupNumber — …………..
Period number — …………
Answer:
Group Number — 1
Period number — 3

→ Is the group 1 element a metal or a nonmetal?
Answer:
Metal

→ Write the electronic configuration of sodium and argon and complete the Table.
Kerala Syllabus 10th Standard Chemistry Chapter 1
Answer:
Sslc Chemistry Chapter 1 Questions And Answers

This article mainly deals with the chemical formula of lithium oxide, the structural lithium oxide formula with its properties and uses.

Periodic Table And Electronic Configuration Class 10 Notes Text Book Page No: 9

→ How many electrons are present in the M shell, the outermost shell of argon?
Answer:
8

→ What is the maximum number of electrons that can be accommodated in the M Shell?
Answer:
18

→ The ‘K’ shell, which is the first shell, has 1 subshell. The next ‘L’ shell has 2, and so on. What will be the number of subshells in the ‘M’ shell and ‘N’
M = ……………… , N = ……………….
Answer:
M = 3, N = 4

→ Which subshell is common to all shells?
Answer:
S

Sslc Chemistry Chapter 1 Notes Kerala Syllabus Text Book Page No: 10

→ Complete the Table 1.3
Periodic Table And Electronic Configuration Class 10 Notes
Answer:
Sslc Chemistry Chapter 1 Notes Kerala Syllabus

→ Complete the Table 1.4
Periodic Table And Electronic Configuration Class 10
Answer:
Chemistry Class 10 Chapter 1 Kerala Syllabus

→ What is the maximum number of electronics that can be accommodated in the ‘s’?
Answer:
2

→ What may be the maximum number of electrons to be filled in the ‘p’ subshell?
Answer:
6

Periodic Table And Electronic Configuration Class 10 Text Book Page No: 11.

The atomic number of hydrogen is 1(1H)

→ How many electrons are present?
Answer:
1

→ In which shell is the electron filled?
Answer:
‘K’ shell

→ In which subshell?
Answer:
S

→ How many electrons are present in helium (2He)?
Answer:
2

Follow along with the alkene reactions cheat sheet.

Chemistry Class 10 Chapter 1 Kerala Syllabus Text Book Page No: 12

→ Complete the subshell electronic configuration?
Answer:
1s2

→ Write the electronic configuration of Lithium (3Li)
Answer:
1s2 2s1

→ Complete the electronic configuration of beryllium?
Answer:
Be[Z=4] -1s2 2s2

→ Write the electronic configuration of Boron
Answer:
B[Z=5] -1s1 2s2 2p1

→ Write the electronic configuration of Carbon
Answer:
C[Z=6] – 1s2 2s2 2p2

→ Complete the Table 1.6
Class 10 Chemistry Chapter 1 Periodic Table And Electronic Configuration
Answer:
Sslc Chemistry Notes Chapter 1 Kerala Syllabus

Class 10 Chemistry Chapter 1 Periodic Table And Electronic Configuration Text Book Page No: 13

→ How was the shell wise electronic configuration of potassium written?
Answer:
2, 8, 8, 1

→ Compare the energies of Is and 2s subshells. Which one has lower energy?
Answer:
1s < 2s

Sslc Chemistry Notes Chapter 1 Kerala Syllabus Question 14.
Among the 3s & 3p subshells which has higher energy?
Answer:
3s < 3p

→Among the 3d & 4s subshells which has higher energy?
Answer:
4s < 3d

→ Write down the subshells in the increasing order of their energies.
Answer:
1s <2s <2p <3s <3p <4s <3d <4p

→ Write the subshell wise electronic configu-ration of potassium.
Answer:
1s2 2s2 2p6 3s2 3p6 4s2

→The electronic configuration of scandium (2lSc) is
Answer:
s2 2s2 2p6 3s2 3p6 3d1 4s2

Periodic Table And Electronic Configuration Kerala Syllabus Text Book Page No: 14

→ Write the electronic configu ration of 22Ti, 23V, the two elements after Sc.
Answer:
22Ti — 1s2 2s2 2p6 3s2 3p6 3d2 4s2
23V — 1s2 2s2 2p6 3s2 3p6 3d3 4s2

→Which is the noble gas preceding sodium (11Na)?
Answer:
Neon(Ne)

→ Write its subshell electronic configuration.
Answer:
10Ne – 1s2 2s2 2p6

→ Subshell electronic configuration of sodium?
Answer:
11Na – 1s2 2s2 2p6 3s1

Periodic Table And Electronic Configuration Class 10 Important Questions Text Book Page No: 15

→ Using the symbol of neon, write the subshell electronic configuration of sodium?
Answer:
[Ne] 3s1

→ Complete the Table 1.7
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 32
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 33

→ Write the subshell electronic configuration of 24Cr
Answer:
24Cr – 1s2 2s2 2p6 3s2 3p6 3d5 4s1

→ On the basis of this, identify the correct electronic configuration of 29Cu from those given below:
Answer:
1s2 2s2 2p6 3s2 3p6 3d9 4s2 – False
1s2 2s2 2p6 3s2 3p6 3d10 4s1 – True

Periodic Table And Electronic Configuration Class 10 Question Paper Text Book Page No: 16

If the subshell wise electronic configuration of an atom is 1s2 2s2 2p6 3s2, find answers to the following:

→ How many shells are present in this atom?
Answer:
3

→Which are the subshells of each shell?
Answer:
K — Is, L — 2s, 2p, M — 3s

→Which is the subshell to which the last electron was added?
Answer:
3s

→ What is the total number of electrons in the atom?
Answer:
12

→ What is its atomic number?
Answer:
12

→ How can the subshell electronic configuration be written in a short form?
Answer:
[Ne]3s2

Periodic Table And Electronic Configuration Class 10 Questions And Answers Text Book Page No: 17

→ Complete the Table 1.8
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 34
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 35

→ Which is the subshell of lithium to which the last electron was added?
Answer:
S

→ What about the subshell to which the last electron of nitrogen was added
Answer:
p

→ What is the relation between the subshell to which the last electron was added and the block to which the element belongs?
Answer:
The subshell in which the last electron enters represent the block in which the element belongs.

→ Write the subshell electronic configuration of the following elements and find the blocks to which they belong.
a. 4Be: ………………..
b. 26Fe……………..
c. 18Ar: ……………
Answer:
a. 4Be : 1s2 2s2 — s block
b. 26Fe : Is2 2s2 2p6 3s2 3p6 3d6 4s2 — d block
c. 18Ar : 1s2 2s2 2p6 3s2 3p6 — p block

Text Book Page No: 18

→ Complete the Table 1.9
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 36
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 37

→ Complete the Table 1.10
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 38
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 39

→ What is the relation between number of electrons present in the last ‘s’ subshell and their group number?
Answer:
The number of electrons in the outermost ‘s’
subshell = The group number

Text Book Page No: 20

→ When the s block elements react, do they donate or accept electrons?
Answer:
They donate electrons.

→ Which type of chemical bond is usually formed?
Answer:
Ionic bonds

→ How many electrons are donated by the first group elements in chemical reactions ?-
Answer:
One

→ How many electrons are donated by the second group elements in chemical reaction?
Answer:
Two

→ Complete the table 1.11
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 40
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 41

→ ‘s’ block elements are present at the extreme left side of the periodic table. Relating to their position, what other characteristics can be listed out?
Answer:

  • More metallic character s
  • Less ionization energy
  • Less electronegativity
  • Lose of electrons in chemical reaction
  • Compounds are mostly ionic
  • Oxides and hydroxides are basic in nature

Text Book Page No: 21

→ Which are the group included in the p block
Answer:
13, 14, 15, 16, 17, 18

→ In which subshell did the filling of the last electron take place?
Answer:
p subshell

→ Complete the table 1.13.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 42
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 43

Text Book Page No: 22

The outermost subshell wise electronic configuration of an element Y (Symbol is hot real) is 3s2 3p4.

→ To which period and group does this element belongs to?
Answer:
Period = 3, Group = 16

→ Write down the outermost subshell electronic configuration of the element coming just below it in the same group?
Answer:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

→ Find out examples of elements in such different states with the help of the periodic tables?
Answer:
Solid – Li, Be, B, C, Na, Mg, Al, Si
Liquid – Br
Gas – H, He, N, O, F, Ne

→ Which element has the highest ionization energy in each period?
Answer:
Group 18 elements.

Text Book Page No: 23

→ The elements having the highest electronegativity is in the p block. Find its name and position?
Answer:
Fluorine F, Period – 2, p block, Group 17

→ Analyze the general characteristics of the p block elements and prepare a note on this?
Answer:

  • The outermost p subshell of the p block elements contains 1 to 6 electrons.
  • Elements showing positive oxidation state and negative oxidation state are members of this block.
  • There are metals and nonmetals in these blocks.
  • Elements in the solid, liquid and gaseous states are present in p block.

→ Complete the table 1.14
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 44
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 45

→ Which element has a valency 1?
Answer:
Y

→Which element shows metallic character?
Answer:
X

→ Which element has the highest ionization energy?
Answer:
Y

→ Write the chemical formula of the compound formed by the combination of X and Y and label the oxidation states?
Answer:
Compound: X Y2
Oxidation state: X2+, Y1-

→ Where is the position of d block elements in the periodic table?
Answer:
3rd Group to 12th Group

→ From which period onwards does the d block begin?
Answer:
4

Text Book Page No: 24

→ Put a tick mark ✓’ against the statements below, which are applicable to d block elements.
Answer:
1. ‘✓’ These are metals.
2. ‘✓’ The last electron is filled in the penultimate shell.
3. ‘✗’ In the case of these elements in the 4th period, the last electron is filled in 4s.
4. ‘✓’ These are found in groups 3 to 12 of the periodic table.

Text Book Page No: 25

→ Complete the table 1.16
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 46
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 47

→ How does Fe change to Fe2+?
Answer:
By losing 2 electrons from 4s valence subshell.

Text Book Page No: 26

→ Write down the subshell electronic configuration of Fe21.
Answer:
1s2 2s2 2p6 3s2 3p6 3d6

There is only a small difference of energy between the outermost s subshell and the penultimate d subshell of transition elements.

→ If so, which will be the subshell from which iron loses the third electron?
Answer:
From 3d sub-shell

→ Write the electronic configuration of Fe3+ on the basis of this.
Answer:
1s2 2s2 2p6 3s2 3p6 3d5

→ Write the subshell electronic configuration of Manganese (Mn).
Answer:
1s2 2s2 2p6 3s2 3p6 3d5 4s2

→ Complete the table 1.17
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 48
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 49

Text Book Page No: 27

→ Examine these compounds available. Find more colored compounds and extend the list.
Answer:

  • Copper sulfate CuSO4.5H2O – blue,
  • Copper nitrate Cu(NO3)2.6H2O – pink.
  • Potassium permanganate KMnO4 – violet.
  • Ferrous sulfate FeSO4.7H2O – Green,
  • Ferrous nitrate (Fe(NO3)2.6H2O) – light green

Text Book Page No: 28
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 50

→ List out the dements of the s block?
Answer:
A, B

→ Which elements shows+2 oxidation state?
Answer:
B, C, D

→ Which elements contains 5 electrons in the outermost shell?
Answer:
E

→ Which is the element that has 5 p electrons in the outermost shell?
Answer:
G

→ Which are the elements in which the last electron enters the d subshell?
Answer:
C, D

→ Which element has the highest ionization energy?
Answer:
H

→ Which is the highly reactive nonmetal?
Answer:
G

→ Which elements show -2 oxidation state?
Answer:
F

Text Book Page No: 29

The outermost electron configuration of an element in this is 2s2 2p6

→ Which is the element?
Answer:
H

→ Write down the complete subshell electronic configuration?
Answer:
Is2 2s2 2p6

→ Write any two characteristics of this element?
Answer:

  • Noble element / gases.
  • The outermost shell is completely filled

→ Write the chemical number of questions, the answer of which is an element in the table
Answer:
A G

Periodic Table and Electronic Configuration Let Us Assess

Question 1.
Based on the hints given, find out the atomic number and write down the subshell electronic configuration of elements (Symbols used are not real).
i. A – period 3 group 17
ii. B – period 4 group 6
Answer:
A17 — 1s2 2s2 2p6 3s2 3p5
B24 — 1S2 2s2 2p6 3S2 3p6 3d5 4S1

Question 2.
When the last electron of an atom was filled in the 3d subshell, the subshell electronic configuration was recorded as 3d8 Answer the questions related to this atom.
1. Complete subshell electronic configuration
2. Atomic number
3. Block
4. Period number
5. Group number
Answer:
1. Complete subshell electronic configuration:
1 s2, 2s2, 2p6, 3s2, 3p6, 3d8, 4s2
2. Atomic number: 28
3. Block : d
4. Period number: 4
5. Groupnumber : 8 + 2 = 10

Question 3.
Pick out the wrong ones from the subshell electronic configuration given below.
a. 1s2 2s2 2p7
b. 1s2 2s2 2p2
c. 1s2 2s2 2p5 3s2
d. 1s2 2s2 2p6 3s2 3p6 3d2 4s2
e. 1s2 2s2 2p6 3s2 3p6 3d2 4s2
Answer:
Wrong electronic configuration
a. 1s2, 2s2, 2p7
(2p maximum 6 electrons only)

c. 1s2, 2s2, 2p5, 3s1 (electrons are filled in 3s only after filling 6 electrons in 2p)

d. 1s2, 2s2, 2p6, 3s2, 3p6, 3d2, 4s1 (electrons are filled in 3d only after filling 2 electrons in 4s)

Question 4.
The element X in group 17 has 3 shells. If so,
a. Write the subshell electronic configuration of the element.
b. Write the period number,
c. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
Answer:
a. Three shells are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s2, 2s2, 2p6, 3s2, 3p5

b. Period-3

c. Y – Third period
∴ shells – 3
1 electron in p – subshell
Total electrons in valence shell 2+1=3 (2 electrons in s + 1 electron in p)
Valency of x – 1(1 electron is recieved – electro negative atom)
Valency of y – 3 (3 electrons are lost – electro positive atom)
Therefore they combine to form compounds with chemical formula YX3
(Symbol of electropositive element first followed by electro negative element).

Question 5.
The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2.
a. Write down the subshell electronic configuration of this ion.
b. Can this element show variable valency? Why?
c. Write down the chemical formula of one compound formed when this element reacts with chlorine (17CI).
Answer:
a. 29Cu — 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1
Cu2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d9

b. Yes. One electron can be lost from 4s subshell and can’exist as Cu+ ion, It is a d-block element.

c. Copper react with chlorine to form two compounds Cu+, Cu2+ ions react with chlorine to form CuCl and CuCl2 respectively.

Question 6.
Certain subshells of an atom are given below. 2s, 2d, 3f, 3d, 5s, 3p
a. Which are the subshells that are not possible?
b. Give the reason.
Answer:
a. Not possible sub-shells are 2d, 3f

b. d – subshell is not possible in 2nd shell
f – subshell is not possible in 3rd shell

Periodic Table and Electronic Configuration Extended Activities

Question 1.
Prepare the comprehensive table which indicates the name, symbol, electron configuration, subshell configuration of elements having atomic number 1 to 36?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 51

Question 2.
Some information related to the elements of the p bllock in the 17th group of the periodic table are given in the table below. Complete the table and analyze the following questions?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 52

a. What is the family names of elements belonging to the 17th group?
Answer:
Halogen

b. What is their common valency?
Answer:
1

c. Which element has the highest electro negativity ?
Answer:
F

d. Which element has the highest ionization energy?
Answer:
F

e. List out the name and chemical formula of the compounds formed by these elements with block elements?
Answer:

  • sodium chloride – NaCl
  • potassium chloride – KCl
  • magnesium chloride- MgCl2
  • calcium chloride- CaCl2
  • magnesium fluoride- MgF2
  • calcium fluoride – Ca F2
  • sodium iodide – Nal
  • potassium iodide – KI
  • potassium bromide – KBr
  • potassium fluoride – KF

Periodic Table and Electronic Configuration Orukkam Questions and Answers

Question 1.
Complete the table of details about shells and subshells.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 53
a. No of electrons in KLMN shell.
b. No of electrons in each shell.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 54
c. Which subshell is common to all sub-shells?
d. Write names of subshells in accordance with increasing energy level,
e. Identify the incorrect subshell electronic configuration.
– 1s2
– ls2 2p6
– ls2 2s2 2p6
– 1s2 2s2 2p6 3s2 3p2
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 55

a, K – 2 ; L – 8 ; M – 18 ; N – 32

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 56

c. s- Subshell

d. 1s <2s <2p <3s <3p <4s <3d <4p <5s <4d <5p <6s <4f <5s

e. 1s2 2p6

Question 2.
Atomic number of iron is 26. It exhibits Fe2+, Fe3+ oxidation state. Write the subshell electronic configuration.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 57
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 58

Question 3.
Manganese, a d-block element exhibits I different oxidation state. Why?
a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 59

b. Write the oxidation number and subshell electronic configuration K, Cl and O.
Answer:
Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions.

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 60

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 61

Question 4.
Find out atomic number, group, block period using subshell electronic configuration and then complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 62
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 63

Question 5.
Write down the characteristics of s,d,p, f block elements
Answer:
s-block elements:
Elements in which last electron enters into s-subshell are called s-block elements. It contains group I elements (Alkali metals) and group II elements (Alkaline earth metals).

1st group elements lose one electron during chemical combination. Therefore its oxidation state is +1.

2nd group elements lose two electrons from valence shell during chemical combination and their oxidation state is +2.

The highest shell number in a sub-shell electronic configuration is the period number of that element.

1. Group number characteristics = no.of electrons in valence sub-shell.
2. s block ionization energy & electro negativity decreases downwards.
3. Metallic character & reactivity increases downwards.
4. Lose electrons during chemical combination j and they form ionic compounds.
Their oxides and hydroxides are basic.
Their atomic radii are high in a period.

p-block elements:

  • Last electron enters into p-subshell.
  • Group 13 -18 elements.
  • Highly reactive elements are non-metals – group 17,
  • These are elements with positive and negative oxidation state.

Group number of p-block elements = electrons in last p-subshell + 12

d-block elements:

  • Last electron enters into penultimate d-subshell
  • Known as transition elements.
  • Metals
  • Shows similarity in group and period.
  • Variable oxidation states.
  • Form coloured compounds.

Group = electrons in ‘d’-subshdl + electrons in s-subshell.

f-block elements:

  • Last electron enters into antepenultimate f sub-shell.
  • Contains Lanthanoids and Actinoids.
  • Variable oxidation state.
  • Most of the Actinoids are radioactive.
  • Most of the elements are artificial.
  • U, Th, Pu are used in nuclear reactors.
  • Some elements are used as catalyst in pet-roleum industry.

Periodic Table and Electronic Configuration Evaluation Questions

Question 1.
Write down subshell electronic configuration of Cu1+ and Cu2+
Answer:
Cu1+ – 1s2 2s2 2p6 3s2 3p6 3d10
Cu2+ – 1s2 2s2 2p6 3s2 3p6 3d9

Question 2.
How many ‘s’ subshell electrons are in 1s2, 2s2, 2p6, 3s2, 3p2
Answer:
6 Electrons

Question 3.
11,17,10 are the atomic number of elements X, Y, and Z.
a. Write down their subshell electronic configuration, group, block, period,
b. Write the molecular formulae of the compound formed when any two of the above elements are combined.
c. Write down the oxidation numbers of the elements in those compounds. Write the subshell electronic configuration of both ions.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 64

b. X Y

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 65

Question 4.
Element ‘X’ is having atomic number 28, it gives two electrons to element ‘Y’.
a Write down the electronic configuration of ‘X’ and its ion
b. In which block ‘X’ belongs?
c. Write down the characteristics of that block
Answer:
a. X28 – 1s2 2s2 2p6 3s2 3p6 3d8 4s2
X2+– 1s2 2s2 2p6 3s2 3p6 3d8

b. d block Compound

c. 1. It exhibits variable oxidation states
2. Forms colored compounds
3. Last electron enters d subshell

Question 5.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 66
a. Write down the group and period of each element.
b. What are the use of writing electronic configuration this fashion?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 67

b. Group and period of the element can be identified easily. In the Same way long electron configuration can be avoided.

Question 6.
24Cr – [Ar] 3d5 4s1
29Cu – [Ar] 3d10 4s1
Why chromium and copper exhibits such electronic configuration ?
Answer:
Half filled and completely filled subshells are most stable. Change in the electronic configuration of 24Cr &, 29Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability.

Periodic Table and Electronic Configuration SCERT Questions and Answers

Question 1.
The electronic configuration of the elements A, B, C, Dare given below.
A – 1s2 2s2 2p6 3s2 3p4
B – 1s2 2s2 2p6 3s2
C – 1s2 2s2 2p6 3s2 3p5
D – 1s2 2s2 2p6 3s1
a. Which of these elements show +2 oxidation state?
b. Which metal belongs to 17th group?
c. Which is the period number of the element A ? What is the basis of your findings?
d. Which of these elements can form basic Oxides?
Answer:
a. B

b. C

c. Period number: 3, Period number = No.of shells

d. B, D

Question 2.
Two compounds of iron are jpven below.
FeSO4 Fe2(SO4)3
(The oxidation state of sulfate radical is-2)
a. Which ofthese compounds show +2 oxidation state for Fe?
b. Which compounds has Fe3+ ion?
c. Write the subshell electronic configuration of Fe3+ ion.
d. Why do transition elements show variable oxidation states?
Answer:
a., FeSO4
b. Fe2(SO4)3
c. Fe3+ – 1s2 2s2 2p6 3s2 3p6 3d5

d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction.

Question 3.
Identify the incorrect electronic configurations and correct them.
i) 1s2 2s2 2p3
ii) 1s2 2s2 2p6 3s1
iii) 1s2 2s2 2p6 2d7
iv) 1s2 2s2 2p6 3s2 3p6 3d4
Answer:
iii). 1s2 2s2 2p6 3s2 3p5 .
iv). 1s2 2s2 2p6 3s2 3p6 4s2 3d2

Question 4.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 68
Answer:
a. – 2
b. 1
c. 17,
d. – 1
e. 12
f. +12

Question 5.
a. Two compounds XY2, XZ4 are given. The oxidation state of Z is 1. What will be the oxidation state of Y ?
b. Write the molecular formula of the compound formed by Y when it combines with aluminum (Al) having oxidation state +3.
Answer:
a. Y= – 2 (oxidation state of X is +4)
b. Al2 Y3

Question 6.
Pick out the statements which suit to f-block elements.
a. All of them are naturally occurring elements.
b. Uranium and Thorium are f block elements.
c. Last electrons is filled in the shell pre-ceding the outermost shell.
d. last electrons are filled up in the antepenultimate shell.
e. Includes some radioactive elements.
f. Some of them are used as catalyst in petroleum industry.
Answer:
b, d, e, f

Question 7.
The atomic number of four elements are given below. (The symbols ore not real)
A – 8
B – 10
C – 12
D – 18
a. Write the sub-shell electronic configuration of the elements,
b. Which of them are inert gases?
c. Write the chemical formula of the compound formed by two elements other than inert gases.
Answer:
a. A – 1s2 2s2 2p4
B – 1s2 2s2 2p6
C – 1s2 2s2 2p6 3s2
D – 1s2 2s2 2p6 3s2 3p6

b. B, D

c. CA, (C2 A2 is simplified and written as CA)

Question 8.
The subshell electronic configuration of two elements ends as follows. (Symbols are not real)
P – 3s2 Q – 3p4
a. Write the complete subshell electronic configuration.
b. Find out the oxidation state of each element.
c. The chemical formula of the compound formed by these elements is PQ. Is this statement correct? Justify your answer.
Answer:
a. P – 1s2 2s2 2p6 3s2
Q – 1s2 2s2 2p6 3s2 3p4

b. P = +2, Q = – 2 :

c. Right, valency of both P and Q is ‘2’

Question 9.
Match the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 69
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 70

Question 10.
The atomic number of two elements are given below.
Si – 14 Ni – 28
a. Write the subshell electronic configu-ration of these elements.
b. Find out the group and period of each element.
Answer:
a. Si – 1s2 2s2 2p6 3s2 3p2
Ni – 1s2 2s2 2p6 3s2 3p6 3d8 4s2
b. Si – Period Number – 3, Group number – 14 Ni – Period Number – 4, Group number – 10

Question 11.
The element ‘X’ has 4 shells and its 3d subshell has 6 electrons. (Symbol is not real)
a. Write the complete electronic configu-ration of the element.
b. What is its group number? Which is the block?
c. Write any two characteristics of the block to which element X belongs to.
d. From which subshell the electrons are lost when the element X shows +2 oxidation state.
Answer:
a. 1s2 2s2 2p6 3s2 3p6 3d6 4s2
b. Group number – 8, Block – d
c. All of them are metals
d – block elements are placed in group 3 to group 12
d. s – Sub shell

Question 12.
The outermost electronic configuration of the element A is 2s2 2p2. (Symbol is not real)
a. Find out the group number and block of the element.
b. Write the chemical formula of the compound formed by A when it combines with chlorine.
c. Write the complete electronic configuretion of the element just below ‘A’ in the j periodic table.
Answer:
a. Group number – 14, Block – P
b. ACl4
c. 1s2 2s2 2p6 3s2 3p2

Question 13.
The figure of an incomplete periodic table is given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 71
a. Which one of these elements shows -2 oxidation state?
b. Which of these elements have 3 electrons in their outermost p subshell?
c. Which element has the highest atomic radius? Which one has the least?
d. Which of these elements show variable oxidation state?
e. Which of these elements has the highest ionization energy?
Answer:
a. G

b.F

c. The element having highest atomic radius – A
The element having lowest Atomic radius – H

d. D, C

e. H

Question 14.
Examine the given electronic configurations.
A – 1s2 2s2 2p6 3s2 3p6 3d10 4s2
B – 1s2 2s2 2p6 3s1
C – 1s2 2s2 2p1 3s2 3p6
D – 1s2 2s2 2p6 3s2
E – 1s2 2s2 2p6 3s2 3p6 4s2
a. Which of these elements belongs to 4th period?
b. Which elements belongs to the same group ?
c. Which element doesn’t participate in chemical reactions generally ?
d. Which element has highest metallic character ?
Answer:
a. A, E
b. B, E
c. C
d. E

Question 15.
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl2, YCl2, YCl3 are formed.
a. What is the specialty of the oxidation number of Y, compared to that of X?
b. Explain the reason for this, on the basis of the subshell based electronic configuration.
Answer:
a. Element X has constant oxidation state. Y shows variable oxidation states.
b. X20 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2
X26 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6
Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part.

Periodic Table and Electronic Configuration Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Arrange the following sub-shells in the in-creasing order of energy 5p, 2s, 4f, 3s, 4s, 3d, 6s
Answer:
2s < 3s < 4s < 3d < 5p < 6s < 4f

Question 2.
Last electron in f-block elements goes to
a. Which shell? Outer shell/Penultimate shell /Antepenultimate shell
b. Which sub-shell? Outer f-subshell Penultimate f-subshell/Antepenulti mate f-subshell.
Answer:
a. Antepenultimate shell
b. Antepenultimate f-sub-shell

Question 3.
Sub-shell electronic configuration of X is given below.
1s2, 2s2, 2p5
a. The element Y is coming just below the element in same group. Then write the sub-shell electronic configuration of Y.
b. Write the sub-shell electronic configuration of the element next to X in same period.
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p2
b. Is2, 2s2, 2p6

Question 4.
A compound of vanadium pentoxide (V20;) is used as catalyst.
a. What is the oxidation state of vanadium in this compound?
b. How vanadium ion is represented?
c. Write the sub-shell electronic configuration of this ion (V – 23)
Answer:
a. +5
b. V5+
c. 1s2, 2s2, 2p6, 3s2, 3p6

Short Answer Type Questions (Score 2)

Question 5.
Find the wrong electronic configurations from the following. What is wrong in these?
a. 1s2,2s2,2p6,3s2,3p6,3d9,4s2
b. 1s2
c. 1s2, 2s1, 2p6
d. 1s2, 2s2, 2p6, 3s2, 3p2
Answer:
(a) and (c) are wrong electronic con figurations.

In (a) one electron from 4s is to be trans-ferred to 3d since completely filled configu-rations are more stable. So the correct electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

In (c) electrons are filled in 2p only after filling electrons in 2s.

Question 6.
Group and period number of two elements are given.
P – group 17, period – 3
Q – group 2, period – 3
a. Write the sub-shell electronic configuration of each.
b Write the chemical formula of the compound formed by their combination.
Answer:
Answer:
a. P – 1s2, 2s2, 2p6, 3s2, 3p5
Q – 1s2, 2s2, 2p6, 3s2

b. Q is electropositive. P is electro negative;
∴Chemical formula QP2

Question 7.
Write the reason for the statement given
below.
a. d-block elements in the same period show similarity.
b. Transition elements show variable oxidation state.
Answer:
a. Valence shell electrons of d-block elements in same periods are almost same. Valence shell electrons are entering in chemical reaction. Therefore they shows similarity.

b. Energy of electrons in s-subshell and inner d- subshells are almost same. Therefore s- electrons or s and d electrons take part in chemical reaction and show variable oxidation state.

Short Answer Type Questions (Score 3)

Question 8.
Write the sub-shell electronic configuration of following elements. Predict the block, group and period. (Symbols are not real)
a. M – 27 b. N – 19 c. P – 15
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p6, 3d7, 4s2
block – d; group – 9; period – 4.
b. 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
block – s; group – 1; period – 4
c. 1s2, 2s2, 2p6, 3s2, 3p3
block – p; group – 15; period – 3

Question 9.
Observe the model of periodic table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 72
a. Which element is having S electrons in valence shell?
b. Which elements are having 2 electrons in valence sub-shell?
c. Which element is having last electron in3p?
d. Which element ends with electronic configuration 4d5, 5s1 ?
Answer:
a. B;
b.A, C;
c. C, D;
d. E

Question 10.
Calculate oxidation state of transition elements in the following compounds.
Answer:
a. KMnO4 – Mn – 7+
b. Cr2 O3 – Cr – 3+
c. K2Cr2O7 – Cr – 6+

Question 11.
Atomic number of some elements are given. A – 15, B – 8, C – 11, D – 18, E – 20, F – 34, G – 10
a. Which are the elements in same period?
b. Which are the elements in same group?
Answer:
A -1s2, 2s2, 2p6, 3s2, 3p3
(group -15 period – 3)
B – 1s2, 2s2, 2p4 (group –16 period – 2)
C – 1s2, 2s2, 2p6, 31 (group – 1 period – 3)
D – 1s2, 2s2, 2p6, 3s2, 3p6 (group – 18 period – 3)
E – 1s2, 2s2, 2p6, 3s2, 3p6, 4s2.(group – 2 period – 4)
F – 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p4 (group – 16 period – 4)
G – 1s2, 2s2, 2p6 (group – 18 period – 2)
a. A, D same period ;
B, G same period 1
b. B, F same group ;
D, G same group

Long Answer Type Questions (Score 4)

Question 12.
Electronic configuration of some elements are given. Write answers to the following questions.
i. [Ne] 3s2
ii. [Ar ] 3d2,4s2
iii. [Xe] 6s2
iv. [Ne]3s2
v [Ne] 3s2,3p5
a. Which metal is having high reactivity?
b. Which is having possibility of formation of colored compounds?
c. Which is the non-metal?
d Which element shows the possibility of +2 oxidation state?
Answer:
a. [Xe] 6s1
b. [Ar ] 3d2, 4s2
c. [Ne] 3s2, 3p5
d. [Ne] 3s2,[Ar] 3d2, 4s2

Question 13.
Pick the wrong statement from the following.
a. Elements with atomic number 5 belong to group 15.
b. Electronic configuration of scandium (Atomic number 21) is 2,8,8,3.
c. d-block elements are known as transition elements.
d. All s-block elements are metals.
Answer:
a. Wrong. It belongs to group 13.
b. Wrong. Electronic configuration
2,8,9,2 (1s2, 2s2, 2p6, 3s2, 3p6,3d1, 4s2)
c. Correct .
d. Correct

Question 14.
Look at the Bohr model of X-atom.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 73
a. Write the sub-shell electronic configuration of this atom.
b. Mention the compounds in which d-subshell electrons are taking part in chemical reaction during their formation.
XCl2, XO2, X2O7
c. Write the electronic configuration of X ions in the above three compounds.
Answer:
a. 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2

b. XCl2 – ion X2+ (electrons in 4s only)
XO2 – ion X+4 (2 electrons in 4s and 2 electrons in 3d)
X2O7 – ion X7+ (2 electrons in 4s and 5 electrons in 3d)
XO2, X2O7 d – subshell electrons are taking part in chemical reaction during the formation of X207

c. X2+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d5
X4+ – 1s2, 2s2, 2p6, 3s2, 3p6, 3d3
X7+ – 1s2, 2s2, 2p6, 3s2, 3p6

Question 15.
Select the suitable one from the following columns A, B, C.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 74
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 1 Periodic Table and Electronic Configuration 75

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

You can Download Chemical Reactions of Organic Compounds Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

Chemical Reactions of Organic Compounds Text Book Questions and Answers

Sslc Chemistry Chapter 7 Kerala Syllabus Question 1.
Complete stages 2, 3 and 4 in the respective order.
Sslc Chemistry Chapter 7 Kerala Syllabus
Answer:
Chemical Reactions Of Organic Compounds Class 10 Kerala Syllabus

The chemical formula calculator is particularly helpful for establishing the percentage of each element

Chemical Reactions Of Organic Compounds Class 10 Kerala Syllabus Question 2.
What are the compounds formed when CH3-CH3 (ethane) undergoes substitution reaction with chlorine? Write them.
Chemistry Chapter 7 Test Answer:
CH3 – CH2 Cl, CH3 – CHCl2,
CH3 – CCl3, CH2Cl – CCl3,
CHCl3 – CCl3, CCl3 – CCl3

Chemistry Class 10 Chapter 7 Kerala Syllabus Question 3.
Write down the structural formulae of ethane and ethene.
Answer:
CH3 – CH3 – Ethane
CH2 = CH2 – Ethene

Sslc Chemistry Chapter 7 Solutions Kerala Syllabus Question 4.
What is the peculiarity of the Carbon- Carbon bond in ethene?
Answer:
In ethene, There is carbon – carbon double bond

Text Book Page No: 121

alkyne reactions cheat sheet summary for organic chemistry reactions.

Class 10 Chemistry Chapter 7 Kerala Syllabus Question 5.
What do we get as the product ?
Answer:
Ethane

Class 10 Chemistry Chapter 7 Notes Kerala Syllabus  Question 6.
Which hydrocarbon is the reactant here? ………..
Answer:
Unsaturated propene

Chemistry Chapter 7 Class 10 Kerala Syllabus Question 7.
Is the product saturated or unsaturated ?
Answer:
Saturated Compounds

10th Class Chemistry 7th Chapter Kerala Syllabus Question 8.
Complete table 7.1
Sslc Chemistry Chapter 7 Notes Kerala Syllabus
Answer:
Chemistry Class 10 Chapter 7 Kerala Syllabus

Text Book Page No: 123

Hsslive Chemistry 10th Kerala Syllabus Question 9.
Complete table 7.2 Suitably.
Sslc Chemistry Chapter 7 Solutions Kerala Syllabus
Answer:

Monomer Polymer Uses
Vinyl Chloride PVC Pipe, Helmet
Ethene Polyethane Carry bags
Isoprene Natural rubber (Poly Isoprene) Tire
Tetra Fluro ethene Teflon Nonstick pan

Text Book Page No: 124

Balance the equation calculator allows you to balance chemical equations accurately.

Sslc Chemistry Chapter Wise Questions And Answers Kerala Syllabus Question 10.
Can you write the balanced chemical equation for the combustion of the fuel butane (C4H10) ?
Answer:
2 C4H10(g) + 13O2(g) → 8CO2 + 10H2O + heat

Text Book Page No: 125

Hsslive Guru 10th Chemistry Kerala Syllabus Question 11.
Complete Table 7.3 and 7.4 contain¬ing chemical reactions of hydrocarbons.
Class 10 Chemistry Chapter 7 Kerala Syllabus
Answer:
Class 10 Chemistry Chapter 7 Notes Kerala Syllabus

Organic Chemistry Class 10 Kerala Syllabus  Question 12.
Match Columns A, B, and C suitably.
Chemistry Chapter 7 Class 10 Kerala Syllabus
Answer:

Reactants (A) Products (B) Name of Reaction (C)
CH3 – CH3 + Cl2 CH3 – CH2 Cl + HCl Substitution Reaction
C2H6+O2 CO2 + H2O Combustion
n CH2 = CH2 [CH2 – CH2]n Polymerisation
CH3– CH2 – CH3 CH2 = CH2 + CH4 Thermal Cracking
CH = CH + H2 CH2 = CH2 Addition Reaction

Class 10 Chemistry Kerala Syllabus Question 13.
CH2 – OH, CH3 – CH2 – OH
Can you write the IUPAC names of these two compounds?
Answer:
CH3 – OH – Methanol
CH3 – CH2 – OH – Ethanol

Text Book Page No: 126

Hss Live Guru 10th Chemistry Kerala Syllabus Question 14.
Complete the following word web including more uses of ethanol.
10th Class Chemistry 7th Chapter Kerala Syllabus
Answer:
Hsslive Chemistry 10th Kerala Syllabus

Chemistry Textbook Class 10 Kerala Syllabus Question 15.
List out the uses of ethanoic acid.
Answer:

  • In the manufacture of rayon
  • In the rubber and silk industry.
  • Vinegar – Impart Sour taste for food item.
  • Used as preservative.

Text Book Page No: 129

Hss Live Guru 10 Chemistry Kerala Syllabus Question 16.
Examine the given structural formulae and select the esters. You may also identify the chemicals required for their preparation.
1. CH3 – CH2 – COO – CH3
2. CH3 – CH2 – COOH
3. CH3 – CH2 – CO – CH3
4. CH3 – OH
5. CH3 – CH2 – CH2OH
6. CH3 – COOH
7. CH3 – COO – CH2 – CH2 – CH3
Answer:
1. CH3 – CH2 – COO – CH3
7. CH3 – COO – CH2 – CH2 – CH3 are esters
1. CH3 – CH2 – COO – CH3
CH3 – CH2 – COOH + OH – CH3 \(\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }\) CH3 – CH2 – COO – CH3+ H2O
7. CH3 – COO – CH2 – CH2 – CH3
CH3 – COOH + OH – CH2 – CH2 – CH3 \(\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }\) CH3 – COO – CH2 – CH2 – CH3 + H2O

Text Book Page No: 130

Chemistry Class 10 Kerala Syllabus Question 17.
Take 10 mL distilled water in a test tube and take the same volume of hard water in another test tube. Add a few drops of soap solution to both the test tubes and shake well. Do both the test tubes contain the same quantity of foam? Which test tube contains more foam? What do you infer?
Answer:
No. Both the test tube does not contain same quantity of foam. Distilled water taken test tube contains more foam. Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather.

Hss Live Guru Chemistry 10 Kerala Syllabus Question 18.
Take 10 mL each of hard water in two test tubes. Add a few drops of soap solution in the first test tube and add the same amount of detergent solution in the second one. Shake both the test tubes well. What do you observe? Which test tube contains more foam?
Answer:
Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather. But detergents do not give insoluble components on reaction with these salts. Hence detergents are more effective than soaps in hard water.

Hsslive Chemistry Class 10 Kerala Syllabus Question 19.
List out the merits and demerits of detergents, compared to soaps.
Answer:
Merit:
Detergents are more effective than soaps in hard water.
Detergents are effective in acidic solutions.
Demerits:
excessive use of the detergents causes environmental problems. The microorganisms in water cannot decompose the components of detergents. Hence the detergents released into water lead to the destruction of aquatic life. For example, the detergents which contain phosphate increases the growth of algae and limits the quantity of oxygen. Therefore, it dej creases the quantity of oxygen for the breath of the organisms in water and causes their destruction.

Let Us Assess

Solve using substitution calculator step-by-step solutions.

10th Chemistry Notes Kerala Syllabus Question 1.
Given below are two chemical equations.
a. CH2 = CH2 + H2 → A
b. \(\mathrm{A}+\mathrm{Cl}_{2} \quad \stackrel{\text { sunlight }}{\longrightarrow} \mathrm{B}+\mathrm{HCl}\)
Identify the compounds A and B. Name these reactions.
Answer:
Sslc Chemistry Chapter Wise Questions And Answers Kerala Syllabus

Question 2.
Name the important chemical reactions of hydrocarbons. Give one example for each.
Answer:
Hsslive Guru 10th Chemistry Kerala Syllabus
e. Thermal cracking:
CH3 – CH2 – CH2 -CH3 → CH2 = CH2 + CH3 – CH3

Question 3.
Write chemical formula of propane. Write the names and structural formulae of two compounds, that may be formed during its substitution reaction with chlorine.
Answer:
CH3 – CH3 – CH3 Propane
Organic Chemistry Class 10 Kerala Syllabus

Question 4.
Complete the equation for the following chemical reaction. Name this reaction.
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}+\ldots \ldots \ldots \ldots \mathrm{O}_{2} \rightarrow–\mathrm{-}+\)
Answer:
CH3 – CH2 – CH2 – CH3 + 13/2 O2
→ 4CO2 + 5 H2O , Combustion.

Question 5.
Which of the given molecules can form po¬lymers ? Butane, Propane, Propane, Methane, Butene.
Answer:
Propene, Butene

Extended Activities

Question 1.
You are familiar with different chemical reactions of hydrocarbons. Identify the situations in daily life in which these are used.
Answer:
a. Substitution Reaction : Chloroform, CCl4 preparation
b. Addition Reaction : Conversion of unsaturated compounds into saturated.
c. Combustion: Preparation of polymers like PVC.
d. Thermal Cracking: Butane (LPG)can be prepared from higher hydrocarbons

Question 2.
List out the different uses of ethanol. Pre-pare an essay on its adverse effects on human body and the related social issues when it is used as a beverage.
Answer:
Uses of ethanol :

  • Fuels
  • Medicines
  • Preservatives
  • Preparation of organic compounds

Health problems :

  • Reason for aneamia .
  • Increases possibility of cancer
  • Problems related with heart

Social problems:

  • Reason for spoiling family relationships
  • May cause financial crisis

Question 3.
You know how to make soap, don’t you? Try to prepare soaps of different colors and fragrance.Prepare a short note on chemistry of soaps.
Answer:
Fats and oils are esters. They react with alkalies such as NaOH, KOH to form Sodium/ Potassium Salts of their carboxylic acids and glycerol.ester formed from fatty acids + NaOH / KOH → Soap + glycerol. Fatty acids such as palmitic acid, stearic acid react with alcohol, glycerol to form esters. Oils and fats are esters formed by the reaction between glycerol with fatty acid and stearic acid. Soaps are the salts formed when these react with alkalies.

Chemical Reactions of Organic Compounds Orukkam Questions and Answers

Question 1.
After completing the chemical reactions write down to which category they belong.
a. CH2Cl + Cl2 → …….. + HCl
b. CH = CH+H2 →…………
c. CH4 + 2O2 → …….. + H2O
d. CH3 – CH2 – CH3 → ……….
Answer:
a. CH2Cl + Cl2 → CH2Cl2 + HCl Substitution reaction
b. CH = CH + H2 → CH2 = CH2 Addition Reaction
c. CH4 + 2O2 → CO2 + H2O Combustion
d. CH3 – CH2 – CH3 → CH2 = CH2 + CH4 Thermal cracking

Question 2.
Rearrange the table suitably.
Class 10 Chemistry Kerala Syllabus
Answer:
Hss Live Guru 10th Chemistry Kerala Syllabus
Question 3.
Methane is reacting with Cl in presence of sunlight. Complete equation of that reaction.
Chemistry Textbook Class 10 Kerala Syllabus
a. Write down the reaction of C3H8 with chlorine.
b. What type of reaction is this?
Answer:
Hss Live Guru 10 Chemistry Kerala Syllabus
C3HCl7+Cl2 → C3Cl8 + HCl
b. Subsititution Reaction

Question 4.
Examples of additon reaction are given below, complete the equation.
a. CH2 = CH2 + H2 → ………
b. CH2 = CH + Cl2 → ……..
c. CH = CH + H2 → ……….
d. CH = CH + Cl2→ ………
Answer:
Chemistry Class 10 Kerala Syllabus

Question 5.
a. Examples for combination of Hydrocarbon are given below complete the equation and balance it.
CH4 + O2 → ……. + ……….
C2H6 + O2 → ……. + ………
C3H8 + O2 → …….. + ………
b. Products formed on combustion of Hydrocarbon are ………….
Answer:
a. CH4 + O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
b. Carbon dioxide and Water

Question 6.
a. Name the product and what type of reaction is this?
nCH2 = CHCl → ………..
b. Write down the names of monomer in it.
c. Give examples for natural polymers
Answer:
Hss Live Guru Chemistry 10 Kerala Syllabus
Polyvinyl chloride, polymerization
b. Vinyl chloride
c. Polyisoprene, Protein

Question 7.
Complete the table.
Hsslive Chemistry Class 10 Kerala Syllabus
Answer:
10th Chemistry Notes Kerala Syllabus

Question 8.
Arrange the points in two separate columns write column heading also.
a. CO and H2 are reacted in presence of a catalyst to form compound.
b. Sugar cane juice is fermented.
c. It is known as wood spirit.
d. It is used to make paint & varnish.
e. Used for drinking.
f. It is used for adding in industrial spirit.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 21
Question 9.
a. Ethanol has very large industrial utility. When it enter into our body it creates large amount of problems in our body as well as in our society. List out the probl-em happening in our body and in the society.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 22
b. In industrial ethanol always Methanol is added to prevent misuse by humans. Name the process and what are the side effects formed after consuming it?
Answer:
a.

In human body In society
Liver problems Economic problems
Cliolestrol Family issues
Kidney problems Loses personality

b. Denatured Spirit:

  • Loses eyesight permanently
  • Can lead to death
  • Vomiting

Chemical Reactions of Organic Compounds SCERT Questions and Answers

Question 10.
Analyze the reactions and answer the following questions.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 23
a. Identify A, B.
b. Write the name of the compound ‘a’.
c. Write the name of the reaction by which ‘b’ is formed.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 24
b. Polythene
c. Addition reaction.

Peptide and Protein Molecular Weight Calculator. Peptide or protein molecular weight is an important parameter for Molecular Biology.

Question 11.
Some reactions of propane are given.
i. Hydrogen atoms are substituted one by one, in presence of sunlight.
ii. When heated in the absence of air, it decomposes to hydrocarbons with lesser molecular mass.
iii. Combines with oxygen to give C02 and H2O.
a. Identify the type of reaction in each case.
b. Write the chemical equation of the reaction (ii).
Answer:
a. i. Substitutional reaction
ii. Thermal cracking
iii. Combustion
b. CH3 – CH2 – CH3 → CH2 = CH2 + CH4

The chemical formula calculator is particularly helpful for establishing the percentage of each element.

Question 12.
Analyse the reactions and answer the following questions.
\(\text { i. } \mathrm{CH}_{3}-\mathrm{OH}+\mathrm{CO} \stackrel{\text { Catalyst }}{\longrightarrow} \ldots \mathrm{A}\)
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 25
\(\text { iii. } \mathrm{A}+\mathrm{B} \stackrel{\mathrm{Con} . \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \quad \ldots . \mathrm{C} \ldots .+\mathrm{H}_{2} \mathrm{O}\)
a. Identify A, B, C.
b. What is the general name/ class to which product ‘C’ belongs? Write the IUPAC name.
Answer:
a. A — CH3 – COOH
B – Methanol/CH3 – OH
C – CH3 – COO – CH3
b. Esters, Methyl ethanoate

Question 13.
Analyze the given reactions and answer the following questions.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 26
a. Identify A and B
b. What is the name of reaction by which ‘B’ is formed?
Answer:
a. A – CH2 = CH2, B – CH3 Cl
b. Substitution reaction

Question 14.
Some reactions regarding the production of ethanol are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 27
a. Identify A and B.
b. Write the name of the ester formed when the product B reacts with propanoic acid,
c. Write the chemical equation for the formation of the ester.
Answer:
a. A- C6 H12 O6 B – C2H5 – OH
b. Ethyl Propanoate
c. CH3 – CH2 – COOH + HO – CH2 – CH3
CH3 – CH2 – COO – CH2 – CH3 + H2O

Question 15.
Acetylene (ethyne) is prepared in the laboratory when calcium carbide reacts with water. Write the chemical equations of the reactions for converting it to ethane.
Answer:
CH = CH + H2 \(\stackrel{N i}{\longrightarrow}\) CH2 = CH2
CH2 = CH2 + H2 \(\stackrel{N i}{\longrightarrow}\) CH3 – CH3

Question 16.
Complete the table
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 28
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 29

Question 17.
a Write the structure of the organic comp¬ound with molecular formula QH.
b. What is the name of the compound formed when one hydrogen atom of benzene is replaced with methyl radical?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 30
b. Methyl benzene (Toluene)

Question 18.
Two equations are given below.
i. CH = CH + HCT → ……… A ………
ii. nA → B
a. Identify A and B.
b. Identify the type of reaction (i)?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 31
b. Addition reaction

Question 19.
Three equations are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 32
a. Identify P, Q, R
b. Identify the name of the chemical reaction (ii) and (iii).
c. Write the IUPAC name of R.
Answer:
a. P – CH2 = CH2
Q – CH3 – CH3
R – CH3 – CH2Cl
b. ii. Addition reaction
iii. Substitution Reaction
c. Chloroethane

Question 20.
Ethanol is an industrially important compound.
a. What is the name of 8-10% solution of ethanol?
b. How is it converted into rectified spirit?
c. What is denatured spirit?
Answer:
a. Wash
b. Fractional distillation of wash
c. Product obtained by adding poisonous (methanol, pyridine) substances.

Question 21.
Uses of some important organic compounds are given. Pick out the suitable compounds from the box.
Power alcohol, Teflon, Polythene,
Ethanoic acid, Ethanol
a. For the preparation of rayon,
b. For making the coating of inner surface of non-stick cookware,
c. Solvent in paint industry,
d. As fuel in motor vehicles
Answer:
a. Ethanoic acid
b. Teflon
c. Methanol
d. Power alcohol

Question 22.
Some reactants, products, and names of reactions are given in the table. Complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 33
Answer:
a. Substitution reaction
b. CH = CH2
c. HBr
d. Addition reaction
e. H2O
f. Combustion
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 34
h. Polymerization

Question 23.
Pick out the suitable compounds from the box for the following reactions.
CH4, C2H4, C3H8, CH3Cl
a Thermal cracking
b. Addition Reaction
Answer:
a. C3H8
b. C2H4

Chemical Reactions of Organic Compounds Exam Oriented Questions & Answers

Very Short Answer Type Questions (Score 1)

Question 24.
Equation of some chemical Reactions are given below. Complete it.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 35
Answer:
a. A → CH3 – CH3
B → CH3 – CH2 – Cl
C → HCl
b. A → CH2 = CH2
B → CH3 – CH2 – Cl

Question 25.
Teflon used as nonstick polymer.
a. Write the structure of the monomer of this. Write the IUPAC name.
b. Write the reaction equation for the preparation of the monomer.
Answer:
a. CF2 = CF2 Tetra fluroethene
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 36

Question 26.
Recognize and write A, B, C from following equation.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 37
Answer:
A. CH3 – CH2 – OH
B. CH3 – COOH – OH
C. H2O

Short Answer Type Questions (Score 2)

Question 27.
Some chemical equations are given below. Write each type of chemical reaction.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 38
CH3 – CH = CH2 + CH3 – CH3
Answer:
a. Polymerization
b. Substitution reaction
c. Addition reaction
d. Combustion
e. Thermal cracking

Question 28.
Look at the following reactions on heating pentane.
i. In the absence of air
ii. In the presence of air
a. Write the name of the reaction (i), (ii).
b. Write the chemical equation for the reaction.
Answer:
a. Reaction (i) – Thermal Cracking
Reaction (ii) – Combustion
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 39
ii. for writing the combustion equation for hydrocarbons we can use the equation.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 40

Question 29.
Write each of the following.
a. Molasses
b.Wood spirit
c. Vinegar
d. Esters
Answer:
a. Molasses is mother liquor left after the crystallization of sugar from sugar cane juice.
b. Poisonous chemical methanol (CH3 – OH) is known as wood spirit.
c. 5-8% ethanoic acid (CH3 – COOH) is known as vinegar.
d. esters are salts formed by the reaction ale ‘ whole and organic acids. They have the smell of fruits and flowers.

Short Answer Type Questions (Score 3)

Question 30.
Chloroform can be prepared from Methane.
a. What is the chemical formula of chlor of or m?
b. What is the name of reaction when chloroform is prepared from methane?
c. Write the chemical equation for the reaction.
Answer:
a. CHCl3
b. Substitution Reaction
\(\mathrm{c.} \mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text {sunlight}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\)
CH2 Cl + Cl2 → CH2Cl2 + HCl
CH2 Cl2 + Cl2 → CHCl3 + HCI

Question 31.
Methyl Ethanoate is an ester,
a. Write the structural formula.
b. Write the structural formula of alcohol and carboxylic acid needed for the preparation.
c. Write the reaction equation for the preparation of this ester.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 41

Question 32.
Fill in the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 42
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 43

Question 33.
Write one use of each of the following,
a. Methanol
b. Power alcohol
c. Butane
Answer:
a. Used as solvent in the preparation of paint
b. Fuel in motor vehicle.
c. Cooking gas (LPG)

Question 34.
Write the reason for the following statements.
a. Hydrocarbons like butane are used as fuel.
b. Drinking denatured spirit is harmful
c. Esters are used in perfumes and fruit juice.
Answer:
a. Combustion of hydrocarbon produces plenty of heat.
b. Ethanol on addition with poisonous material is called denatured spirit.
c. Esters have the pleasant smell of flowers and fruits.

Question 35.
Write structural formula of following com-pounds!
a. Benzene
b. Phenol
c. Toluene
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 44

Long Answer Type Questions (Score 4)

Question 6.
Ethyne is a compound that belongs to the class of alkynes.
a. Write chemical formula of ethyne.
b. Write the chemical equation for the preparation of following compounds from ethyne.
i. PVC
ii. 1, 2 – dichloroethane
iii. Chloro ethane
Answer:
a. CH = CH
b. i. CH = CH + HCl →
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 45
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds 46

Question 37.
Molecular formula of some compounds are given in the box.
C2H4 C6H14 CH3 – CH2 – CI
CH3 – COOH C6H6
a. Which is aromatic compound?
b. Which can be prepared by substitution reaction?
c. Which monomer is used in preparation of polythene?
d. Which compound can be used in food?
Answer:
a. C6H6
b. CH3 – CH2 – Cl
c. C2H4
d. CH3COOH

Question 38.
Answer the following:
a. Name the reaction for the conversion of sugar solution into ethanol.
b. Which enzymes are used in this reaction
c. Chemical involved in grape spirit.
d. Name the monomer of P VC.
e. Products formed during cracking of propane.
Answer:
a. Fermentation
b. Invertase, Zymase
c. Ethanol (CH3 – CH2 OH)
d. Polyvinyl chloride (CH2 = CH – Cl)
e. Ethene(CH2 = CH2), Methane (CH4)

Kerala Syllabus 9th Standard Chemistry Solutions Chapter 4 Periodic Table

You can Download Periodic Table Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 4 Periodic Table

Periodic Table Textual Questions and Answers

Earlier Attempts for Classification of Elements

Kerala Syllabus 9th Standard Chemistry Notes Chapter 4 Question 1.
Explain the earlier attempt of classification by Lavoiser?
Answer:
Antoine Lavoisier classified the known elements into metals and nonmetals. But he was not able to duly classify metalloids.

9th Class Chemistry Periodic Table Kerala Syllabus Question 2.
Explain Newland’s law of octaves?
Answer:
Newlands arranged elements in the increasing order of atomic mass. He noticed that every eighth element has properties similar to those of the first elements. But this peculiarity could be noticed in elements upto calcium only.

Octaves of Newlands
Kerala Syllabus 9th Standard Chemistry Notes Chapter 4

9th Class Chemistry 4th Chapter Kerala Syllabus Question 3.
Define Mendeleev’s periodic law?
Answer:
In 1869 Mendeleev arranged the known 63 elements in horizontal and vertical columns and gave shape to the periodic table. He found that the chemical and physical properties of elements repeat at a regular intervals when they were arranged in the increasing order of atomic masses. Based on this Mendeleev proposed the periodic law of elements. The law states that physical and chemical properties of elements are periodic function of their atomic masses.

Chemistry Notes For Class 9 Periodic Table Kerala Syllabus Question 4.
What is meant by groups and periods in the periodic table?
Answer:
The vertical columns in the periodic table are known as groups and the horizontal rows are called periods.

Kerala Syllabus 9th Standard Chemistry Notes Question 5.
Evaluate the Mendeleev’s periodic table and find the following.
a) Total number of periods
b) Total number of Groups
c) Are the same elements showing similar properties arranged in the same group or same period?
Answer:
a) 6
b) 8
c) Same group

Class 9 Chemistry Notes Kerala Syllabus Question 6.
Advantages of Mendeleev’s periodic table?
Answer:
1. For the first time elements were comprehensively classified in such a way that elements of similar properties were placed in the same group. This has made the study of chemistry easy.
2. When the classification was made in such a way that the elements of similar properties came in the same group. It was noticed that certain their proper group. The reason for this was wrongly determined atomic masses and consequently, those wrong atomic masses were corrected.
Eg. The atomic mass of beryllium was known to be 14. Mendeleev reassessed it as a and assigned beryllium a proper place.
3. Columns were left vacant for elements which were not known at the time and their properties were predicted also. This gives an impetus to experiments in chemistry.

Ex Mendeleev give names Eka aluminum and Eka silicon to those elements which were to come below aluminum and silicon respectively in the periodic table and predicted their properties. Later when these elements gallium and germanium were discovered the prediction of Mendeleev turned out to be true.

Hsslive Guru 9th Chemistry Kerala Syllabus Question 7.
Explain the limitation of Mendeleev’s Periodic table?
Answer:
1. Elements with large difference in properties were included in the same group.
eg. Hard metal like copper [Cu], Silver [Ag] were included along with soft metals like sodium [Na] potassium [K],
2. No proper position could be given to element hy¬drogen. Non-metallic hydrogen was placed along with metals like sodium [Na] and potassium [K]
3. The increasing order of atomic mass was not strictly followed throughout.
eg. Co and Ni, Te and I
4. As isotopes are atoms of same element having different atomic masses, they should have been given different position while arranging them in the order of atomic mass. But this was not done.

Counting Atomic Calculator is a free online tool that displays the atomic mass for the given chemical formula.

Modern Period Table

Periodic Table 9th Class Pdf Kerala Syllabus Question 8.
State and explain modern periodic table and mod-ern periodic law?
Answer:
In 1913 Mosely through his x-ray diffraction experiments proved that the properties of elements depended on the atomic number not on the atomic mass.

According to this the periodic law of Mendeleev and the periodic table were modified consequently the modern periodic table was prepared by arranging elements in the increasing order of atomic number. The modern periodic law states that the physical and chemical properties of elements are periodic function of their atomic number.
9th Class Chemistry Periodic Table Kerala Syllabus

Periodic Table Chapter Class 9 Kerala Syllabus Question 9.
How many periods in the modern periodic table?
Answer:
7

Labour India Class 9 Chemistry Kerala Syllabus Question 10.
Which is the shortest period?
Answer:
I period

9th Class Chemistry Chapter 4 Kerala Syllabus Question 11.
Number of elements in the third period?
Answer:
8

Class 9 Chemistry Periodic Table Kerala Syllabus Question 12.
Total number of groups?
Answer:
18

Periodic Table In 9th Class Kerala Syllabus Question 13.
Explain representative elements?
Answer:
Elements of group 1 and 2 also those in groups of 13 – 18 are called representative elements it belongs to metals, nonmetals, and metalloids.

Periodic Table Chemistry Class 9 Kerala Syllabus Question 14.
Do in representative elements do they include metalloids [eg. Si, Ge, As, Sb…) exhibiting the characteristics of metals and non-metals?
Answer:
Yes

Periodic Table Notes Pdf Class 9 Kerala Syllabus Question 15.
As there elements existing in solid, liquid and gaseous state find examples?
Answer:

  • In solid-state- sodium, aluminum, carbon
  • In liquid state – Bromine
  • In gaseous state – oxygen, neon, argon

9th Class Chemistry Chapter 4 Notes Kerala Syllabus Question 16.
Write the electronic configuration of elements with atomic number 1-10
Answer:

Element Atomic number Electronic configuration
Hydrogen 1 1
Helium 2 2
Lithium 3 2, 1
Beryllium 4 2, 2
Boron 5 2, 3
Carbon 6 2, 4
Nitrogen 7 2, 5
Oxygen 8 2, 6
Fluorine 9 2, 7
Neon 10 2, 8

The atom of the elements of these group show the periodically in electron filling they contain 1 -8 electron in their outermost shell. The elements of these groups are called representative elements.

Noble Gases

Kerala Syllabus 9th Standard Notes Chemistry Question 17.
List the elements in group 18
Answer:
Helium, Neon, Argon, Krypton, Xenon, Radon

9 Class Chemistry Chapter 4 Kerala Syllabus Question 18.
Now try to write their electronic configuration
Answer:
2He – 2
10Ne – 2, 8
18Ar – 2, 8, 8
36Kr – 2, 8, 18, 8
54Xe – 2, 8, 18, 18, 8
86Rn – 2, 8, 18, 32, 18, 8

Question 19.
How many electrons are there in the outermost shell of each element?
Answer:
8

Question 20.
The elements do not normally take part in chemical reactions. Find the reason?
Answer:
They have a stable configuration in the outermost shell.

Transition Elements

Question 21.
Which group of elements belong to transition elements?
Answer:
Elements of group 3-12 in the periodic table are transition elements.

Question 22.
Find out whether elements familiar to you are present in these groups?
Answer:
Copper, silver, gold, iron

Question 23.
Aren’t transition elements metals?
Answer:
Yes

Question 24.
What are the characteristics of transition elements?
1. They from coloured compound,
2. They show similarity in properties as well as in a period.
3. In compounds, they exhibit different oxidation state
eg. Fe2+ and Fe3+

Lanthanides and Actinoids

Question 25.
Which element is next to lanthanum with atomic number 57 of group 6 in the periodic table?
Answer:
Cerium with atomic No. 58

Question 26.
Find out the position allotted to the elements with atomic number 58-71?
Answer;
Separate position at the bottom of the periodic table.

Question 27.
Is the same way aren’t the elements with atomic number 90 to 103 of period 7 give separate positions at the bottom of the periodic table?
Answer:
Yes. These elements are called inner transition elements.

Question 28.
What is meant by inner transition elements?
Answer:
Inner transition elements from Cerium [Ce] to Lutecium [Lu] of period 6 are called lanthanides. Inner transition elements from Thorium (Th] to Lewrencium [Lr] of period 7 are called actinoids. Lanthanoids are also called rare earth. Actinoids are man-made artificial elements (except thorium and uranium).

Periodic trends in the periodic table

Question 29.
Electronic configuration of group I elements of the periodic table are given
Answer:

Element Atomic number Electron configuration Group Period
H 1 1 1 1
Li 3 2, 1 1 2
Na 11 2, 8, 1 1 3
K 19 2, 8, 8, 1 1 4
Rb 37 2, 8, 18, 8, 1 1 5
Cs 55 2, 8, 18, 18, 8, 1 1 6
Fr 87 2, 8, 18, 32, 18, 8, 1 1 7

Question 30.
What is the peculiarity seen in the electronic configuration of the outer most shell of these elements?
Answer:
All these elements we can see one electron in the outermost shell.
Hence elements of group I exhibit similarity in chemical properties.

Question 31.
Which are the electrons shows the chemical properties of elements?
Answer:
Outermost electrons.

Question 32.
Is there any relationship between the group number and the number of electrons present in the outermost shell? What is it?
Answer:
Same, group number equal tot he number of election in the outermost shell for the elements in groups 1 and 2.

Question 33.
Observe figure the electronic configuration of the second-period elements of the group from 13-18 given below.
9th Class Chemistry 4th Chapter Kerala Syllabus

(i) Won’t we get the group number of these elements by adding 10 to the number of elements by adding 10 to the number of electrons in the outermost shell?
Answer:
Yes

(ii) Analyze table 3.1 and find whether there is any relation between the number of shells in an atom and the number of periods?
Answer:
Number of shells in an atom and the period number is same.

Size of an Atom in Group

Question 34.
Are you familiar with the Bohr model of an atom? See the Bohr model of atoms of certain elements, in group I.
Chemistry Notes For Class 9 Periodic Table Kerala Syllabus
Kerala Syllabus 9th Standard Chemistry Notes

(i) Which among them is the biggest?
Answer:
Potassium (K)

(ii) Which one is the smallest?
Answer:
Hydrogen [H]

(iii) What happens to the size of an atom when we move down the group?
Answer:
Increases

(iv) What is the reason for this?
Answer:
Number of shells increases.
As we move from top to bottom of a group in the periodic table the size of the atom increases as there is an increase in the number of shells.

Atomic Size in Period

See (Fig 3.3) the representation of Bohr model of elements with atomic numbers 3 to 9 in the second period of the periodic table.
Class 9 Chemistry Notes Kerala Syllabus

Question 35.
Is there are in the number of shells with the increase in atomic number?
Answer:
No.

Question 36.
What happens to the nuclear charge with increase in atomic number?
Answer:
On moving from left to right in a period, as nuclear charge increases, the force of attraction on the outer-most electrons increases and consequently the size of atom decreases.

Ionisation Energy

Question 37.
You have understood how sodium chloride is formed by combining sodium and chlorine atoms. The Bohr model of sodium and chlorine are given below
Answer:
Hsslive Guru 9th Chemistry Kerala Syllabus

(i) Which among these atoms lose electrons?
Answer:
Sodium atom

(ii) Which one gains electrons
Answer:
Chlorine atom

Question 38.
How the ions are formed?
Answer:
Atom becomes charged when there is transfer of electrons [Lose or gain electrons] they are called ions.

Question 39.
Define ionization energy?
Answer:
The amount of energy required to liberate the most loosely bound electrons from the outermost shell of an isolated gaseous atom of an element is called ionization energy.

Question 40.
What are the factors affecting the ionization energy? Nuclear charge
Answer:
Size of the atom

Question 41.
When the size of an atom increases, does the attraction of the nucleus on the outermost electron increase or decrease?
Answer:
Decrease

Question 42.
Then what is the change in ionization energy?
Answer:
As the size of atom increases ionization energy decreases.

Question 43.
Can you find out how ionization energy changes as we move from top to bottom in a group?
Answer:
Ionization energy decreases.

Question 44.
What is the general trend in the variation of ionization energy on moving across a period from left to right?
Answer:
Ionization energy increases

Question 45.
Find how ionization energy changes with increase in nuclear charge?
Answer:
On moving from left to right in a period, as nuclear charge increases, the size of the atom decrease hence ionization energy increase.

Question 46.
Define electronegativity?
Answer:
In the case of two atoms joined by a covalent bond, electronegativity is the ability of each atom to attract the bonded electrons.

Question 47.
How size of an atom influence the electronegativity?
Answer:
As the size of an atom increases the distance between the nucleus and the outermost electron increases, hence the electronegativity decreases. As we move in the same period form left to right size of atom decrease hence electronegativity increases.

Question 48.
What is the basis for the chemical properties of metals and non-metals?
Answer:
Metals are the elements which give away the electrons and those that accept electrons are generally non-metals. Metals are electropositive elements because they lose electrons to form positive ions. Non-metals are called electronegative elements because they gain electrons in chemical reactions to form negative ions.

Question 49.
What is relationship between metallic character and the size of an atom?
Answer:
As the size of the atom increases metallic character also increases.

Question 50.
How do the metallic character and nonmetallic character vary while moving from left to right in a period? Arriving at a conclusion by assessing the size of atom?
Answer:
In the periodic table, while moving from to top to bottom in groups metallic character generally in-creases while non-metallic character decreases.
In a period as we move form left to right metallic character generally decreases while non-metallic character increases.

Question 51.
Don’t you think that there is a relationship between ionization energy and metallic -non-metallic character? Is the element with highest ionization energy metallic or non-metallic?
Answer;
Non-metallic

Question 52.
Then what about those having the low ionization energy?
Answer:
Metals

Question 53.
Isn’t there a relationship between electronegativity and metallic, non-metallic character? Explain the relationship?
Answer:
Non-metals are more electronegative.

Metalloids

Question 54.
Explain metalloids?
Answer:
Elements exhibiting the properties of both metal as well as nonmetal are called metalloids, eg. Silicon [Si], germanium [Ge] Arsenic [As], Antimony [Sb] and Tellurium [Te] belongs to this category.

Question 55.
You must understand certain periodic trends in the periodic table? Based on these (✓)the correct option given below in table 3.7.
Periodic Table 9th Class Pdf Kerala Syllabus
Answer:

Trends In a group from  top to bottom In period from  left to right
Size of atom ✓ Increases/  decreases Increases/  decreases ✓
Metallic character ✓ Increases/ decreases Increases/ decreases ✓
Non-metallic character Increases/ decreases ✓ ✓ Increases/ decreases
Ionization energy Increases/ decreases ✓ ✓ Increases decreases
Electronegativity Increases/ decreases ✓ ✓ Increases/ decreases

Let Us Assess

Question 1.
The table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Fill in the blanks.
Periodic Table Chapter Class 9 Kerala Syllabus
Answer:

Contribution/Findings Name of Scientist
Triads Dobereiner
Law of octaves Newlands
Classification of elements into metals and non-metals Antonie Lavoisier
Modern periodic law Henry Moseley

Question 2.
Complete the table
Answer:

Element Atomic number Electronic  configuration Group  number Period  number
Lithium 3 2,1 1 2
Oxygen 8 2J3 16 2
Argon 18 2,8,8 18 3
Calcium 20 2,8,8,2 2 4

Question 3.
Symbols of certain elements are given. Write their electronic configuration and find the period and group in which they are included.
Labour India Class 9 Chemistry Kerala Syllabus
Answer:
a) \(_{6}^{12} C\)
Electronic configuration 2, 4
Period – 2
Group – 14
b) \(_{12}^{24} \mathrm{Mg}\)
Electronic configuration 2,8,2
Period-3
Group – 2
c) \(_{17}^{35} \mathrm{Cl}\)
Electronic configuration 2,8,3
Period – 3
Group – 17
d) \(_{13}^{27} \mathrm{Al}\)
Electronic configuration 2,8,3
Period – 3
Group – 13
e) \(\begin{array}{l}{20} \\ {10}\end{array} \mathrm{Ne}\)
Electronic configuration 2,8
Period – 2
Group -18

Question 4.
There are three shells in the atom of element ‘X’, 6 electrons are present in its outermost shell.
a) Write the electronic configuration of the element.
b) What is its atomic number?
c) In which period does this element belong?
d) In which group is this element included?
e) Write the name and symbol of this element.
f) To which family of element does is this element belong to?
g) Draw and illustrate the Bohr atom model of this element.
Answer:
a) 2, 8, 6
b) 16
c) 3
d) 16
e) Sulphur, ‘S’
f) Oxygen family
9th Class Chemistry Chapter 4 Kerala Syllabus

Question 5.
Electronic configurations of elements P, Q, R, and S are given below. (These are not actual symbols).
P – 2, 2
Q -2, 8, 2
R – 2, 8, 5
S – 2, 8
a) Which among these elements are included in the same period?
b) Which are those included in the same group?
c) Which among them is a noble gas?
d) To which group and period does the element R belong?
Answer;
a) P and S, Q and R – belongs to same period
b) P and Q, belongs to same group
c) S
d) R belongs to 3rd period and 15th group

Question 6.
An incomplete form of the periodic table is given below. Write answers to the questions connecting the position of elements in it.
Class 9 Chemistry Periodic Table Kerala Syllabus
a) Which is the element with the biggest atom in group 1?
b) Which is the element having very lowest ionization energy in group 1?
c) Which element has the smallest atom in period 2?
d) Which among them are transition elements?
e) Which of the elements L and M has the lowest electronegativity?
f) Among B and I which has higher metallic character?
g) Which among these are included in the halogen family?
h) Which is the element that resembles E the most in its properties?
Answer:
a) D
b) D
c) M
d) G, H
e) L
f) B
g) M, N
h) F

Periodic Table Model Questions and Answers

Question 1.
Symbols of certain elements are given write down the electronic configuration and find the period and group in which they are included.
Periodic Table In 9th Class Kerala Syllabus
Answer:
a) \(\begin{array}{l}{23} \\ {11}\end{array} \mathrm{Na}\)
Electron configuration 2, 8, 1
Period – 3
Group – 1
b) \(_{17}^{35} \mathrm{Cl}\)
Electron configuration -2, 8, 7
Period – 3
Group-17
c) \(_{9}^{19} \mathrm{F}\)
Electron configuration -2, 7
Period – 3
Group – 17

Question 2.
A, B, C, D are four elements. The electronic configuration is given below and find the answers in the following (Hint. The symbols are not real)
A – 2, 2
B – 2, 8, 5
C – 2, 7
D – 2, 8, 2
a) Find the elements belongs to same period?
b) Find the elements belongs to same group.
c) ‘C’ belongs to which period and group?
Answer:
a) B, D, and A, C because the number of shells are same.
b) A, D because the number of electrons in the outermost shell is same.
c) ‘C’ belongs to second period and 17th group

Question 3.
Table given below lists the contributions and names of scientists who made earlier attempts in the classification of elements. Make them in the correct order.

Contribution/Findings Name of scientist
Octet Rule John Dalton
Triads New Lands
Modern periodic table Lavoisier
Classify into metals Henry Moseley
Non-metals
Atomic theory Dobereiner

Answer:

Contribution/Findings Name of Scientist
Triads Dobereiner
Law of octaves Newlands
Classification of elements into metals and non-metals Antonie Lavoisier
Modern periodic law Henry Moseley

Question 4.
An incomplete form of the periodic table is given below.
write answers in the question connecting the position of elements in it.
Periodic Table Chemistry Class 9 Kerala Syllabus
1. Which element has the largest atomic size in group I?
2. Write the transition elements?
3. Which element has the lowest ionization energy in the 2nd period?
4. Which element belongs to Noble gas elements?
5. Compare L, M which element has the lowest electronegativity?
6. Write the element belongs to Halogen family?
Answer:
1. D
2. F, G, H
3. C
4. N
5. L
6. M

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept

Gas Laws Mole Concept Text Book Questions and Answers

Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th Text Book Page No: 33

→ Complete the table 2.1
Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th
Answer:
Sslc Chemistry Chapter 2 Kerala Syllabus

→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter

→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases

→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases

Use the Molarity Calculator Chemistry to calculate the mass, volume or concentration required to prepare a solution of compound of known molecular weight.

Sslc Chemistry Chapter 2 Kerala Syllabus  Text Book Page No: 34

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.

→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy

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→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased

→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase

→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.

→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)

→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature

Sslc Chemistry Chapter 2 Questions And Answers Text Book Page No: 35
Sslc Chemistry Chapter 2 Questions And Answers

→ Is there any change in the number of molecules?
Answer:
No

→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases

→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions

Mass to Moles Calculator — The quantity of substance n in moles is equal to the mass m in grams divided by the molar mass M in g/mol.

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→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.

→ What about decreasing the pressure?
Answer:
Volume increasing

→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.

Gas Laws And Mole Concept Kerala Syllabus 10th Text Book Page No: 36
Gas Laws And Mole Concept Kerala Syllabus 10th

→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

→ What do you observe?
Answer:
Ink rises through the tube

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→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.

→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

→ Complete the table 2.2
Sslc Chemistry Chapter 2 Notes Kerala Syllabus
Answer:
Kerala Syllabus 10th Standard Chemistry Chapter 2
Sslc Chemistry Chapter 2 Notes Kerala Syllabus Text Book Page No: 37

→ In which unit is the temperature stated?
Answer:
Kelvin (K)

→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.

→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.

→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased

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→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.

→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases

→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.

→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.

Kerala Syllabus 10th Standard Chemistry Chapter 2 Text Book Page No: 38

→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g

→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000

→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.

→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes

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→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.

→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12th mass of carbon atom.

Enter the formula and press “calculate” to work out the molar mass calculator with steps, the number of moles in 1 g and the percentage by mass of each element.

Chemistry Class 10 Chapter 2 Kerala Syllabus Text Book Page No: 39

→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms

→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.

→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 1023 oxygen atoms

→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g

Gas Laws And Mole Concept Extra Questions 10th Text Book Page No: 40

→ Complete the table 2.5
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 7
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 8
→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 1023 sodium atoms.

Gas Laws And Mole Concept Notes Pdf Kerala Syllabus 10th Text Book Page No: 41

→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 1023

2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 1023

→ Complete the table 2.6
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 9
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 10

→ Calculate the molecular mass of glucose (C6 H12 O6) and sulphuric acid (H2 SO4)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g

Gas Laws And Mole Concept Questions And Answers Pdf 10th Text Book Page No: 42

→ Complete the table 2.7
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 11
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 12
→ What is the molecular mass of oxygen?
Answer:
32 g

→ How many GMM is present in 32g oxygen?
Answer:
1 GMM

→ How many molecules are present in it?
Answer:
6.022 × 1023 oxygen molecules

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→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM

→ How many molecules are present in N2?
Answer:
6.022 × 1023 N2, molecules

→ How many GMM is present in 18 gm water?
Answer:
1 GMM

→ How many H2O molecules arf present in it?
Answer:
6.022 × 1023 H2O Molecules

→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM

Sslc Chemistry Chapter 2 Notes Pdf Kerala Syllabus Text Book Page No: 43

How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?

→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 1023

→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 1023

→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 1023 water molecules

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→ What is its mass?
Answer:
18 g

→How many GMM is present in it?
Answer:
1 GMM

Sslc Chemistry 2nd Chapter Notes Kerala Syllabus Text Book Page No: 44

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 13

22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole

Gas Law And Mole Concept Kerala Syllabus 10th Text Book Page No: 45

Complete the flow chart given below, related to one mole of substance.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 14
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 15

Gas Laws Mole Concept Let Us Assess

Gas Laws And Mole Concept Notes Kerala Syllabus 10th Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law

Gas Laws And Mole Concept Pdf Kerala Syllabus 10th Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law

Gas Laws And Mole Concept Class 10 Kerala Syllabus Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 16
a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 17
b. Avogadro’s law

Hss Live Guru 10th Chemistry Kerala Syllabus Question 4.
a. Calculate the mass of 112 L CO2 gas kept at STP (molecular mass = 44)
b. How many molecules of CO2 are present in it?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 18

Sslc Chemistry Chapter 2 Gas Laws And Mole Concept Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L

Hsslive Chemistry 10th Kerala Syllabus Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N2=28g, H2O= 18g)
a. 56g N2
b. 90g H2O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2

b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5

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10th Chemistry 2nd Chapter Kerala Syllabus Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM

b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10

c. Number Of molecules = Mole × 6.022 × 1023
= 10 × 6.022 × 1023

Class 10 Chemistry Chapter 2 Kerala Syllabus Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O2
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O2 =2 × 16 = 32 g

b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 1023
= 2 × 6.022 × 1023
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O2) =2
∴ total number of atoms = 2 × 6.022 × 1023 × 2
=4 × 6.022 × 1023

Gas Laws Mole Concept Extended activities

Hss Live Guru Chemistry 10 Kerala Syllabus Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 1023
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 1023
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)

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Kerala Syllabus 10th Standard Chemistry Guide Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH3 at STP
c. 67.2 L of N2 at STP
d. 1 mol of H2SO4
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 19
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 20

Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 1023 = 5 × 6.022 × 1023 18

b. ∴ Number of atoms = 3 × 5 × 6.022 × 1023 = 15 × 6.022 × 1023

c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H2O molecule? 10
Total electrons in 90 g H2O = 10 × 5 × 6.022 × 1023 = 50 × 6.022 × 1023

Gas Laws Mole Concept Orukkam Questions and Answers

Question 1.
a. Complete the table based on the data given in the box.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 21
b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 22
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 23

b.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 24

Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 25
a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H2SO4 = …………. g ………… Molecules
1/2 Mole of Al2O3= …………. g ………….. Molecules
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 26
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 27

a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 1023 Molecules
5 mole of CaO = 280g, 5 × 6.022 × 1023 Mol-ecules
2 Mole of H2SO4 = 196g, 2 × 6.022 × 1023 Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 1023

Question 3.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 27
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 29
Question 4.
Complete the data.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 30
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 31
Question 5.
Based on the reaction given below, write the answers for the questions.
N2 + 3H2 → 2NH3 ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen

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Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH4+O2 → CO2 + H2O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C2H6 + 7O2 → 4CO2 + 6H2O
b. Based on the equation above, How many moles of CO2 is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO2 gas is formed.
When 20 moles methane burn 20
moles of CO2 are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g

b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO2 when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g

Question 7.
Based on the given equation write down the answers.
2H2+ O2 → H2O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g

b. 0.5 Mole

Gas Laws Mole Concept Evaluation Questions

Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC1 contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC1 = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol

Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.

Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO3?
b. How many moles of Oxygen present in 1000gms of CaCO3?
Answer:
a. Mass of 1 mole of CaCO3 = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO3 = \(\frac { 1000 }{ 100 }\) = 10 mol

b. No of moles of Ca in 10 moles CaCO3 =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO3 = 10 × 3 = 30mol.

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This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.

Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H2O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol

Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.

b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.

Finally, you encounter how to find molar concentration step-by-step manually, and if your preference indulges with instant calculations.

Question 6.
How many moles of Cl2 present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.

Question 7.
Find out the mass of Oxygen atom in 44.8L of CO2 in STP.
Answer:
No of moles in 44.8 litre of CO2 \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO2 contains 1 mole of C and 1 mole of O2
∴ 2 Mole of CO2 contains 2 mole of O2 or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g

Question 8.
Find out the amount of CO2 formed when the burning of one mole of Ethane.
Answer:
2 moles of CO2 is formed when 1 mole of ethane burns.
Mass of 2 moles of CO2 = 2 × 44 = 88g

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Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)

Gas Laws Mole Concept SCERT Questions and Answers

Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 1023
b. i.6.022 × 1023
ii. 2 x 6.022 × 1023
iii \(\frac { 32 }{ 12 }\) × 6.022 × 1023

Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen

b. A, D – 2 × 6.022 × 1023
B,C – 6.022 × 1023

Question 3.
N2 + 3H2 → 2NH3
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 32

Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 33
Answer:
a. 1:3
b. a – 2 NH3,
b – l H2,
c – 12H2,
d – 4NH3

Question 4.
2H2 + O2 → 2H2O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O2 molecules are required to react 100 H2 molecules completely?
c. How many water molecules are formed when 1000 H2 molecules are reacted completely
Answer:
a. 2:1
b. 50 O2 molecules
c. 1000 H2O molecules

Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 34
Answer:
a. 6.022 × 1023
b. 6.022 × 1023
c. 71 g
d. 14 g

Question 6.
A sample of substances are given.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 35
Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH3, 128gO3
b. 49 g H2SO4

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Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 1023 chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 1023 H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 1023
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 1023 H2O molecules.

Question 8.
Complete the following.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 36
Answer:
a. 2 × 6.022 × 1023
b. 1GMM
c. 6.022 × 1023

Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO2 present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO2=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO2 at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 1023

Question 10.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 37
Answer:
a. 2
b. 2 × 6.022 × 1023
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L

Question 11.
CH4 + 2O2 → CO2 + 2H2O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH4?
b. Calculate the amount of CO2 formed when 100g of CH4 is completely burnt?
Answer:
a. 2 mol
b. Amount of CO2 produced by the combustion of 16g CH4 = 44g
Amount of CO2 produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO2 produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g

HSSLive.Guru

Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose

Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 38
a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO2 = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO2 = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 1023

Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H2(g) + O2(g) → 2H2O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g

Question 15.
Complete the table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 39
Hint: (MM – CO2 = 44, CH4 = 16, SO2 = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.

Question 16.
Analyse the following equation
2NO(g) + O2(g) → 2NO2(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO2 formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O2(g) → 2NO2(g) (2 : 1: 2)
No. of moles in 112L O2 = 5 mol
According to equation no. of moles of NO2 obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO2 obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO2 = 10 × 46 = 460g

HSSLive.Guru

Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO3→ CaO + CO2
(HintMM: CaCO3 – 100, CaO – 56, CO2 – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO3 → CaO + CO2
100g 56g 44g
I I : I
Amount of CaCO3 required to get 56g of CaO = 100g
Amount of CaCO3 required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO3 required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 1023

Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g

Gas Laws Mole Concept Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 1023 atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g

Question 2.
Number of molecules of substance is 3.011 × 1024.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 1024 molecules.
Answer:
a. 6.022 ×1023
b. Number of moles of molecules = \(\frac { Number of molecules }{ NA }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)

Short Answer Type Questions (Score 2)

Question 3.
Fill the patterns.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 40
Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass

Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct

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Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, PV = a constant
Therefore, P1 V1 = P2 V2
Here, P1 = 2atm V1 = 20L P2 = 3atm V2=?
∴ 2 × 20 = 3 × V2
Thus, V2 = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L

Charles Law Calculator is a free online tool that displays the volume of gas that tends to expand when heated.

Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 41

Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g

Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L

Question 9.
Look at the balanced equation given.
2NaOH + CO2 → Na2 CO3 + H20
a. Find out the mass of NaOH needed for 264 g CO2 to react completely.
b. Find out the total number of moles of water molecules when CO2 reacts.
Answer:
a. GMM of CO2 = 44 g
∴Number of moles in the molecule of 264 g CO2 = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO2= 2 moles
∴ NaOH needed for the reaction of 6 moles CO2 = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g

b. H2O formed when lmole CO2 reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO2 reacts = 6 moles

HSSLive.Guru

Question 10.
2C4H10 + 13O2 → 8CO2 + 10H2O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO2 in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C4H10) = 14 kg = 1400g
GMM of C4H10 = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO2when 2 moles of C4 H10 ignites = 8 moles of C4H10 ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO2 formed in STP
= 965.52 × 22.4 L = 21627.65 L

Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.

Question 12.
The molarity of250 ml of Na2CO3 solution is 0.5 M. Find the mass of Na2CO3.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 42
Question 13.
63 g HNO3 is in the dilute solution of 200 ml HNO3 (Nitric acid). Find the molarity.
Answer:
a.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 43

Short Answer Type Questions (Score 3)

Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. Pv = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct

b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law

Question 15.
In 100 g of CaCO3
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 44

Question 16.
Fill the blanks in the given table.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 45
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 46
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 47
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 48

Long Answer Type Questions (Spore 4)

Question 17.
See CO2 gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO2 molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)

HSSLive.Guru

Question 18.
The molecular formula of ammonium sulfate is (NH4)2SO4.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH4)2SO4
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 49
Question 19.
Fill in the blanks of the table given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 50
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 51

Question 20.
See the diagram given below.
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 52
Answer:
a. 196g
b. 2 × 6.022 × 1023
c. 2 GMM
d. 2 × 6.022 × 1023

Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 53
Question 22.
Certain compounds and its masses are given,
i) 68 g NH3
ii) 28 g N2
iii) 9 g H2O
iv) 128 g O2
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 54
Kerala Syllabus 10th Standard Chemistry Solutions Chapter 2 Gas Laws Mole Concept 55

Question 23.
368 g NO 2gas is given. Find the answers of each one given below,
a. GMM of NO2
b. Number of moles of molecules of 368 g NO2
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO2
= 14 + 2 × 16 = 46 g

b. Number of mole = \(\frac { 368 }{ 46 }\) = 8

c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 1023

d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 1023 = 24 × 6.022 × 1023

e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L

HSSLive.Guru

Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H2)
b. 88.75 g Chlorine (Cl2)
c. 4 g Calcium (Ca2) ,
d. 7.75 g phosphorus (p4)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 1023
Total number of atoms = 2 × 10 × 6.022 × 1023
= 20 × 6.022 × 1023

b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 1023
Total number of atoms = 2 × 1.25 × 6.022 × 1023
= 2.5 × 6.022 × 1023

c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 1023
Total number of atoms = 1 × 0.1 × 6.022 × 1023
= 0.1 × 6.022 × 1023

d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 1023

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Students can Download Chapter 2 Solutions Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Plus Two Chemistry Solutions One Mark Questions and Answers

Question 1.
Which of the following is a liquid in solid type solution?
(a) glucose dissolved in water
(b) Camphor in N2
(c) amalgam of Hg with Na
(d) Cu dissolved in Au
Answer:
(c) amalgam of Hg with Na

Question 2.
The concentration term used when the solute is present in trace quantities is _______
Answer:
ppm (parts per million)

Question 3.
A binary solution of ethanol and n-heptane is an example of
(a) ideal solution
(b) Non-ideal solution with +ve deviation
(c) Non-ideal solution with -ve deviation
(d) unpredictable behaviour
Answer:
(b) Non-ideal solution with +ve deviation

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 4.
A solution which has higher osmotic pressure as compared to other solution is known as _____
Answer:
Hypertonic

Question 5.
Which of the following solutions will have the highest boiling point at 1 atm pressure?
(a) 0.1M FeCl3
(b) 0.1MBaCl2
(c) 0.1MNaCl
(d) 0.1Murea
Answer:
(a) 0.1M FeCl3

Question 6.
1 kilogram of sea water sample contains 6 mg of dissolved O2. The concentration of O2 in the sample in ppm is
Answer:
6.0

Question 7.
The amount of solute (molar mass 60 g mol-1) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer:
60 g

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Pure Aqua provides users with an osmotic pressure calculator to gain better insight on your osmotic pressure requirements.

Question 8.
The correct equation for the degree of association (a) of an associating solute ‘n‘ molecules of which undergoes association in solution is
Answer:
a = \(\frac{n(i-1)}{1-n}\)

Question 9.
A solution is prepared by dissolving 10 g NaOH in 1250 ml of a solvent of density 0.8 ml/g. The molarity of the solution is _______
Answer:
0.25

Question 10.
If the elevation in boiling point of a solution of non volatile, non electrolyte in a solvent (Kb = xk. kg mol-1) is 7 K, then the depression in freezing point would be kf = ZK kg mol-1
Answer:
\(\frac{Y Z}{x}\)

Plus Two Chemistry Solutions Two Mark Questions and Answers

Question 1.
Match the terms of list A with those in list B.

A B
Raoult’s Law. Colligative property.
Henry’s Law. Ideal solution.
Elevation of boiling point. Solution of gases in liquids.
Benzene + Toluene. Vapour pressure of solutions.

Answer:

A B
Raoult’s Law. Vapour pressure of solutions.
Henry’s Law. Solution of gases in liquids.
Elevation of boiling point. Colligative property.
Benzene + Toluene. Ideal solution.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 2.
At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm of Hg respectively. Two moles of A and 3 Moles of B are mixed to form an ideal solution. What is the vapour pressure for the solution?
Answer:
A = 120mm of Hg
B= 180 mm of Hg
χA = 2/5
χB =3/5
PA = P°A × χA
= 120 × 2/5 = 48 mm of Hg
PB = P°B × χB
= 180 × 3/5 = 108 mm of Hg
Ps = PA + PB
= 108 + 48 = 156 mm of Hg

Question 3.
Find the volume of H2O that should be added to 300 mL of 0.5 M NaOH so as to prepare a solution of 0.2 M.
Answer:
M1V1 = M2V2
300 × 0.5 = V2 × 0.2
V2 \(\frac{300 \times 5}{2}\) = 750 mL
H2O to be added = 750 mL – 300 mL = 450mL

Question 4.
Calculate the osmotic pressure of 5% solution of urea at 27°C?
Answer:
Mass of urea, WB = 5 g
R = 0.0821 L atm K-1 mol-1
T = 273 + 27°C = 300 K
Molecular mass of urea, MB = 60 g mol-1
V = 100 mL = 0.1L
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions two marks q4 img 1
= 20.53 atm

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 5.
Osmotic pressure of 1M solution of NaCl is approximately double than that of 1M sugar solution. Why?
Answer:
Osmotic pressure is a colligative property and it depends on the number of solute particles present in the solution. In solution, each NaCl unit undergoes dissociation to form two particles (NaCl → Na+ + Cl) and hence osmotic pressure of 1M NaCl is twice that of 1M sugar solution. Sugar molecules does not undergo association or dissociation in solution.

Question 6.
What do you mean by ideal solution?
Answer:
Ideal solution is a solution which obeys Raoult’s law over the entire range of concentration and temperature, i.e., for an ideal solution having two volatile components A and B.
PA = P°A χA, PB = P°B χB,
Ps = PA + PB =P°A χA + P°B χB

Question 7.
Many countries use desalination plants to meet their potable water requirements. Comment on the phenomenon behind it.
Answer:
This is based on reverse osmosis. When a pressure more than osmotic pressure is applied, pure water is squezed out of the sea water through a semi-permeable
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions two marks q7 img 2

Question 8.
Movement of solvent molecules through a semipermeable membrane from pure solvent to the solution side is called osmosis.
What are the following

  1. Isotonic solution
  2. Hypertonic solution?

Answer:
1. Isotonic solution:
If the osmotic pressure of the two solutions are equal, they are called isotonic solutions.

2. Hypertonic solution:
A solution having higher osmotic pressure than another solution is called hypertonic solution.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 9.
What is meant by azeotrope?
Answer:
Liquid mixtures which distil without change in composition are called azeotropic mixtures or azeotropes.

Plus Two Chemistry Solutions Three Mark Questions and Answers

Question 1.
A student was asked to define molality. Then he answered that it is the number of gram moles of the solute dissolved per litre of the solution.

  1. Is it correct?
  2. Can you help the student to define molality?
  3. Calculate the molality of a solution containing 20 g of NaOH in 250 g of H2O.

Answer:

  1. It is not correct.
  2. It is the number of moles of the solute present in 1000 g of the solvent. Molality can be determined by using the formula.
    Molality, m = \(\frac{\text { Mass of the solute in gram } \times 1000}{\text { Molar mass of the solute } \times \text { Mass of the solvent in gram }}\)
  3. Molality, m = \(\frac{20 \times 1000}{40 \times 250}\) = 2 m

Question 2.
The graph of non-ideal solution showing -ve deviation as drawn by a student is given below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 3

  1. Is this diagram correct?
  2. Substantiate your argument with a suitable example.

Answer:

  1. No.
  2. Because this graph shows the non-ideal solution showing +ve deviation and not -ve deviation. Consider a solution obtained by mixing chloroform and acetone. Here chloroform molecule is able to form hydrogen bond with acetone molecule as shown below:

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 4
As a result of this, vapour pressure of the solution decreases. Due to this, boiling point increases. The volume of the solution is less than the expected value. The mixing process is exothermic. So the actual graph is as given below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 5

Question 3.
Some words are missed in the following paragraph. Add suitable words in the blanks:
If osmotic pressure of 2 solutions are equal they are called ______(a)_____ solution. The solution which is having ______(b)______ osmotic pressure is called hypertonic solution and the solution which is having lower osmotic pressure is called ______(c)_____ solution.
Answer:

  1. Isotonic
  2. Higher
  3. Hypotonic

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 4.
The solubility of gases depends upon some factors.

  1. Can you suggest the factors?
  2. Which is the law behind it?
  3. What are the limitations of this law?

Answer:

  1. Nature of the gas, Nature of the solvent, Pressure, Temperature
  2. Henry’s law. It states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure applied on it.
  3. Henry’s law is valid only under the following conditions:
    • The pressure of the gas is not too high.
    • The temperature is not too low.
    • The gas is not highly soluble.

Question 5.
A solution of 12.5 g of an organic solute in 170g of H2O a boiling point elevation of 0.63 K. Calculate the molecular mass of the solute (K2=0.52 K/m).
Answer:
Kb for water = 0.52 K Kg mol-1
WA = 170g
ΔTb = 0.63 K
WB = 12.5 g
MB = ?
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q5 img 6

Question 6.
Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H20. (Kf for H2O = 1.86 K/m).
Answer:
Mass of glucose, WB = 3.6 g
Molecular mass of glucose = 180 g mol-1
Mass of solvent, WA = 50 g
Kf for H2O = 1.86 K/m = 1.86 K kg-1 mol-1
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q6 img 7
i.e., ΔTf = T°f – Tf= 0.744 K
∴ Freezing point of the solution, Tf = T°f – ΔT
= 273 K – 0.744 K = 272.3 K

Question 7.
Raw mangoes shrivel when pickled in brine solution.

  1. Name the process behind this.
  2. Define that process.

Answer:

  1. Osmosis
  2. When a solution is separated from its solvent by a semipermeable membrane, the solvent flows into the solution through the semipermeable membrane. This process is called osmosis.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 8.
200 cm3 of an aqueous of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q8 img 8

Question 9.
What type of deviation from Rauolt’s law is exhibited by a mixture of phenol and aniline? Explain with the help of graph.
Answer:
Negative deviation.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q9 img 9
In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules. Thus, escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

Question 10.
Wilted flowers revive when placed in fresh water.

  1. Identify and define the phenomenon.
  2. Mixing of acetone and CHCl3 results in a reduction in volume. What type of deviation from Raoult’s law is observed here? Give reason.
  3. Benzoic acid dissolved in benzene shows double of its molecular mass. Explain.

Answer:
1. Osmosis. It is the phenomenon of flow of solvent from pure solvent into a solution or from a solution of lower concentration into a solution of higher concentration through a semi-permeable membrane.

2. Negative deviation. Here chloroform molecule is able to form hydrogen bond with acetone molecule.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q10 img 10
Thus escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

3. Benzoic acid undergoes association in solution.
2C6H5COOH \(\rightleftharpoons \) (C6H5COOH)2 Thus, the number of particles as well as colligative properties decreases. So molecular mass increases.

Question 11.
A solution is obtained by mixing 300 g of 25 % solution and 400 g of 40 % solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
Mass of solute in 300 g of 25 % solution
\(\frac{300 \times 25}{100}\) = 75 g
Mass of solute in 400 g of 40 % solution
\(\frac{400 \times 40}{100}\) = 160 g
Total mass of solute = (75 + 160) g = 235 g
Total mass of solution = (300 + 400) g = 700 g
Mass % of solute in resulting solution = \(\frac{235 \times 100}{700}\) = 33.57%
Mass % of solvent (water) in resulting solution
= 100 – 33.57 = 66.43%

Question 12.
A graph showing vapour pressure against mole fraction of an ideal solution with volatile components A and B are shown below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q12 img 11

  1. Which law is supported by the graph?
  2. State the law.
  3. Resketch the graph if the attraction between A and B is greater than that between A-A and B-B.

Answer:

  1. Raoult’s law.
  2. Raoult’s law states that vapour pressure of a volatile component in a solution is the product of vapour pressure of that component in the pure form and mole fraction of that component in the solution.
  3. If the A-B attraction is greater than A-A and B-B attractions the liquid mixture behaves as a non-ideal solution with negative deviation.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q12 img 12

Question 13.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q13 img 13

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 14.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Answer:
Dissolution of gases is exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent \(\rightleftharpoons \) Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution.

Question 15.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= \(\frac{1000}{18}\) = 55.5
∴ χH2O = \(\frac{55.5}{55.5+1}\)
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 16.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer:
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
\(\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}\)
= 5.27 × 10-3 atm

Question 17.
Solution of sucrose is prepared by dissolving 34.2 g of it in 1000 g of water. Find out the freezing point of the solution, if Kf of water is 1.86 K/kg/mol? (Molecular mass of sucrose is 342 g/mol).
Answer:
ΔTf = kf × m
=1.86 K kg mol-1 × \(\frac{34.2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{342 \mathrm{g} \mathrm{mol}^{-1} \times 1000 \mathrm{g}}\) = 0.186K
Tf = OK – 0.186 K
Freezing point of the solution = – 0.186 K

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 18.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
In the first case, π1 = C1RT
i.e., 4.98 bar = \(\frac{36 \mathrm{g} \times \mathrm{R} \times 300 \mathrm{K}}{180 \mathrm{g} \mathrm{mol}^{-1}}\) ——– (1)
In second case, π2 = C2RT
i.e., 1.52 bar = C2R × 300 K ———- (2)
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q18 img 14

Question 19.
A solution containing 12.5 g of non-electrolytic substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molecular mass of the substance? (Kb for water = 0.52 K kg mol-1)
Answer:
Molecular mass of the solute, MB = \(\frac{1000 \mathrm{K}_{\mathrm{b}} \mathrm{W}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \Delta \mathrm{T}_{\mathrm{b}}}\)
\(\frac{1000 \mathrm{g} \mathrm{kg}^{-1} \times 0.52 \mathrm{K} \mathrm{kg}^{-1} \mathrm{mol}^{-1} \times 12.5 \mathrm{g}}{175 \mathrm{g} \times 0.70 \mathrm{K}}\)
= 53.06 mol-1

Question 20.

  1. “For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. Name the law stated above.
  2. Study the graph. What phenomenon it denotes? Based on your observation predict the reason for the greater volatility of a mixture of carbon disulphide and acetone?

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q20 img 15

Answer:

  1. Raoult’slaw
  2. Positive deviation from Raoult’s law. In a solution formed by adding CS2 to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules. Thus, the escaping tendency of the particles increases and the solution shows +ve deviation.

Question 21.

  1. To get hard boiled eggs, common salt is added to water during boiling. Give reason.
  2. Which colligative property is more suitable for the determination of molecular mass of polymers? Give the expression to determine molecular mass by this method.

Answer:

  1. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg and thus get hardened.
  2. Osmotic pressure method.

Question 22.
18 g of glucose is dissolved in 1kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1)
Answer:
Number of moles of glucose = 18/180 = 0.1 mol.
Mass of solvent = 1 kg.
Morality of glucose , m = \(\frac{n_{8}}{W_{A}}\)
= \(\frac{0.1 \mathrm{mol}}{1 \mathrm{kg}}\) = 0.1 mol/Kg
Elevation of boiling point ΔTb = Kb × m
= 0.52 K kg/mol × 0.1 mol/kg = 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 K + 0.052 K = 373.202 K

Plus Two Chemistry Solutions Four Mark Questions and Answers

Question 1.
Colligative properties are exhibited by dilute solutions.

  1. What do you mean by colligative properties?
  2. Which are the four colligative properties?

Answer:

  1. Colligative properties are those properties of dilute solutions of non-volatile solutes whose value depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.
  2. The four colligative properties
    • Relative lowering of vapour pressure
    • Elevation of boiling point
    • Depression of freezing point
    • Osmotic pressure

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 2.
PA = P°A χA
PB = P°B χB
ΔmixV = O

  1. Arun said to Subin that it is the condition for a type of solutions.
  2. Identify the type of solutions.
  3. What are the differences between ideal and non-ideal solutions?

Answer:
1. Ideal solutions.

2. Vapour pressure of a volatile component in the solution is the product of vapour pressure of pure component and mole fraction of that component in the solution.

3.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q2 img 16

Question 3.

  1. What are the characteristics of a non-ideal solution?
  2. Explain +ve and -ve deviations from Raoult’s law with suitable examples.

Answer:
1. The characteristics of a non-ideal solution

  • Does not obey Raoult’s law over the entire range or concentration. Fora non-ideal solution having two volatile components A and B,
    PA ≠ P°A χA
    PB ≠ P°B χB
  • Volume of mixing not equal to zero, DmixV10
  • Enthalpy of mixing not equal to zero, DmixH10

2.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q3 img 17

Question 4.
Osmosis and osmotic pressure are two important terms related to solutions.

  1. Explain these terms.
  2. How will he determine the molar mass of a substance by this method?

Answer:
1. The phenomenon of the spontaneous flow of a solvent from a solution of lower concentration to higher concentration, separated by a semipermeable membrane is called osmosis.

The excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semipermeable membrane is called osmotic pressure.

2. Osmotic pressure (p) is proportional to the molar concentration/molarity (C) of the solution at a given temperature (T K).
π = CRT, where R is the gas constant.
π = \(\frac{n_{\mathrm{B}}}{V}\) RT, where nB is the number of moles of the solute and V is the volume of the solution in litres.
π V = nBRT
π V = \(\frac{w_{B}}{M_{B}}\)RT, where WB is the mass of the solute and MB is the molar mass of the solute.
Or MB = \(\frac{\mathrm{W}_{\mathrm{B}} \mathrm{RT}}{\pi \mathrm{V}}\)
Osmotic pressure measurement is widely used to determine molar mass of proteins, polymers and other macro molecules.

Question 5.
The value of molecular mass determined by colligative property measurement is sometimes abnormal.

  1. Explain these abnormalities in the case of benzoic acid in benzene and KCl in water.
  2. What is van’t Hoff factor?

Answer:
1. This is caused by dissociation in the case of KCl and association in the case of acetic acid. KCl in aqueous solution undergoes dissociation as KCl → K+ + Cl
Molecules of ethanoic acid (acetic acid) dimerises in benzene due to hydrogen bonding. As a result of dimerisation the actual number of solute particles in solution is decreased. As colligative property decreases molecular mass increases.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q5 img 18

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 6.
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. This is the commonly used law for expressing the solubility of gas in liquid.

  1. Name the law. Write its mathematical expression.
  2. What are the factors affecting the solubility of a gas in a liquid? Explain.

Answer:
1. Roult’s law.
For an ideal solution containing two volatile components A and B,
PA = P°A χA,
PB = P°B χB and
P[Total] = PA + PB = P°A χA + P°B χB

2. The factors affecting the solubility of a gas in a liquid:

  • Nature of the gas and the liquid – Each gas has a characteristic solubility in a particular liquid at a particular temperature and pressure.
  • Temperature – solubility of a gas in a liquid is an exothermic process. Hence according to Le Chatelier’s principle solubility of gases in liquids decreases with rise in temperature.
  • Pressure – According to Henry’s law, solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Question 7.
Concentration of solution may be expressed in different ways.

  1. Mention any one of the concentration terms.
  2. What are colligative properties?
  3. Show that elevation in boiling point is a colligative property.

Answer:
1. Molarity – It is the number of moles of the solute present in one litre of the solution.

2. Colligative properties are those properties which depends only on the number of solute particles.

3. ΔTb = Kbm
= \(\mathrm{k}_{\mathrm{b}} \frac{\mathrm{n}_{\mathrm{B}} \times 1000}{\mathrm{W}_{\mathrm{A}}}\)
i.e., ΔTb α nB i.e., elevation of boiling point depends on number of moles of solute. Hence, it is a colligative property.

Plus Two Chemistry Solutions NCERT Questions and Answers

Question 1.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q13 img 13

Question 2.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Answer:
Dissolution of gases is an exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent \(\rightleftharpoons \) Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 3.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= \(\frac{1000}{18}\) = 55.5
∴ χH2O = \(\frac{55.5}{55.5+1}\)
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 4.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer:
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
\(\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}\)
= 5.27 × 10-3 atm

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

You can Download The Last Leaf (Story) Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

A The Trio (Story) Textual Questions and Answers

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5 Question 1.
Observe figure try to write down the activities shown in them.
Answer:

  1. A man pushes a trolly.
  2. Batting of a cricket ball.
  3. Pushing a wall.

Kerala Syllabus 9th Standard Physics Notes Chapter 5 Question 2.
Write down more activities familiar to you
Answer:

  1. A man carrying a load
  2. Throwing a ball
  3. Lifting the bag on to the shoulder
  4. Pushing a car into motion

Hsslive Guru 9th Physics Kerala Syllabus Chapter 5 Question 3.
You have understood that a force is to be applied on a body to do an activity. Find out the source of applied force for every activity and note them down in the table.
Kerala Syllabus 9th Standard Physics Notes Chapter 5
Hsslive Guru 9th Physics Kerala Syllabus Chapter 5
Answer:

Activity Source of applied force
Falling of a mango The earth
A trolley being pushed The person pushing
Batting of a cricket ball Batsman
Pushing a wall The person pushing
Lifting a load The person lifting

Objects undergo displacement only when the force is applied on it them.

Displacement takes place in the direction of force applied No displacement for the body in the direction of force applied
1. A cricket ball when hit by a bat 1. A wall is pushed
2. A trolley is being pulled. 2. A car is pushed by sitting inside the car
3. Climbing a ladder with a load on head.
4. Falling of a mango from mango tree.

Work :

Work is said to be done only when a body under¬goes displacement in the direction of the applied force.
Kerala Syllabus 9th Physics Notes Chapter 5

work energy and power Question 4. Observe figure and write down the situation where work is said to be done
Answer:
The person loaded on head and goes upward.

Kerala Syllabus 9th Physics Notes Chapter 5 Question 5.
A boy pushed an object of mass 30Kg horizontally across the floor through 50m. Another boy pushed an object of mass 50kg across the same floor through 50m. Both of them gave the same speed for moving the objects.
a) Who applied greater force here?
b) In which case was the work greater?
c) Write down a factor influencing work.
Answer:
a) Second boy applied greater force
b) More work is done in the second situation
c) The Factor influencing work is force.

Learn Mass Transfer MCQ questions & answers are available for a Chemical Engineering students.

Hss Live Guru Physics 9th Kerala Syllabus Chapter 5 Question 6.
A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.
a) Who pushed a greater distance here?
b) What about the force applied
c) Who did the greater work?
d) Which is the factor influencing work here?
Answer:
a) Second boy
b) Force is same
c) Second boy done greater work
d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)
Equation for calculating work done Work W = Fs
Unit of work is Joule (J), Kilojoule (KJ)
1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Get the free “Nuclear Equation Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Physics Notes For Class 9 Kerala Syllabus Chapter 5 Question 7.
When a force of 10N is applied continuously on a body it undergoes a displacement of 2m find out the magnitude of the work done?
Answer:
F = 10N, s = 2m
Work W= Fs = 10 × 2 = 20 Nm
Look at the Figure
Hss Live Guru Physics 9th Kerala Syllabus Chapter 5

Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5 Question 8.
A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer:
Weight of the body applies downwards and the table applies an equal force upwards.

Hss Live 9th Physics Kerala Syllabus Chapter 5 Question 9
In which directions do these forces act?
Answer:
Forces acting on both directions, upward direction and downward direction.

9th Physics Notes Kerala Syllabus Chapter 5 Question 10.
A book of mass 100g is raised to the top of a table of height 1 m. Find the magnitude of the work done by the force applied against the gravitational force(g = 10m/s2)
Answer:
m = 100g = 0.1kg
g = 10m/s2
h = 1m
W = mgh
= 0.1 × 10 × 1 = 1J
1 J is the amount of work done to raise a body of mass 100 g through a height of 1m.

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Question 11.
If a force of 50N is applied on a body and it under¬goes a displacement of 2m in the direction of the force, calculate the amount of work done.
Answer:
F = 50N
s = 2m
W = Fs = 50 x 2 = 100J

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5 Question 12.
a) If a force of 200N is applied on a table of mass 50kg, it undergoes a displacement of 0.5m in the direction of force. Calculate the amount of work done.
b) If the same table is raised by 3 m, what would be the work done against the gravitational force?
Answer:
a) F= 200N, s = 0.5m
W = Fs = 200 x 0.5 = 100J
b) m = 50 kg, g = 10m/s2
h = 3 m
W = mgh = 50 x 10 x 3 = 1500J
Observe Figure
Physics Notes For Class 9 Kerala Syllabus Chapter 5
Let a body mass m be pulled by a force F. If the body has a displacement s in the direction of the force, then the work done by the force F, Wf = Fs Here the displacement produced is in the direction of the force itself.

Kerala Syllabus Class 9 Physics Solutions Chapter 5 Question 13.
Write whether this work is negative or positive.
Answer:
Work is positive

9th Class Physics Notes Kerala Syllabus Chapter 5 Question 14.
The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer:
If the displacement is in opposite direction work done by frictional force is negative.

Class 9 Physics Kerala Syllabus Chapter 5 Question 15.
In which direction is the force of gravity on the body?
Answer:
Direction of the gravitational force will be in the downward direction

9th Standard Physics Notes Kerala Syllabus Chapter 5 Question 16.
Is there a displacement for the body in the direction of the gravitational force?
Answer:
No displacement occurs in the direction of the gravitational force.
When a body on a floor is pulled and if it is displaced in the direction of the applied force, the work done by the applied force will be positive and the work done by the frictional force exerted by the floor will be negative.

Energy

Question 17.
What is the work to be done to raise a body of mass m kg through h meter?
Physics chapter 5 work and energy Answer:
w = mgh
Energy is the capacity to do work. Unit is Joule (J)

work energy book Question 18. In daily life, we use different forms of energy for various activities. List the forms of energy familiar to you
Answer:

  1. Mechanical energy
  2. Heat energy
  3. Electrical energy
  4. Chemical energy
  5. Light energy
  6. Nuclear energy

There are two type of Mechanical energy Kinetic energy and potential energy

Kinetic Energy

Pulling the toy car backwards a little and allow it to hit the plastic ball
Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5

Question 19.
What happened to the ball when the moving car hit it?
Answer:
Ball moves forward

10 to the 9th power Question 20.
How did the car get the energy to move the ball forward?
Answer:
The energy is obtained from the motion of toy car
Conclusion:

  • Moving objects possess energy
  • The energy possessed by a body by virtue of its motion is the Kinetic energy

Experiment:
Let’s do another activity. Allow a powder tin to slide down a polished, inclined plane as shown in the figure and let it hit a toy car. Try to measure the displacement of the toy car. Repeat the experiment by increasing the height of the inclined plane and filling the tin with sand.
Hss Live 9th Physics Kerala Syllabus Chapter 5
Observation:

  • The displacement of the toy car is greater When it is hit by the powder filled with sand
  • Also the displacement is greater when it is dropped from a greater height

Conclusion: Kinetic energy depends on mass (m) and velocity (v)
Derivation of the equation for calculating K.E
Work W = Fs
As per II law of motion
F = ma
∴ W = mas
According to third equation of motion
v2 = u2 + 2as = 0 + 2as (u = 0) = 2as
∴ as = \(\frac{v^{2}}{2}\)
W = mas , if we put \(\frac{v^{2}}{2}\) instead of as
9th Physics Notes Kerala Syllabus Chapter 5
Work is equal to the magnitude of kinetic energy
i.e, Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2,
M – mass, V – Velocity
When a body of mass m moves with a velocity v, its kinetic energy will be K = \(\frac { 1 }{ 2 }\) mv2

Question 21.
A man having a mass of 70kg is riding a scooter of mass 80kg. What is the total kinetic energy if the velocity of the scooter is 10m/s?
Answer:
m = 70kg + 80kg = 150kg
v = 10m/s
Kerala Syllabus 9th Standard Physics Solutions Chapter 5

Question 22.
A car of mass 1500 kg is moving at a velocity of 20m/s. What is its kinetic energy?
Answer:
m = 1500 kg
v = 20m/s
K = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 1500 × 202
= 300000J = 300kJ

Question 23.
A boy of mass 50kg is riding a bicycle with a speed 2m/s. The bicycle has a mass of 10kg. Calculate the total kinetic energy?
Answer:
m = 50kg + 10kg = 60kg
v = 2m/s
K = \(\frac { 1 }{ 2 }\) mv2
1/2 × 60 × 22 = 120J

Potential Energy

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5
When we lift a body in perpendicular direction, the work is said to be done against gravitational force. According to figure, maximum work is done when it reaches at a height fb from the group, That means as the height increases work done also increases.

  • The energy received by the body is equal to the work done on it.
  • The body attains more energy when the height from the ground level increase

The energy possessed by a body due to its position is the potential energy.
ie, Potential energy
U = mgh
m – mass, g – acceleration due to gravity
h = height from the ground

Question 24.
Identify more situation in which potential energy is acquired by virtue of position
Answer:

  • Coconut in a coconut tree
  • Water stores in huge reservoirs
  • Objects placed above the buildings,

Inference:
Height increases, potential energy increases

Question 25.
Write down situations in which potential energy varies.
Answer:

  • Falling of a coconut from a coconut tree
  • Pumping water to a tank at a height
  • Climbing a ladder

Question 26.
Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground?
Answer:
Mass m = 1 kg, Acceleration due to gravity
g = 10m/s2
h = 6 m,
U = mgh = 1 × 10 × 6 = 60J

Question 27.
A bird of mass 0.5 kg is flying at the same speed at the same height of 5m. In this state, if its Kinetic energy and potential energy are equal.
a) What is the potential energy of the bird?
b) What is the velocity of the bird?
Answer:
a) m = 0.5kg
g = 10m/s2
h = 5m
u = mgh
= 0.5 × 10 × 5 = 25J
b) Kinetic energy K = 25J (∴U = K)
Kerala Syllabus Class 9 Physics Solutions Chapter 5

Question 28.
Write other examples for getting potential energy due to strain
Answer:
9th Class Physics Notes Kerala Syllabus Chapter 5
Class 9 Physics Kerala Syllabus Chapter 5

  • A Stretched bow
  • An elongated rubber band
  • Compressed spring
  • Compressed spring in a toy car

Conclusion: The factor which gives potential energy are position and strain

Law Of Conservation Of Energy

9th Standard Physics Notes Kerala Syllabus Chapter 5
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 16

Question 29.
What form of energy does the flower pot have when it is on the sunshade of a building?
Answer:
Potential energy

Question 30.
While the flower pot is falling down, what forms of energy does it possess?
Answer:
potential energy and kinetic energy

Question 31.
Answer:
Does its potential energy increase/decrease when the pot falls down.
Answer:
Potential energy decreases

Question 31.
Will the kinetic energy increase/decrease at that time?
Answer:
Kinetic energy increases

Question 33.
What energy transformation takes place just before the flower pot reaches the ground?
Answer:
Potential energy is converted completely into kinetic energy

Question 34.
Let the mass of the flower pot 15kg and the height of the sunshade 4m.
a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2).
b) When it is on the sunshade, what is its kinetic energy?
c) If so, what is its total energy?
Answer:
U = mgh
= 15 × 10 × 4
= 600J
b) Kinetic energy will be zero
c) Total energy = PE + KE
= 600J + 0 = 600J

Question 35.
While falling, when the flower pot is at a height of 2m from the ground, what will be its kinetic energy?
Answer:
K = \(\frac { 1 }{ 2 }\) mv2
u = 0, g = 10m/s2 ,
s = 4 – 2 = 2m
v2 = u22 + 2as
=0 + 2 × 10 × 2 = 40
K = 1/2 × 15 × 40
= 300J

Question 36.
What is the potential energy when it is at a height of 2m from the ground? What is the total energy now?
Answer:
Potential energy at 2m height
U = mgh = 15 × 10 × 2 = 300J
∴ Total energy = 300 + 300 = 600J

Question 37.
What is the kinetic energy of the flower pot just before it touches the ground?
Answer:
K = \(\frac { 1 }{ 2 }\) mv2
v2 = u2 + 2as
= 0 + 2 x 10 x 4 = 80
Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 15 × 80 = 600J

Question 38.
The potential energy
U = mgh
= 15 x 10 x 10 x = 0.
What will be the total energy?
Answer:
Total energy = 600 + 0 = 600J

Question 39.
To sum up the amount of energy at each situation:
a) When on the sunshade
b) When at a height of 2m from the ground
c) Just before hitting the ground
Answer:
a) When on the sunshade = 600J
b) At a height of 2m from the ground = 600J
c) Just before hitting the ground = 600J
Energy can neither be created nor be destroyed. It can be transformed from one from to another
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 17
Sun is the major source of energy. We utilize solar energy in different ways. It is renewable source of energy Plants prepare food by using sunlight. From these fossil fuels, firewood are formed. Windmills works by utilizing kinetic energy of wind energy from infra¬red rays. Tidal energy is obtained due to gravitation. When we consider any energy sources, they were originated from solar energy.

Power

Question 40.
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s2).
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 18
Answer:
b) 150000J
c) 150000J

Question 41.
Is the amount of work done by the pump to fill water in the three tanks equal?
Answer:
Amount of work done is equal

Question 42.
Find the amount of work done per second by each pump.
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 19
Answer:

Pump Work done Time (s) (J) Work done per second (J/S)
A 150000 100 1500
B 150000 200 750
C 150000 400 375

Amount of work done per second is referred as power of the pump.
Work done per unit time or rate of doing work is power
Power = \(\frac{\text { work }}{\text { time }}, P=\frac{w}{t}\)
Units of power
watt (W), kilowatt (KW), Horse power (H.P)
1 kW= 1000W
1 HP = 746W

Question 43.
If a man of mass 70kg climbs up a mountain of height 30m in 5 minutes, what is his power?
Answer:
m = 70kg, g = 10m/s2
h = 30m
Work W = mgh = 70 × 10 × 30 = 21000J
Time t = 5mt = 5 × 60 = 300s
Power p = \(\frac { w }{ t }\) = \(\frac { 21000 }{ 300 }\) = 70W

Question 44.
If a man of mass 50kg takes 60s to climb up 20 steps, each 15cm high, calculate his power.
Answer:
m = 50kg, g = 10m/s2
h = 15cm × 20 = 300cm = 8m
t = 60s
work W = mgh = 50 × 10 × 3 = 1500J
Power p = \(\frac { w }{ t }\) = \(\frac { 1500 }{ 60 }\) = 25W

Let Us Assess

Question 1.
A boy is trying to push the concrete pillar of the building using a force of 300N. Calculate the amount of work done by the boy.
Answer:
F= 300N, s = 0
w = Fs = 300 × 0 = 0

Question 2.
From what you have learnt of potential energy and kinetic energy write down the from of energy possessed by the bodies given below.
a) water in a dam
b) Stretched rubber band
c) Mango falling from a tree
Answer:
a) Potential energy
b) Potential energy
c) Potential energy lesser. Kinetic energy greater

Question 3.
Calculate the kinetic energy of an athlete of mass 60kg running with a velocity 10 m/s
m = 60kg
v= 10m/s
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 20

Question 4.
A stone of mass 2kg is thrown upwards from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.
Answer:
u = 3m/s v = 0
a = g = -10m/s2 (Velocity of a body moving upwards is decreasing so acceleration is negative)
v2 = u2 + 2as
0 = 32 + 2 × -10 × s
0 = 9 – 20s
20s = 9, s = \(\frac { 9 }{ 20 }\)
U = mgh = 20 × 10 × \(\frac { 9 }{ 20 }\) = 90J

Question 5.
The heart of a healthy person beats 72 times per minute and each beat uses up about 1J of energy. Calculate the power of the heart.
Answer:
W= 1 x 72 = 72J
t = 1mt = 60s
P = \(\frac { W }{ t }\) = \(\frac { 72 }{ 60 }\) = 1.2w

Question 6.
Which among the following is a vector quantity?
a) work
b) momentum
c) power
d) energy
Answer:
momentum

Question 7.
If the velocity of an object is doubled, its kinetic energy becomes
a) 2 times
b ) 1/2
c) 4 times
e) 1/4
Answer:
4 times

Question 8.
An object of mass 1 kg is falling from a height 10 m. What be the work done while falling?
a) 10 J
b) 1J
c) 100 J
d)1000 J
Answer:
w = mgh = 1 × 10 × 10 = 100J

Question 9.
Which one among the following is correct?
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 21
Answer:
W = p × t

Question 10.
A roller weighing 1 tonne is being dragged along a road. What is the work done against gravity? Why?
Answer:
Zero, No displacement in the direction of force.

Question 11.
Is it possible for an object to possess energy without momentum? Give one example for such a situation.
Answer:
Yes, possible. A coconut on a coconut tree has potential energy, but it has no momentum.

Question 12.
Say whether the following are positive work or negative work.
1) The work done by a person drawing water from a well using a rope without a pulley.
2) Work done by gravitational force in this situation
3) Work done by the frictional force while and object is sliding down along an inclined plane.
4) Work done by the force is moving along a plane surface.
Answer:

  1. Positive work
  2. Negative work
  3. Negative work
  4. Positive work

Question 13.
How much joule is 1 kWh?
Answer:
1 kWh = 1000 × 60 × 60 = 3600000J

Question 14.
Find out the work done against the gravitational force in the situations given below.
1. A child is standing still with a bundle of books of mass 5 kg
2. With the same bundle of books, she travels 1m along a plane surface with a speed 5 m/s
3. The bundle of books is lifted onto the top of a cupboard having 1m height (g=10m/s2).
Answer:
1. 0
2. 0
3. W = mg = 5 × 10 × 1 = 50 J

Question 15.
A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m/s. Calculate its kinetic energy and potential energy after 1 s.
(Hint: v + at, s = ut + 1/2at2)
Answer:
v = u + at
= 14 + 10 × 1
= 24 m/s
KE = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 0.4 × 24
= 115.2 J
S = ut + 1/2 at2
= 14 × 1 + 1/2 × 10 × 12
= 14 + 5 = 19 m
E = mgh = 0.4 × 10 × 19
= 76 J

Question 16.
An object of mass 1000 kg is travelling with a velocity 72 km/h. Calculate the work done to bring it rest.
Answer:
m = 1000 kg
v = 72 km/h = 20 m/s
Work = difference of kinetic energy
= \(\frac { 1 }{ 2 }\) mv2
= \(\frac { 1 }{ 2 }\) × 1000 × 20 × 20 = 400000J

Question 17.
Estimate the work done on a object of mass 80 kg to change its velocity from 5 m/s to 10 m/s
Answer:
m = 80 kg
Work = difference of kinetic energy
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 22

Work, Energy and Power More Questions

Question 1.
Fill in the blanks
a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ……….
b) The two factors related to potential energy to body are………….
c) 1HP = ……… Watt
Answer:
1 Joule
b) Position, Strain
c) 746W

Question 2.
Classify the following into work done and work is not done
1. A mango falling from a mango tree
2. Pushing a table while sitting on it
3. Pushing a wall
4. Kicking a football
5. A trolley is moving forward
6. Standing with a load above the head
Answer:
Work done:
1. A mango falling from a mango tree
2. Kicking a football
3. A trolling is moving forward
Work is not done:
1. Pushing a table while sitting on it
2. Pushing a wall
3. Standing with a load on the head

Question 3.
When Lekshmi is applied a force of 50N on an object it undergoes a displacement of 2m. When Vinitha applied the same force on itundergoesadisplacementof3m if so
a) Which person done more work?
b) Give reason
c) Calculate the work done by each person?
Answer:
a) Vinitha
b) Displacement is greater
c) Work done by Lekshmi
W = Fs = 50N × 2m = 100J
Work done by Vinitha
W = Fs = 50N × 3m = 150J

Question 4.
Calculate work done when a boy of mass 40kg climb¬ing a staircase of height 50cm, g = 10m/s2
Answer:
W= Fs
F = mg = 40 × 10 = 400N
S = h = 0.5m
W= 400 × 0.5 = 200J

Question 5.
a) What is mean by kinetic energy?
b) How mass and velocity affect kinetic energy?
c) A body of mass 20kg is moving with a velocity of 5m/s. Calculate the kinetic energy?
Answer:
a) The energy possessed by a body by virtue of its motion is called kinetic energy
b) Mass increases, kinetic energy increases mass . is doubled kinetic energy also doubled. Velocity increases, kinetic energy increases, Velocity is doubled kinetic energy becomes four times.
c) Kinetic energy KE = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 20 × 52 = 1/2 × 20 × 25 = 250J

Question 6.
a) Write the equation for finding potential energy?
b) indicate the representation of each letter
Answer:
a) U = mgh
b) M – mass,
g – acceleration
h – height from the ground

Question 7.
State law of conservation of energy?
Answer:
Energy can neither be created nor be destroyed. It can be transformed from one form to another.

Question 8.
A store of mass 5kg was raised from the ground to the second floor of height 7m and from there to the third floor of height 3m from the second floor. Calcu¬late the potential energy of the store with respect to the ground floor and the second floor?
Answer:
Mass m = 5kg
Height from ground to third floor
h = 7m + 3m
g = 10m/s2
Potential energy with respect to the ground
U = mgh = 5kg × 10m/s2 × 10 = 500J
Height from second floor to third floor h = 3m
Potential energy with respect to the second Floor
U = mgh = 5kg × 10m/s × 3 = 150J

Question 9.
a) What is power?
b) The time takes to move an object at a distance of 5m is 10s is the force applied is 30N, what is the power?
Answer:
a) Rate of doing work is power
b) Power P = E
W = Fs = 30 × 5 = 150J
t = 10sec.
Power P = \(\frac { 150 }{ 10s }\) = 15W

Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion

Students can Download Chapter 5 Law of Motion Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion

Plus One Physics Law of Motion One Mark Questions and Answers

Plus One Physics Laws Of Motion Questions Chapter 5 Question 1.
Which one of the following is not a force?
(a) Impulse
(b) Tension
(c) Thrust
(d) Weight
Answer:
(a) Impulse
Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.
Impulse = Force × Time duration.

Plus One Physics Laws Of Motion Questions And Answers Chapter 5 Question 2.
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is an example for
(a) Inertia of motion
(b) Second law of motion
(c) Third law of motion
(d) Inertia of rest
Answer:
(a) Inertia of motion
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.

Plus One Physics Important Questions And Answers Pdf Chapter 5  Question 3.
Which one of the following is not a contact force?
(a) Viscous force
(b) Magnetic force
(c) Friction
(d) Buoyant force
Answer:
(b) Magnetic force

Plus One Physics Chapter Wise Questions And Answers Chapter 5 Question 4.
A jet engine works on the principle of
(a) Conservation of linear momentum
(b) Conservation of mass
(c) Conservation of energy
(d) Conservation of angular momentum
Answer:
(a) Conservation of linear momentum
A jet engine works on the principle of linear momentum.

State And Prove Impulse Momentum Theorem Chapter 5 Question 5.
Newton’s second and third laws of motion lead to the conservation of
(a) linear momentum
(b) angular momentum
(c) potential energy
(d) kinetic energy
Answer:
(a) linear momentum
Newton’s second and third laws lead to the conservation of linear momentum.

Hsslive Plus One Physics Chapter Wise Questions And Answers Chapter 5 Question 6.
A large force is acting on a body for a short time. The impulse imparted is equal to the change in
(a) acceleration
(b) momentum
(c) energy
(d) velocity
Answer:
(b) momentum
If a large force F acts for a short time dt, the impulse imparted is
I = F.dt, = \(\frac{d p}{d t}\).dt
I = dp = change in momentum.

Laws Of Motion Class 11 Questions With Solutions Pdf Chapter 5 Question 7.
When a shell explodes, the fragments fly apart though no external force is acting on it. Does this violate Newton’s first law of motion?
Answer:
No. The explosion takes place due to the internal force. The internal force does not change the position of centre of mass.

Prove Impulse Momentum Theorem Chapter 5 Question 8.
In taking a catch, a cricket player moves his hands backward on holding the ball. Why?
Answer:
We know F = \(\frac{\Delta P}{\Delta t}\)
When ∆t increases, the force acting on hand decreases.

350 degrees f to c … T C T F 32 x 59 is the method for converting degrees Fahrenheit to degrees Celsius.

Plus One Physics Important Questions And Answers Chapter 5 Question 9.
Name the factor on which inertia depends.
Answer:
Mass

Laws Of Motion Class 11 Test Paper Chapter 5 Question 10.
Why does a swimmer push the water backwards?
Answer:
A swimmer pushes the water backward in order to be pushed forward (Newton’s third law).

Laws Of Motion Previous Year Questions Chapter 5 Question 11.
Rocket works on the principle of conservation of_______.
Answer:
Momentum

Motion Questions And Answers Pdf Chapter 5 Question 12.
A man experience a backward jerk, while firing bullet from gun. Which law is applicable here? Answer:
Conservation of momentum.

Plus One Physics Laws Of Motion Notes Chapter 5 Question 13.
If you jerk a piece of paper under a book quick enough, the book will not move. Why?
Answer:
This is due to inertia of rest.

Class 11 Physics Chapter 5 Important Questions Chapter 5 Question 14.
Why it is difficult to walk on a slipper road?
Answer:
We will not get required reaction from slippery road.

Laws Of Motion Class 11 Important Questions Chapter 5 Question 15.
A stone, when thrown on a glass window, smashes the window pan to pieces. But a bullet fired from the gun passes through it making a hole why?
Answer:
This is due to inertia of rest of glass window.

Question 16.
Why an athlete runs some distance before taking a jump?
Answer:
An athletic runs some distance before taking a jump to gain some initial momentum. It helps the athlete to jump more.

Question 17.
Why a horse can not pull a cart and run in empty space?
Answer:
The horse-cart system moves forward due to reaction of ground on the feet of horse. In free space, there is no reaction. So it can not pull cart.

Question 18.
Why parachute descends slowly?
Answer:
Parachute has large surface area. This increases fluid friction and slows down the motion of parachute.

Question 19.
Sand is thrown on tracks with snow. Why?
Answer:
The presence of snow on tracks reduces friction and driving is not safe. If sand is thrown, friction will be increased and driving becomes safe.

Question 20.
It is difficult to move a cycle along a road with its brakes on. Explain.
Answer:
When the cycle is moved with its brakes on, wheels can only skid. There will be sliding friction. The sliding friction is more compared to rolling friction. Hence it is difficult to move a cycle with its brakes on.

Plus One Physics Law of Motion Two Mark Questions and Answers

Question 1.
Two masses are in the ratio 1:5

  1. What is inertia.
  2. What is the ratio of inertia of above case?

Answer:

  1. The inability of a body to change it’s state of rest or uniform motion is called inertia.
  2. Mass is a measure of inertia. Hence ratio of inertia is 1:5.

Question 2.
More force is required to push a body than pull to get same speed on a ground with some friction. Why?
Answer:
When we push, the action on the surface and normal reaction on the body increases. (Friction is directly proportional to normal reaction).

As a result more force is required to push the body. When we pull, normal reaction decreases. Hence friction decreases. Hence less force is required to pull the body.

Question 3.
A lift in a multistoried building is moving from ground floor to third floor. What will happen to weight of a person sitting in side of the lift.

  1. A When starts to move up from ground floor.
  2. When the lift moves with constant speed.

Answer:

  1. A weight increases weight w = mg + ma
  2. weight is constant ie. w = mg

Question 4.
Why it is advisable to hold a gun tight to one’s shoulder when it is being fired?
Answer:
The recoiling gun can hurt the shoulder. If gun is held tightly against the shoulder, the body and gun act a system. This will reduce recoil velocity as it is inversly proportional to mass of system.

Question 5.
Why shockers are used in vehicles?
Answer:
When there is a jerk or jump, the time for which force acts (∆t) increases. As the product of force and time for which force acts (F∆t) remains constant, increase in At will reduce the force. This provide smooth motion.

Plus One Physics Law of Motion Three Mark Questions and Answers

Question 1.
Give the magnitude and direction of net force on

  1. a drop of rain falling down with a constant velocity.
  2. a stone of mass 0.1 kg just after it dropped from the window of a tram accelerating at 1 ms-2.

Answer:
1. Net force is zero

2. When stone is dropped, gravitational force will act on the stone.
Gravitational force F = mg
= 0.1 × 10
= 1 N downward.

Question 2.
An external force is always required to break the inertia of a body which is either in the state of rest or state of uniform motion.

  1. Which law governs this statement?
  2. Can all forces produce acceleration? Why?
  3. A boy holding a spring balance in his hand suspend a mass 2kg from it. If the balance slips from his hand and falls down, find the reading of the balance while it is in the air.

Answer:

  1. Newtons first law of motion.
  2. No. If resultant force acting on the body is zero, the body will move with constant velocity or remain at rest.
  3. Zero

Question 3.
A man weighs 70 kg. He stands on a weighing scale in a lift which is moving.

  1. upward with a uniform speed of 10 m/s.
  2. downward with an uniform acceleration of 5 m/s2.
  3. upward with an uniform accelerate of 5 m/s2. (Take g = 10m/s2). Find weight in each case.

Answer:
1. Weight W = mg
= 70 × 10 = 700 N.

2. W = mg – ma
= 70 × 10 – 70 × 5
= 700 – 350
= 350 N

3. W = mg + ma
= 70 × 10 + 70 × 5
= 700 + 350
= 1050N.

Question 4.
A body of mass ‘m’ is placed on a rough inclined plane having coefficient of friction µs. The inclination of plane is given as ‘θ’.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 1

  1. Which component of weight brings the body towards the bottom along the plane.
  2. Find how much force is required to pull the body along the plane.

Answer:
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 2

  1. mg Sinθ brings the body downwards
  2. When the body moves upwards the frictional force (Fs) acts downwards
    Total pulling Force = mg Sinθ + Frictional force (Fs) (u, mgCosθ).

Question 5.
Four person sitting in the back seat of a car at rest, is pushing on the front seat.

  1. Does the car move. Why?
  2. State the law which help you to answer above question.
  3. Long jumpers take a long run before the jump. Why?

Answer:

  1. No. Action and reaction cancel each other.
  2. Newtons third law of motion.
  3. To get large inertia of motion.

Question 6.
A Cricket player lowers his hands while catching a Cricket ball to avoid injury.

  1. What do you mean by impulsive force?
  2. Prove impulse – momentum theorem.

Answer:
1. The forces which acton bodies for short time are called impulsive forces.
Example:

  • In hitting a ball with a bat
  • In firing a gun

2. F = \(\frac{d p}{d t}\)
F∆t = dp
impulse = change in momentum.

Plus One Physics Law of Motion Four Mark Questions and Answers

Question 1.
A bead sliding on a wire A moves to C through B as shown in the figure. The bead at A has a speed of200cms
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 3

  1. what is speed at B?
  2. To what height will it rise before it returns?
  3. Why the ball moves up even after reaching the bottom most point B?

Answer:
1. mgh = 1/2 mv2
m × 10 × 0.8 = 1/2 mv2
V2 = 2 × 10 × 0.8
V = \(\sqrt{2 \times 10 \times 0.8}\)
V = 4 m/s.

2. 80 cm (if friction is neglected).

3. when the ball reaches at B, the potential energy is converted into kinetic energy. Due to this kinetic energy the ball raises to the point c.

Question 2.
Figure shows a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 4

  1. Obtain formula for the acceleration of the system and tension in the cord.
  2. If m1 and m2 interchanges its position, will it affect the tension of the string?
  3. What is the acceleration of the system if m1 = 5 kg and m2 = 2kg?

Answer:
1. When the body m2 moves in down ward direction.
m2g – T = m1 a
T = m2g – m1a.

2. New tension can be found from the relation
m1g – T = m2a
T = m1g – m2 a.

3. Acceleration of system, a
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 5

Question 3.
The collision of two ice hockey players are shown in figure.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 6
Analyse the data given in the figure and answer the following questions.

  1. Which conservation law is applicable in this case.
  2. In which direction and at what speed do they travel after they stick together.
    [Hint – towards right can be taken us +ve direc¬tion and vice versa]
  3. If we assume the friction of playing ground is zero, predict the nature of motion and the point at which they come to rest.

Answer:
1. Conservation of linear momentum.

2. Total momentum before collision = Total momentum after collision.
110 × 4 + 90 × -6 = (110 + 90)v
v = 0.5 m/s
-ve direction, (in the direction of man mass 90 kg).

3. Uniform motion They will not stop.

Question 4.
A circular track of radius 300m is kept with outside of track raised to make 5 degree with the horizontal.

  1. Name the process in which outside of the road is raised little above the inner.
  2. Obtain an expression for the optimum speed to avoid skidding (considering to friction)

Answer:
1. Banking of roqd

2.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 7
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 8
Consider a vehicle along a curved road with angle of banking θ. Then the normal reaction on the ground will be inclined at an angle θ with the vertical.
The vertical component can be divided into N Cosθ (vertical component) and N sinθ (horizontal component). The frictional force can be divided into two components. Fcosθ (horizontal component) and F sinθ (vertical component).
From the figure
N cos θ = F sinθ + mg
N cosθ – F sinθ = mg ______(1)
The component Nsin0 and Fsinθ provide centripetal force. Hence
N sinθ + F cos θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) ______(2)
eq (1) by eq (2)
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 9
Dividing both numerator and denominator of L.H.S by N cosθ. We get
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 10
This is the maximum speed at which vehicle can move over a banked curved road.
Optimum speed:
Optimum speed is the speed at which a vehicle can move over a curved banked road without using unnecessary friction. Putting µ = 0 in the above equation we get
v0 = \(\sqrt{\mathrm{Rg} \tan \theta}\).

Question 5.
A circular track of radius 400m is kept with outer side of track raised to make 5° with the horizontal (coefficient of friction 0.2)
(a) Name such track?
(b) What is optimum speed to avoid wear and tear of type?
(c) What is the maximum permissible speed to avoid skidding?
Answer:
(a) Banking.

Plus One Physics Law of Motion Five Mark Questions and Answers

Question 1.
A horse pulls a cart with constant force so that the cart moves with a constant speed.

  1. Does it violate Newtons second law of motion?
  2. If not, how will you account for the non acceleration of the cart?
  3. Will the speed of the cart increase, decrease or remain the same if the horse applied more force?
  4. A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.

Answer:
1. No.

2. The force applied by the car is balanced by the frictional force. Hence the cart moves with constant velocity.

3. If the horse is applied more force, the speed of the cart increases.

4. The resultant force,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 11
F = 10N
We know, F = ma
10 = 5 × a
acceleration, a = \(\frac{5}{10}\) = 2 m / sec2
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 12
The angle of resultant force,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 13
θ = tan-1 6/8
θ = 36°521
The angle of acceleration θ = 36°521.

Question 2.

  1. Friction is the force which opposes the relative motion between two surfaces in contact with each other. What is a limiting static friction? State the laws related to this.
  2. Show that the coefficient of friction is equal to the tan of the angle between the resultant and normal reactions.
  3. For a body of mass 5kg on a plane at a limiting static friction of 30 degrees. What is the force of friction?

Answer:
1. The maximum value of static friction is called limiting static friction.

  • The magnitude of the limiting friction is independent of the area of contact between the surfaces.
  • The limiting static friction is directly proportional to the normal reaction R.

ie f α R
fs = µsR.

2. Angle of friction is the angle whose tangent gives the coefficient of friction.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 14
Consider a body placed on a surface. Let N be the normal reaction and limit is the limiting friction. Let ‘θ’ be the angle between Resultant vector and normal reaction. From the triangle OBC,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 15
∴ tanθ = µ.

3. Tangent of the angle cient of friction.
µs = tanθ
µs = tan 30
µs = \(1 / \sqrt{3}\)
Friction F = µsmg
= \(1 / \sqrt{3}\) × 5 × 10
F = \(\frac{50}{\sqrt{3}}\)N.

Question 3.
The rate of change of linear momentum of a body is directly proportional to the external force applied on it, and takes place always in the direction of force applied.

  1. Name this law.
  2. Using this law obtain the expression for force.
  3. The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\) gt2. Find the force acting on it.

Answer:
1. Newton’s Second Law.

2. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 16
Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 17

3.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 18
Hence force F = mg.

Question 4.
Recoil of gun is based on the principle of conservation of momentum.

  1. State the principle of conservation of momentum.
  2. Explain the reoil velocity of gun.
  3. A bullet of mas 100g is fired from a rife of mass 200 kg with a spped of 50 m/s. Calculate the recoil velocity of the rife.

Answer:
1. According to law of conservation of linear momentum, if the external force acting on a body is zero, total linear momentum remains constant. According to Newton’s second law.
F = \(\frac{d p}{d t}\)
If F = 0, \(\frac{d p}{d t}\) = 0 i.e; P is constant.

2. Let M, m be the mass of gun and bullet respectively. Let V and ν be the velocities of gun and bullet after firing.
According to consevation of momentum
Total momentum before firing = Total momentum after firing
∴ O = MV + m ν
-MV = mν
The above equation shows that when bullet moves in forward direction, the gun moves in back direction. This motion of gun is called recoil of gun.

3. M = 200kg, m = 100g = 0.1kg
ν = 50 m/s, V = ?
MV = mν
200 × V = 0.1 × 50
V = \(\frac{0.1 \times 50}{200}\)m/s.

Question 5.
While firing a bullet, the gun must be held tight to the shoulder.

  1. Which conservation law helps you to explain this
  2. “In the firing process, the speed of the gun is very low compared to the speed of the bullet.” Substantiate the above statement using mathematical expressions.
  3. A shell of 20kg moving at 50m/s bursts in to two parts of masses 15kg and 5kg. If the larger part continues to move in the same direction at 70 m/s. What is the velocity and direction of motion of the other piece.

Answer:
1. Conservation of momentum.

2. Total momentum is conserved
∴ mu + MV = 0
V = \(\frac{-m u}{M}\) M is very large. Hence v is small

3. MV = m1 u1 + m2 u2
20 × 50 = 5u1 + 15 × 70
5u1 = 50
u1 = 10m/s.

Question 6.
While firing a bullet, the gun must be held tight to the shoulder.

  1. This is a consequence of______
  2. Show that recoil velocity is opposite to the muzzle velocity of the bullet.
  3. A gun of mass 5 kg fire a bullet of mass 5g, vertically upwards to a height of 100m. Calculate the recoil velocity of gun.

Answer:
1. Conservation of linear momentum.

2. Let M be the mass of gun and m be the mass of bullet. When gun fires, the gun and bullet acquire velocities V and v respectively.
According to conservation of momentum.
Total momentum before firing = Total momentum afterfiring
m × o + M × o = mu + MV
O = mv + MV
ie. – MV = mv
V = \(\frac{-m v}{M}\)

3. M = 5kg, m = 5 × 10-3 kg, h = 100m
v2 = u2 + 2as
0 = u2 + 2 × 10 × 100
Velocity of bullet,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 19

Question 7.
A standing passenger falls backwards when the bus starts suddenly.

  1. Explain why this happens?
  2. Which Newtons law gives the above concept. State the law.
  3. Obtain an expression for force using Newtons law.

Answer:
1. Due to inertia of rest, the body continues in the state of rest.

2. Newtons first law:
Everybody continues in its state of rest or of uniform motion along a straight line unless it is compelled by an external unbalanced force to change that state:

3. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 20
Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 21

Question 8.
According to Newton’s law of motion rate of change of momentum is directly proportional to applied force.
a. Impulse has the unit similarto that of

  1. Momentum
  2. force
  3. time
  4. Energy

b. A man falling from certain height receives more injuries when he falls on a marble floor than when he falls on a heap of sand. Explain. Why?
c.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 22
Force – time graph for a body starting from rest is shown in the figure. What is the velocity of the body at the end of 12 second? (Mass of the body is 5 kg)
Answer:
a. 1. Momentum.

b. When a man falls on a marble floor, the momentum is reduced to zero in lesser time. Due to this, the rate of change of momentum is large. So greater force acts on a man falls on marble floor.

c. The area of force – time graph gives change in momentum.
ie. change in momentum,
mv = 1/2 × (12 – 4) × (20 -10)
mv = 40

Plus One Physics Law of Motion NCERT Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed.
(b) a cork of mass 10g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30km h-1 on a rough road
(e) a high – speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
Applying Newton’s first law of motion, we find that no net force acts in any of the situations, (a) to (d). Again, no force in situation (e). This is because electron is far away from all material agencies producing electromagnetic and gravitational forces.

Question 2.
A constant retarding force of 50 N is applied to a body of mass 20kg moving initially with a speed of 15ms-1. How long does the body take to stop?
Answer:
Acceleration, a = –\(\frac{50 \mathrm{N}}{20 \mathrm{kg}}\) = -2.5ms-2
[Negative sign indicates retardation]
u = 15ms-1, v = 0, t = ?
v = u + at
0 = 15 – 2.5t or 2.5t = 15 or
t = \(\frac{15}{2.5}\)s = 6.0s.

Question 3.
A constant force acting on a body of mass 3.0kg changes its speed from 2.0ms-1 to 3.5 ms-1 in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
m = 3kg; u = 2ms-1; v = 3.5 ms-1;
t = 25s ; F = ?
v = u + at
3.5 = 2 + 25a or a = 0.06 ms-2
F = ma = 3kg × 0.06 ms-2 = 0.18N.
The direction of force is along the direction of motion.

Question 4.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the string is cut when the bob is

  1. at one of its extreme positions.
  2. at its mean position.

Answer:

  1. At the extreme position, the speed of the bob is zero. If the string is cut, it will fall vertically down wards.
  2. At the mean position, the bob has a horizontal velocity. If the string is cut, it will fall along a parabolic path.

Question 5.
A man of mass 70kg stands on a weighing scale in a lift which is moving

  1. upwards with a uniform speed of 10ms-1
  2. downwards with a uniform acceleration of 5ms-2
  3. upwards with a uniform acceleration of 5ms-2 What would be the readings on the scale in each case?
  4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

  1. a = 0, R = mg = 70 × 10 N = 700N
  2. mg – R = , ma ; R – mg – ma = (g – a)
    = 70(10 – 5) N = 350N
  3. R – mg = ma or R = m(g + a)
    = 70(10 + 5)N = 1056 N
  4. In the event of free fall, it is a condition of weight lessness.

Question 6.
A nucleus is at rest in the laboratory frame of reference. Show that if it dist integrates into two smaller nuclei, the products must move in opposite directions.
Answer:
Applying principle of conservation of momentum,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 23
The negative sign indicates that the products move in opposite directions.

Question 7.
A shell of mass 0.020 kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms-1, what is the recoil speed of the gun?
Answer:
m = 0.02kg, M = 100kg, v = 80ms-1, V = ?
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 24
= -0.016ms-1 = -1.6cm s-1
Negative sign indicates that gun moves in a direction opposite to the direction of motion of the bullet.

Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images

You can Download Moving Images Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 9 Moving Images

Moving Images Questions and Answers

Section -1

Question 1.
The animation software in IT @ School GNU/Linux is
a) Synfig studio
b) Tupi
c) Gimp
d) Paint
Answer:
a) Synfig studio

HSSLive.Guru

Question 2.
Which type of animation software is synfig studio.
a) Proprietary
b) not able to take copy
c) pay and use
d) free software
Answer:
d) free software

Question 3.
To prepare a film, How many times images appear one after the other continuously in front of our eyes in one second?
a) 48
b) 12
c) 24
d) 36
Answer:
c) 24

Question 4.
Say the important stage of an animation film?
a) character designing
b) background
c) characters
d) action
Answer:
a) character designing

Question 5.
Construction of a storyboard is a preparation of
a) Drawing a picture
b) finding time zones
c) producing an animation
d) character designing
Answer:
c) Producing an animation

Question 6.
Synfig Studio is a free-animation software
a) Drawing pictures
b) Studio
c) Three dimensional
d) Two dimensional
Answer:
d) Two dimensional

Question 7.
Synfig studio software is designed by
a) Stall men
b) Robert B Quattlebaum
c) Leslie Lamport
d) Donald Knuth
Answer:
b) Robert B Quattlebaum

Question 8.
What versions of animation software can run in GNU/Linux and Microsoft Windows?
a) Grass
b) Notepad
c) Synfig studio
d) Pencil
Answer:
c) Synfig studio

Question 9.
Which tool is used to fill colors to the objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 1
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 2

Question 10.
Which tool is used to draw rectangular objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 3
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 4

Question 11.
Which tool is used to move the objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 5
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 6

Question 12.
Expand FPS
a) Frames per system
b) Frames per second
c) File properties settings
d) Frames per hour
Answer:
b) Frames per second

Question 13.
Animation is done by fast and continuous movements of images in a two-dimensional canvas. The images are known as
a) characters
b) FPS
c) Frames
d) Scene
Answer:
c) Frames

Question 14
Which menu is used to studio
a) File
b) view
c) window canvas
d)canvas
Answer:
d)canvas

Question 15.
To change the FPS in Synfig studio, by clicking
a) canvas → properties → time
b) view → pause
c) window → toolbox
d) canvas properties → image
Answer:
a) canvas → properties → time

Question 16.
We can import Images into synfig studio and use them
a) odf
b) vector
c) Pdf
d) Bitmap
Answer:
d) Bitmap

Question 17.
FPS = 24, Time =5, for an animation. Find the total number of frames in that animation
a) 24,
b) 48
c) 120
d) 920
Answer:
c) 120

Question 18.
The frames that represent important positions are known as
a) Tweening
b) current time
c) key
d)keyframe
Answer:
d) keyframe

Question 19.
What happens when you press
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 7
play button
a) Animation works
b) Animation stops
c) no change
d) to save
Answer:
a) Animation works

Question 20.
The software fills up the frames in between two keyframes is called.
a) Interpolation
b) keyframe
c) Tweening
d) Edit
Answer:
c) Tweening

Question 21.
Rey frame utility is set in one of the panels in synfig audio. Give the names of that panel
a) Time track panel
b) layers panel
c) parameters panel
d) panel
Answer:
c) parameters panel

Question 22.
Name the extension of project file in synfig studio?
a). svg
b) .pdf
c) .sifz
d) .ods
Answer:
c) .sifz

Question 23.
In which panel to give the number of frames in current time?
a) Time track panel
b) layers panel
c) meters panel
d) panel
Answer:
a) Time track panel

HSSLive.Guru

Question 24.
What is the current time to animation from first frame
a) 0 f
b) 60 f
c) 120f
d) 121 f
Answer:
a) 0 f

Question 25.
What is the name for the frame with current time is ‘o’ f
a) first keyframe
b) last keyframe
c) middle keyframe
d) keyframe
Answer:
a) first keyframe

Question 26.
Give the order of activity to export a project file on synfig studio
a) File → Render
b) File → Save
c) File → Export
d) File → Save as
Answer:
a) File → Render

Question 25.
Which is not a member related to synfig studio?
a) parameters panel
b) time track panel
c) panel
d) layers panel
Answer:
c) panel

Question 28.
The utility to animate the flap its wings of a bird is
a) joining wings
b) adding time loop layer
c) copy of the first wing
d) adding the layer
Answer:
b) adding time loop layer

Question 29.
Write the activity to include an image to synfig studio
a) file → open
b) file → new
c) file → import
d) file → save
Answer:
c) file → import

Question 29.
in which menu import facility is available”?
a) Edit
b) file
c) canvas
d) windows
Answer:
b) file

Question 30.
Select ofie not group
a) . dv
b) .flv
c) . mpeg
d). svg
Answer:
d) .svg

Question 32.
What is the maximum value of current time?
a) 20 f
b) 60 f
c) 120 f
d) 1 f ,
Answer:
c) 120 f

Question 33.
What is the least value of current time?
a) 60 f
b) 120 f
c) 40 f
d) Of
Answer:
d)0f ‘

Question 34.
Keyframe are constructed in different type or same type
a) same type
b) different type
c) not constructing a keyframe
d) Equal type
Answer:
b) different type

Question 35.
Select any two free animation software’s
a) Anim studio
b) pencil
c) synfig studio
d) adobe flash
Answer:
b) pencil
and
c) synfig studio

Question 36.
An important stage in preparing animation is character designing. What is the meaning of character designing?
a) characters
b) Draw the characters
c) bringing character to life with humanity & personality
d) Giving life to the story
Answer:
b) Draw the characters
and
c) bringing character to life with humanity & personality

Question 37.
What are the uses of multi-colored buttons on the handle of a image?
a) To switch on the Animate editing mode
b) To adjust size
c) To take copy
d) To rotate if needed.
Answer:
b) To adjust size
and
d) To rotate if needed

Question 38.
Give the two activities done before starting animation from first keyframe
a) Current time is 60 f
b) Current time is 0 f
c) Before editing the motion, animate the edit mode button is active
d) press the play button to see the animation
Answer:
b) Current time is 0 f
and
c) Before editing the motion, animate the edit mode button is active

Question 39.
Sky and one star are drawn in synfig studio canvas. In the layers panel we can see a rectangle 001, and star 001. What are the tools use dot draw them
a) Circle tool
b) Fill tool
c) Star tool
d) Rectangle tool
Answer:
c) Star tool
and
d) Rectangle tool

Question 40.
Animation became much easier in film industry, with the arrival of new technologies. What are they?
a) computers
b) Radio
c) Animation software’s
d) office software.
Answer:
a) computers
and
c) Animation software’s

HSSLive.Guru

Question 41.
Create animations is one of the stages in animation film. There are two other stages before creating an animation. What are they?
a) To save animation project
b) Character designing
c) Playing an animation
d) Preparation of storyboards
Answer:
b) Character designing
and
d) Preparation of storyboards

Question 42.
Utilities of synfig studio windows are given below. Select one of each set and make a list.

Set-I

a) Database panel
b) Work area
c) Layers Panel
d) Task area
Answer:
c) Layers Panel

Set -II

a) Audio track
b) Video track
c) Time track panel
d) Fill
Answer:
c) Time track panel

Set-III

a) Parameters panel
b) Table
c) Storyboard
d) Path
Answer:
a) Parameters panel

Set -IV

a) Database window
b) tooIX
c) Editor window
d) Window
Answer:
b) tooIX

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Question 43
Select one video formats from each set

Set I

a) PNG
b) gif
c) xcf
d) sifz
Answer:
b) gif

Set II

a) dv
b) ods
c) odf
d) odp
Answer:
a) dv

Set III

a) jpg
b) flv
c) jpg
d) py
Answer:
b) flv

Set -IV

a) Html
b) ph
c) htm
d) mpeg
Answer:
d) mpeg

Question 44.
There are a lots of job opportunities related to animation in Government/public sectors of India and abroad. A list of job opportunities are given below. Select one from each set.

Set-I

a) Cinema production
b) Still photograph
c) Newspaper
d) Shops
Answer:
a) Cinema production

Set-II

a) Schools
b) advertising agencies
c) building construction
b) advertising agencies
Answer:
d) Library

Set-III

a) Radio
b) FM radio
c) TV
d) Tape recorder
c) TV

Set-IV

a) mobile
b) computer
c) calculator
d) computer games
Answer:
d) computer games

Question 45.
Images are included in synfig studio through import menu what are the uses of handles on the image? Select one form each set.

Set-1

a) To join the parts of an image
b) To arrange its position
c) To animation
d) Frame
Answer:
b) To arrange its position

Set – II

a) To arrange its size
b) To reduce its size
c) To increase its size
d) Can’t change its size
Answer:
a) To arrange its size

Set-II

a) To rotate
b) Can’t rotate
c) To remove the image
d) To include and image
Answer:
a) To rotate

Set-IV

a) To add a background image
b) Creating a scene
c) To adjust the views of the images
d) To change its background-color
Answer:
c) To adjust the views of the images

HSSLive.Guru

Question 46.
The layers are seen in the layers panel of a synfig studio some activities related to the layers are given in sets. Select correct activities from each set.

Set-1

a) Can’t change its order
b) can change its order
c) Arrange the order before including the layer
d) Layers are not visible
Answer;
b) can change its order

Set -II

a) Can’t group together
b) Objects
c) Can group together
d) Can’t change its color
Answer:
c) Can group together

Set-III

a) To take a copy of a layer
b) Layer is fixed
c) Can edit animation
d) Can’t delete the layer
Answer:
a) To take a copy of a layer

Set-IV

a) Removing the layer may effect he other
b) Can delete a layer
c) Can edit animation
d) Can’t delete the layers
Answer:
b) Can delete a layer

Plus One Physics Model Question Paper 4

Kerala Plus One Physics Model Question Paper 4

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

  • There is a ‘cool off time’ of 15 minutes in addition to the writing time.
  • Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
  • Read the instructions carefully.
  • Read questions carefully before you answering.
  • Calculations, figures, and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Plus One Physics Previous Year Question Papers and Answers 2018 1

Answer any four questions from question numbers 1 to 5. Each carries one score.
Plus One Physics Model Question Papers Paper 4 1

Engineering Physics MCQ with answers in PDF format.

Question 1.
Name the weakest force among the fundamental forces.
Plus One Physics Model Question Papers Paper 4 2

Question 2.
The work done during an isochoric process is …………….
Plus One Physics Model Question Papers Paper 4 3

The harmonic sequence formula is a sort of average calculator that is estimated by dividing the number of utilities.

Question 3.
Highway police detect over speeding vehicles by using ……………….
a. Magnus effect
b. Pascals law
c. Doppler effect
d. Bernoulli’s theorem
Plus One Physics Model Question Papers Paper 4 4

Question 4.
Two forces 3N and 4N are acting perpendicular to each other. The magnitude of the resultant force is
Plus One Physics Model Question Papers Paper 4 5

Question 5.
Say true/false: “Trade winds are produced due to conduction.”
Plus One Physics Model Question Papers Paper 4 6

Answer any five questions from question numbers 6 to 11. Each carries two scores.
Plus One Physics Model Question Papers Paper 4 7Plus One Physics Model Question Papers Paper 4 8

Question 6.
The displacement (S) of a body in time ‘t’ is given by S = at2 + bt. Find the dimensions of a and b.
Plus One Physics Model Question Papers Paper 4 9

Question 7.
Give the magnitude and direction of the net force on a stone of mass 0.1 kg.
a. Just after it is dropped from the window of a train accelerating 1 ms2.
b. Lying on the floor of a train which is accelerating with 1 ms-2, the stone being at rest relative to the train.
Plus One Physics Model Question Papers Paper 4 10

Question 8.
A body is rolling on a horizontal surface. Derive an equation for its kinetic energy,
Plus One Physics Model Question Papers Paper 4 11

Question 9.
The stress-strain graphs for two materials A and B are shown below (the graphs are drawn using the same scale) Which one is more elastic? Why?
Plus One Physics Model Question Papers Paper 4 12
Plus One Physics Model Question Papers Paper 4 13

Question 10.
“A heavy and a light body have the same kinetic energy.” Which one has greater momentum? Why?
Plus One Physics Model Question Papers Paper 4 14

Question 11.
The following figures refer to the steady flow of a nonviscous liquid. Which of the two figures is correct? Why?
Plus One Physics Model Question Papers Paper 4 15

Answer any five questions from question num 1.5 m numbers 12 to 17. Each carries three scores.
Plus One Physics Model Question Papers Paper 4 16

Question 12.
The side of a cube is measured as 3.405 cm.
a. How many significant figures are there in the measurement?
b. If the percentage error in the measurement of the side of the cube is 3%, find; the percentage error in its volume.
Plus One Physics Model Question Papers Paper 4 17

Question 13.
According to the conservation of energy “energy can neither be created nor be destroyed”
a. Prove law of conservation of mechanical energy in the case of a freely falling body.
b. The bob of a pendulum of length 1.5 m is released from the position A shown in the figure. What is the speed with which the bob arrives at the lowermost point B, given that 5% of its initial energy is dissipated against air resistance?
Plus One Physics Model Question Papers Paper 4 18

Question 14.
Acceleration due to gravity on earth changes with depth and height.
a. What is the weight of a body placed at the center of the earth? Why?
b. Find the height at which the acceleration due to gravity is 1/4th that at the surface of the earth.
Plus One Physics Model Question Papers Paper 4 19
Plus One Physics Model Question Papers Paper 4 20

Question 15.
A metal sphere of density ‘p’ and radius ‘a is falling through an infinite column of liquid of density ‘o’ and coefficient of viscosity Ty
a. Name any two forces acting on the; sphere.
b. With the help of Stokes theorem, derive an equation for the terminal velocity of I the sphere.
Plus One Physics Model Question Papers Paper 4 21

Question 16.
Conduction is the mode of transfer of heat in solids. of Write the unit of thermal conductivity.
b. “Burns produced by steam is severe than that produced by boiling water”Why?
Plus One Physics Model Question Papers Paper 4 22

Question 17.
A gas has ‘f’ degrees of freedom.
a. Calculate its Cp, Cv, and γ.
b. Define the mean free path.
Plus One Physics Model Question Papers Paper 4 23

Answer any five questions from question numbers 18 to 22. Each carries two scores.
Plus One Physics Model Question Papers Paper 4 24

Question 18.
A satellite moves in a circular orbit of radius ‘r’ with an orbital velocity.
a. Derive an equation for the orbital velocity of a satellite.
b. The time taken by Saturn to complete one orbit around the Sun is 29.5 times the earth year. If the distance of the earth from the Sun is 1.5 × 108km, then what will be the distance of the Saturn from the Sun?
Plus One Physics Model Question Papers Paper 4 25

Question 19.
In the simple harmonic motion, force is directly proportional to the displacement from the mean position.
a. Give an example of a harmonic oscillator.
b. Derive equations for the kinetic and potential energies of a harmonic oscillator.
c. Show graphically the variation of kinetic energy’ and potential energy of a harmonic oscillator.
Plus One Physics Model Question Papers Paper 4 26

Question 20.
A stretched string can be used as a musical instrument.
a. What is the fundamental frequency of a stretched string?
b. With neat diagrams, derive equations for the second and third harmonics of a stretched string.
Plus One Physics Model Question Papers Paper 4 27
Plus One Physics Model Question Papers Paper 4 28

Question 21.
A body having an initial velocity ‘v0’ has an acceleration ‘a’.
a. Using the velocity-time graph, derive an equation for displacement of the above body.
b. Draw the velocity Time graph and speed Time graph of a body thrown vertically in the air.
Plus One Physics Model Question Papers Paper 4 29

Question 22.
A javelin is thrown with an initial velocity ‘ V0‘ at an angle ” with the horizontal.
a. What are the horizontal and vertical velocities of the body
i. At the point of projection
ii. At maximum height
b. Find the angle of projection at which the maximum height attained by the javelin is equal to the horizontal range.
Plus One Physics Model Question Papers Paper 4 30

Answer any three questions from question numbers 23 to 26. Each carries five scores.
Plus One Physics Model Question Papers Paper 4 31

Question 23.
a. What is meant by ‘banking of roads’?
b. With a neat diagram, derive an equation for the maximum velocity of a car on a banked road.
c. What is the optimum speed of the car along the banked road?
Plus One Physics Model Question Papers Paper 4 32

Question 24.
The moment of inertia of a thin rod of mass M and length 1 about an axis perpendicular to the rod at its midpoint is \(\frac { { Ml }^{ 2 } }{ 12 }\).
a. What is the radius of gyration in the above case?
b. A student has to find the moment of inertia of the above rod about an axis (AB) perpendicular to the rod and passing through one end of the rod. Name and state the law used for this case.
c. Using the theorem, find the moment of inertia of the rod about AB.
Plus One Physics Model Question Papers Paper 4 33

Question 25.
Small drops of water assume spherical shape due to surface tension.
a. Define surface tension.
b. Derive an equation for the excess pressure inside a liquid drop of radius ‘R’ having surface tension σ.
c. Why do farmers plow the fields before summer?
Plus One Physics Model Question Papers Paper 4 34

Question 26,
Carnot engine is considered as an ideal heat engine.
a. Draw the PV graph of Carnot’s cycle.
b. Derive an equation to find the work done during an adiabatic process.
c. Calculate the efficiency of a heat engine working between ice point and steam point.
Plus One Physics Model Question Papers Paper 4 35

Answers

Answer 1.
Doppler effect

Answer 2.
Zero

Answer 3.
Doppler effect

Answer 4.
7N

Answer 5.
False

Answer 6.
[S] = [L]
[at2] = [L]
a = [LT-2]
[bt] = [L]
[b] = [LT-1]

Answer 7
a. Only force is gravitational. F = mg = 0.1 × 9.8 = 9.8 N downward j
b. Gravitational force is cancelled by normal I reaction.
∴ F2 = ma = 0.1 × 1 = 0.1 N, direction of motion of train.

Answer 8.
Plus One Physics Model Question Papers Paper 4 36

Answer 9.
In the two graphs, the slope of a graph of material A is greater than the slope of a graph of material B. So material A is more elastic than B. For material A the break-even point (D) is higher.

Answer 10.
Plus One Physics Model Question Papers Paper 4 37
Momentum is greater for a heavy body.

Answer 11.
Figure b is correct. According to an equation of continuity, the speed of liquid is larger at a smaller area. From Bernoulli’s theorem due to larger speed, the pressure will be lower at a smaller area and therefore the height of liquid column will also be at lesser height, while in Fig(a) height of liquid column at the narrow area is higher.

Answer 12.
Plus One Physics Model Question Papers Paper 4 38

Answer 13.
a. Law of conservation of energy. Energy can neither be created nor be destroyed, but it can be transformed from one form into another. Consider a body of mass’s’ placed at
Plus One Physics Model Question Papers Paper 4 39
b. Changing in PE after dissipation.
Plus One Physics Model Question Papers Paper 4 40

Answer 14.
Plus One Physics Model Question Papers Paper 4 41
Plus One Physics Model Question Papers Paper 4 42

Answer 15.
a. i. Weight, F, = mg acting downward
ii. Viscous force, F2 acting upward,
b. By strokes, formula F = 6πrηV Viscous force = Apparent weight of sphere in the solid
Plus One Physics Model Question Papers Paper 4 43

Answer 16.
a. W m-1K-1
b. Boiling water contains only a specific amount of heat energy required for it to boil. However, as steam is formed from boiling water, it contains the heat energy of boiling water, along with the latent heat of vaporization.i.e., 1kg of steam at 100°C contains 22.6 × 105 J more heat than 1 kg of water at 100°C. Hence, as steam has more heat energy, it can cause more severe burns than boiling water.

Answer 17.
Plus One Physics Model Question Papers Paper 4 44
b. Mean free path is an average distance between two successive collisions.

Answer 18.
a. It is the velocity required to put the satellite into its orbit around the earth.
Plus One Physics Model Question Papers Paper 4 45
The gravitational force on the satellite
Plus One Physics Model Question Papers Paper 4 46
The centripetal force required by the satellite to stay in this orbit is
Plus One Physics Model Question Papers Paper 4 47
in this orbit is In equilibrium the centripetal force is given by the gravitational force
Plus One Physics Model Question Papers Paper 4 48

Answer 19.
a. Oscillation of simple pendulum Oscillation of loaded spring
b. Let m be the mass of the particle executing SHM. Let v be the velocity at any instant,
Plus One Physics Model Question Papers Paper 4 49
Potential energy is the work required to take a particle against the restoring., force. Let a particle be displaced through a distance x from the mean position. Then restoring force, F = – kx, where k is the force constant. Now if we displace the particle further through a distance dx, Small work done, dw = – Fdx = kx dx Total work done from 0 to x
Plus One Physics Model Question Papers Paper 4 50

Answer 20.
a. Fundamental mode (or) First harmonic: If the string is plucked in the middle and released, then it vibrates in one segment with nodes at its ends and an antinode in the middle.
Plus One Physics Model Question Papers Paper 4 51
This is the lowest frequency with which string vibrates.
b. Second harmonic If the string is pressed in the middle and plucked at one-fourth of its length, then the string vibrates in two segments.
Plus One Physics Model Question Papers Paper 4 52
Plus One Physics Model Question Papers Paper 4 53
Third harmonic If the striping is pressed at one-third of its length from one end and plucked at one-sixth its length, it will vibrate in three segments.
Plus One Physics Model Question Papers Paper 4 54
Thus a collection of all possible mode is called harmonic series and n is called harmonic number.

Answer 21.
Plus One Physics Model Question Papers Paper 4 55
The area under the velocity-time graph gives the displacement of the body. Displacement, x = area OABD x = area of triangle ABC+ area of rectangle OACD.
Plus One Physics Model Question Papers Paper 4 56

Answer 22.
a.
i. Horizontal Vx = V0 cosθ Vertical Vy = V0sinθ
ii. Horizontal V’x = VO cosθ
Plus One Physics Model Question Papers Paper 4 57

Answer 23.
a. To avoid skidding and damage to tires of vehicles, the outer part of a road is slightly raised than the inner part. This is known as banking of roads.
Plus One Physics Model Question Papers Paper 4 58
The forces on the car are:
1. The weight of the car vertically downwards.
2. Normal reaction Racing normal to the road.
3. Frictional force acting parallel to the road.
Since there is no vertical acceleration,
R cosθ = mg + F sinθ
or R cosθ – F sinθ = mg …(1)
Now for maximum speed, F = μ, R
The centripetal force is provided by horizontal components of Rand Fas shown in the figure.
Plus One Physics Model Question Papers Paper 4 59
Plus One Physics Model Question Papers Paper 4 60

Answer 24.
a. The radius of gyration (k). It is the defined as the distance from an axis of rotation at which, if the whole mass of the body was concentrated, then its moment of inertia about that point would be the same as the moment of inertia of actual distribution of mass. l = Mk2
The radius of gyration (k) of a body is the square root of a ratio of the moment of inertia and a total mass of the body.
ie., a radius of gyration, k= \(k=\sqrt { \frac { l }{ M } }\)

b. Theorem of parallel axes: This theorem is good for any shape. The moment of inertia of the body about any axis is equal to the sum of a moment of inertia of a.parallel axis passing through the center of mass and product of its mass of the body and square of the distance between the two parallel axes.
Plus One Physics Model Question Papers Paper 4 61
where I am the moment
c. Using parallel axes theorem, the moment of inertia about AB,
Plus One Physics Model Question Papers Paper 4 62

Answer 25.
a. Surface tension (a) is the property due to which the free surface of a liquid at rest behaves like an elastic stretched membrane tending to contract so as to occupy a minimum surface area.
Plus One Physics Model Question Papers Paper 4 63
Thus it is measured as the force acting per unit length of an imaginary line drawn on the liquid surface, the direction of force being perpendicular to this line and tangential to the liquid surface.

b. Consider a liquid drop of radius R and surface tension o. Let P be the excess pressure inside the drop. The work done by the force due to excess pressure is
Plus One Physics Model Question Papers Paper 4 64
c. On plowing, the gap between sand particles act as a capillary tube, so that groundwater reaches the surface easily due to capillary rise.

Answer 26.
Plus One Physics Model Question Papers Paper 4 65
b. Work was done in the adiabatic process: We have a small amount of work done when volume changes through at pressure P.
Plus One Physics Model Question Papers Paper 4 66
(volume changes from v1 to v2 diabolically)
Plus One Physics Model Question Papers Paper 4 67
Plus One Physics Model Question Papers Paper 4 68
Plus One Physics Model Question Papers Paper 4 69

Plus One Physics Previous Year Question Papers and Answers

Kerala SSLC IT Theory Questions and Answers Chapter 2 Publishing

You can Download Publishing Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 2 Publishing

Publishing Questions & Answers

Question 1.
Clickthrough Insert → fields → more fields in writer software. We get a field window contains utility.
a) File
b) Edit
c) Database
d) None
Answer:
c) Database

HSSLive.Guru

Question 2.
Formattings of a text is copped to another text is using tool
a) Clone formatting
b) style
c) Paragraph formatting
d) styles and formatting
Answer:
a) Clone formatting

Question 3.
Which is the Clone formatting tool?
Kerala SSLC IT Theory Questions and Answers Chapter 2 Publishing 1
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 2 Publishing 2

Question 4.
A word can define headings, subheadings and para-graph separately is called
a) Style
b) Insert
c) Formatting
d) Clone
Answer:
a) Style

Question 5.
In which menu lies the Index and Tables
a) File
b) Insert
c) Edit
d) Data
Answer:
b) Insert

Question 6.
Where does contain the table of contents in your text?
a) Middle of the text
b) Not entered
c) At the beginning
d) At the end
Answer:
c) At the beginning

Question 7
The Software developed by Leslie Lamport is
a) Linux
b) Windows
c) Android
d) LaTex
Answer:
d) LaTex

Question 8.
The program for type setting technical documents is developed by Donald Knuth is
a) LaTex
b) Tex.
c) Linux
d) Windows
Answer:
b) Tex

Question 9.
If you click at the word preface on the table of contents in writer. What message appears?
a) Ctrl-click to follow link
b) Click to follow link
c) Follow link
d) Click to follow link to close the window
Answer:
a) Ctrl-click to follow link

Question 10.
Which menu lies the field’s utility?
a) Edit
b) Insert
c) File
d) formatting
Answer:
b) Insert

HSSLive.Guru

Question 11.
Which software having the mail merge facility?
a) Libre office writer
b) Python
c) Firefox
d) Calculator
Answer:
a) Libre office writer

The Full form of DSLR is Digital Single Lens Reflex Cameras.

Question 12.
To include the names and address one by one in a table in the letters using a feature called
a) Style
b) Mail merge
c) Database
d) Indexable
Answer:
b) Mail merge

Question 13.
Which style is used to the headings more attractive?
a) Character style
b) Page style
c) Paragraph style
d) List style
Answer:
c) Paragraph style

Question 14.
Anu was prepared a style named main head. Then he selects the all headings in a report and clicks on the style named main head. What happens for the headings?
a) All headings are changed to same type
b) No change to the headings
c) Headings are changed to different sizes
d) Headings are seen in different colors
Answer:
a) AI1 headings are changed to same type

Question 15.
The use of organizer tab in styles and formattings window for creating a new style is?
a) No use at all
b) To applying the style
c) Giving the name for the style
d) Don’t give a name for the style
Answer:
c) Giving the name for the style

Questions 16.
Which software includes the styles and formattings utility
a) Cale
b) Python
c) Writer
d) Sun clock
Answer:
c) Writer

Question 17.
Pick out the correct statement related to a frame
a) Can’t able to type inside the frame
b) Frame can move at any place on the page
c) Frame can’t move on the page
d) Can’t able to insert an image in a frame
Answer:
b) Frame can move at any place on the page

Question 18.
Where did you get the fields window?
a) Insert →Fields → More fields
b) Insert → Media → Fields
c) Format → More fields
d) Tools → More fields
Answer:
a) Insert → Fields → More fields

Question 19.
A correct statement related to clone formatting is
a) Use clone formatting for a large report
b) Formattings of headings can be changed by changing the formatting of each heading separately
c) Formattings of headings can be changed by changing the formatting of one heading
d) Formattings of heading cannot be copped to another
b) Formattings of headings can be changed by changing the formatting of each heading separately

Section – II

Question 20.
Right-click on the heading (styles name) on a paragraph style, we get the utilities
a) Modify
b) Colour
c) New
d) Pattern
Answer:
a) Modify

Question 21.
Libre office writer file prepared in Malayalam fonts and English fonts, on which fonts are changed to change the styles of the files?
a) If the file is in Malayalam fonts, then change the CTL font
b) If the file is in English fonts, then change the CTL font
c) If the file is in Malayalam fonts, then change western text font
d) If the file is in English fonts, then change the Western text font
Answer:
a and b

Question 22.
Select the correct statements
a) The styles in a writer are changed and use them
b) The styles in a writer cannot be changed
c) New styles cannot be created
d) New styles can prepared in a writer.
Answer:
a and d

Question 23.
Select two correct statements related to a frame
a) Can’t move from one part of a page
b) To place text or images within a document separated from the main contents
c) To place anywhere in the page
d) Not separated from the main contents
Answer:
b and c

Question 24.
Participant’s cards are prepared by mail merge utility and select print from file menu. Then you get a window to select two utilities to save the output file what are they?
a) Save as single document
b) Save as image
c) Save as individual documents
d) Save picture as
Answer:
a and c

Question 25.
Using mail merge to preparing the letters for the parents about the Youth Festival. What are the preparations we make for this?
a) Making a list of parents address in Libre Office writer
b) Making a list of parents addresses in Libre Office Calc
c) Preparing the letters to the parents in Libre office writer
d) Preparing the letters to the parents in Libre office calc
Answer:
b and c

HSSLive.Guru

Question 26.
Why did scientific articles and research reports are prepared by LaTex Software?
a) Facility to use English fonts
b) Facility to use Malayalam fonts
c) Several features for typesetting Symbols
d) Several features for typesetting equations.
Answer:
c and d

Question 27.
What are the difficulties to prepare a text /document using clone formatting?
a) Separate formattings the contents of a text is difficult
b) Formattings of heading can be changed by changing the formattings of each heading separately
c) Formattings of headings/ paragraph can be changed by changing one
d) One by one formatting is very easy
Answer:
a and b

Question 28.
The order of preparing a table of contents to the school report are following. Select one from each set.

Set -1

a) Close the report
b) Type the school report
c) Correct the report
d) Open the prepared school report
Answer:
d) Open the prepared school report

Set – II

a) Click on the last page of the report
b) Click on the place where the table of contents to be inserted
c) Insert the table of contents on the cover page
d) Click on the blank space of the report
Answer:
b) Click on the place where the table of contents to be inserted

Set – III

a) Select Index and Tables from Insert Menu
b) Select fields from Insert menu
c) Select modify
d) Select styles and formatting from insert menu
Answer:
a) Select Index and Tables from Insert Menu

Set – IV

a) Close the window
b) A new window appears, give a heading and background color for the table of contents and click OK.
c) Click cancel on the new window
d) Leave the window without any change
Answer:
b) A new window appears, give a heading and background color for the table of contents and click OK.

Question 29.
What are the styles included in styles and formatting window?

Set -1

a) Paragraph style
b) Western-style
c) Formatting style
d) More style
Answer;
a) Paragraph style

Set – II

a) Apply style
b) Character styles
c) Style Box
d) Clone formatting
Answer:
b) Character styles

Set – III

a) Heading style
b) Report style
c) Frame style
d) Life style
Answer:
c) Frame style

Set-IV

a) Page style
b) Heading 1
c) Head
d) Address
Answer:
a) Page style

Question 30.
The uses of mail merge utility in LibreOffice writer.

Set-I

a) Preparing Notice
b) Preparing the letter for parents about youth festival
c) Writing the mark list
d) To prepare address
Answer:
b) Preparing the letter for parents about youth festival

Set – II

a) Creating a style
b) Taking a print
c) To taking a copy
d) To prepare participant’s card
Answer:
d)To prepare participant’s card

Set – III

a) To prepare certificate
b) To prepare a letter to a friend
c) To prepare a document
d) To prepare a study report
Answer:
a) To prepare certificate

Set – IV

a) To calculate the achievement
b) To prepare electricity bill
c) To apply style
d) To prepare the table of contents
Answer:
b) To prepare electricity bill

HSSLive.Guru

Question 31.
Which are the styles we are using from paragraph style? Make a list using from each set

Set – I

a) Table
b) Report
c) Heading
d) Contents
Answer:
c) Heading

Set – II

a) Caption
b) Font
c) Footer
d) Color
Answer:
a) Caption

Set-III

a) Find
b) Link
c) Object
d) Index
Answer:
d) Index

Set-IV

a) Body
b) Media
c) Text body
d) Frame
Answer:
c) Text body

Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids

You can Download Solids Questions and Answers, Activity, Notes, Kerala Syllabus 10th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Maths Solutions Chapter 8 Solids

Solids Textbook Questions & Answers

Textbook Page No. 191

Solids Class 10 Kerala Syllabus Kerala Syllabus Chapter 8 Question 1.
A square of side 5 centimetres, and four isosceles triangles!of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper is needed?
Answer:
Area of base = 5 × 5 = 25 cm2 Area of one triangle 1/2 × 5 × 8 = 20 cm2 Curved surface area = 4 × 20 5cm = 80 cm2
Paper is needed to make a square pyramid = 25 + 80 = 105 cm2
Solids Class 10 Kerala Syllabus Kerala Syllabus Chapter 8

Solids in Maths SSLC Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Surface Area of the toy
= 16 × 16 + 4 × 1/2 × 16 × 10
=256 + 320 = 576 cm2
Surface Area of 500 toys = 500 × 576 = 288000 cm2
Sslc Maths Chapter 8 Kerala Syllabus
Sslc Maths Solids Kerala Syllabus Chapter 8

Kerala Syllabus 10th Standard Maths Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Lateral faces are equilateral surfaces Surface area
= Base Area + Curved surface area
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 85
= 900 + 900 √3
=900 + 1558.8 = 2458.8 cm2 = 2459 cm2
Sslc Maths Solids Equations Kerala Syllabus Chapter 8

The Chebyshev’s theorem calculator counts the probability of an event being far from its expected value.

Sslc Maths Chapter 8 Kerala Syllabus Question 4.
The perimeter of the base of square^»yra- mid is 40 centimetres and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Base perimeter = 4a = 40
a=10cm
Total length of edges = 92 cm
Length of total laterals edge = 92 – 40 = 52
Length of one laterals edge = \(\frac { 52 }{ 4 }\) = 13 cm
Surface area of pyramid = a 2 + 2 al
= 102 + 2 × 10 × 13
= 100 + 260 = 360 cm2

KBPS full form, KBPS stands for, meaning, what is KBPS, description, example, explanation, acronym for, abbreviation, definitions, full name.

Surface Area of Kerala Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Curved surface area = 2al
Area of base = a2
a2=2al
a = 21 ⇒ 1 = a/2
For making a square pyramid first we must determine its base, one side of the lateral will be the base. Other two sides make half of base by reducing the angle. That is angle at apex will be less than 90°.

Textbook Page No. 193

Sslc Maths Solids Kerala Syllabus Chapter 8 Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made.
Sslc Maths Chapter 8 Solids Kerala Syllabus
What is the height of this pyramid?
What if the square and triangles are like this?
Solids Chapter Class 10 Kerala Syllabus
Answer:
Sslc Maths Chapter Solids Kerala Syllabus Chapter 8

Sslc Maths Solids Equations Kerala Syllabus Chapter 8 Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Base edge of a square pyramid = 10 cm
Let h be the height
Base edge a = 10 cm
Height h= 12 cm
Slant height =
Solids Class 10 Kerala Syllabus Chapter 8

Sslc Maths Chapter 8 Solids Kerala Syllabus Question 3.
Prove that in any square pyramid, the squares of the height, slant height and lateral edge are in arithmetic sequence.
Answer:
Height = h,
Slant height = l,
Lateral edge = e
Solids In Maths Sslc Kerala Syllabus Chapter 8

Solids in Maths Question 4. A square pyramid is to be made with the triangles shown here as a lateral face. What I would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?
Class 10 Maths Solids Kerala Syllabus Chapter 8
Answer:
Sslc Maths Chapter 8 Solutions Kerala Syllabus
It is impossible to make a square pyramid of base edge 40cm.

Textbook Page No. 195

Solids Chapter Class 10 Kerala Syllabus Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
a = 10, l = 15
Volume of pyramid =
Sslc Maths Solids Questions Kerala Syllabus Chapter 8
Maths Solids Class 10 Kerala Syllabus Chapter 8

Sslc Maths Chapter Solids Kerala Syllabus Chapter 8 Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times the height of the second pyramid is the height of the first?
Answer:
Sslc Solids Solutions Kerala Syllabus Chapter 8
The height of the second pyramid is 4 times the height of the first pyramid.

Solids Class 10 Kerala Syllabus Chapter 8 Question 3.
The base edges of two square pyramids are in the ratio 1:2 and their heights in the ratio 1:3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:
Solids Maths Questions Kerala Syllabus Chapter 8

Solids In Maths Sslc Kerala Syllabus Chapter 8 Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Length of base edge a = 18 cm
Sslc Maths Solutions Kerala Syllabus Chapter 8

Class 10 Maths Solids Kerala Syllabus Chapter 8 Question 5.
The slant height of a square pyramid is 25 centimetres and its surface area is 896 square centimetres. What is its volume?
Answer:
l = 25 cm
Surface area = 896 cm
a2 + 2al = 896
a2 + 2a × 25 = 896
a2 + 50a – 896 = 0
Kerala Syllabus 10 Maths Solutions Kerala Syllabus Chapter 8

Sslc Maths Chapter 8 Solutions Kerala Syllabus Question 6.
All edges of a square pyramid are of the same length and its height is 12 centimetres. What is its volume?
Answer:
Std 10 Kerala Syllabus Maths Solutions Chapter 8

Sslc Maths Solids Questions Kerala Syllabus Chapter 8 Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base perimeter 4a = 64
a= 16 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 19
Surface area = a2 + 2al
= 162 + 2 × 16 × 17 = 256 + 544 = 800 cm2

Textbook Page No. 198

Maths Solids Class 10 Kerala Syllabus Chapter 8 Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
R = Radius of the circle
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 20
Radius of cone = 1.66 cm
Radius of circular part = slant height of cone = 10 cm.
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 21

Sslc Solids Solutions Kerala Syllabus Chapter 8 Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Central anglejof the sector
(x) = \(\frac { r }{ l }\) × 360,
r = 10 cm, l = 25 cm
\(=\frac{10}{25} \times 360\)
= 144°

Solids Maths Questions Kerala Syllabus Chapter 8 Question 3.
What is the ratio of the base-radius and slant height of a cone made by rolling up a semicircle?
Answer:
Radius of bigger circle = R
Radius of smaller circle = r
Radius of circular base of the pyramid = r = \(\frac { R }{ 2 }\)
Ratio between radius and slant height
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 22

Textbook Page No. 199

Sslc Maths Solutions Kerala Syllabus Chapter 8 Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 23
r = 12cm l = 25cm
Curved surface area = πrl
π × 12 × 25 = 300π = π × 12 × 25
= 300 × 3.14 = 314 × 3
= 942 cm2

Kerala Syllabus 10 Maths Solutions Kerala Syllabus Chapter 8 Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 40 centimetres?
Answer:
Radius = \(\frac { 30 }{ 2 }\)= 15 cm =r
Height = h = 40 cm
Total surface area = Base area + Curved surface
area = πr2 + πrl
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 24
= 865 πr = 2718.6 m2

Std 10 Kerala Syllabus Maths Solutions Chapter 8 Question 3.
A Conical firework is of. base diameter 10 centimetres and height 12 centimetres, 10000 such fireworks are to be wrapped in colour paper. The price of the colour paper is 2 rupees per square metre. What is the total cost?
Answer:
r = 5 cm
h =12 cm
t = 13 cm
Total surface area of one firework = Base area + Curved surface area
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 25

Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
Perimeter of base of a cone is equal to half of perimeter of large cone.
Radius of pyramid = R/2
Perimeter of base
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 26
= 2 × Base area , that is twice

Textbook Page No. 200

Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
r = 15cm, h = 40cm
Volume of cone = \(\frac { 1 }{ 3 }\) πr² h
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 27
= \(\frac { 1 }{ 3 }\) π × 15 × 15 × 40
= 3000π = 3000 × 3.14
= 314 × 30
= 9420 cm3

Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 28
No. of cones = Volume of cylinder/Volume of cone Volume of cylinder
= π × 12 × 12 × 20
Volume of cone = \(\frac { 1 }{ 3 }\) π × 4 × 4 × 5
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 29
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 30

Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 31

Question 4.
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 32

Question 5.
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 33

Textbook Page No. 203

Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area 0f each hemisphere?
Answer:
Surface area of the solid sphere
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 34
Surface area of one hemisphere = 3 πr2
3 πr2 = 3 π × \(\frac { 30 }{ π }\) = 90 cm2

Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Ratio of volumes
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 35

Question 3.
base radius and length of a metalder are 4 centimetres and 10 centimetres, If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 36
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 37
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 38

Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres. What is the radius of each sphere?
Answer:
Radius = 12cm
Volume of bigger sphere
\(=\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \times 12^{3}\)
If the radius of smaller sphere is ‘r’
Volume of 27 smaller spheres = Volume of the bigger sphere
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 39

Question 5.
From a solid sphere of radius 10 centimetres, a cone of height 16 centimetres is carved out What fraction of the volume of the sphere is the volume of the cone?
Answer:
Radius of sphere = 10 cm
Radius of cone
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 40

Question 6.
The picture shows the dimensions of a petrol tank.
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 41
How many litres of petrol can it hold?
Answer:
Length of circular cylinder = 4 m
Height = 4 m
Radius = 1m
Volume of circular cylinder = π × 12 × 4 = 4π
= 4 × 3.14 = 12.56 cm3
Volume of two hemisphere = Volume of a sphere
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 42
Litres of petrol the tank can hold = 12.56 + 4.19 = 16.73 m3 = 16750 litre

Question 7.
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 43

Solids Orukkam Questions & Answers

Worksheet 1

Question 1.
The base edge of a square pyramid is Stem, height 3cm. Calculate slant height and lateral edge
Answer:
Base edge = 8 cm, height = 3 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 44

Question 2.
Slant height of a square pyramid is 10cm, height 6cm .Calculate total length of the edges.
Answer:
Slant height =10 cm, height = 6 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 45

Question 3.
The slant height of a square pyramid is 12 cm, lateral edge 13 cm. Calculate height
Answer:
Slant height = 12 cm
lateral edge = 13 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 46

Question 4.
The length of base edge is 24 cm, slant height 13 cm. Find height and lateral edge
Answer:
Base edge = 24 cm
Slant height = 13 cm
Height \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{25}=5 \mathrm{cm}\)
Length of lateral edge = \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{313} \mathrm{cm}\)

Worksheet 2

Question 5.
A sector is folded in such a way as to get a cone. Radius of the sector is 12 cm, central angle 120°.Calculate radius and slant height
Answer:
Slant height of cone = radius of sector = 12 cm
Radius of cone = \(\frac { 120 }{ 360 }\) of radius of sector = 12 × \(\frac { 120 }{ 360 }\) = 4 cm

Question 6.
The central angle of a sector is 90°, radius 16cm, calculate slant height and radius
Answer:
slant height of cone = 16 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 47

Question 7.
Slant height of a cone is 20 cm, radius 10 cm. What should be the radius and central angle of the sector?
Answer:
Radius of sector = slant height of cone = 20 cm
Central angle of the sector = radius of cone \(\times \frac{360}{R}=10 \times \frac{360}{20}=18^{\circ}\)

Question 8.
Radius of a cone is 4cm, slant height is 5/2 times radius. Calculate the radius and central angle of the sector.
Answer:
Radius of sector = slant height of cone =
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 48

Worksheet 3

Question 9.
The base edge of a square pyramid is 6cm, height 4cm, calculate slant height and total surface area.
Answer:
Length of base edge = 6 cm, height = 4 cm,
slant height = \(\sqrt{(3)^{2}+(4)^{2}}=\) \(\sqrt{9+16}=\sqrt{25}=5 \mathrm{cm}\)
Total surface area = base area + curved surface area=(6)2 + 2 × 6 × 5 = 36 + 60 = 96 cm2.

Question 10.
The height of a square pyramid is 12cm, slant height 15cm , calculate total surface area and volume
Answer:
height =12 cm. slant height = 15 cm
\(a=2 \sqrt{15^{2}-12^{2}}\) = 2 × 9 = 18 cm2
Total surface area = base- area + curved surface area
= (18)2 + 2 × 18 × 15 = 324 + 540 = 864 cm2
Volume = \(\frac{1}{3}(18)^{2} \times 12=1296 \mathrm{cm}^{3}\)

Question 11.
The base perimeter of a square pyramid is 48cm. Slant height is 10cm. Calculate lateral surface area and volume.
Answer:
Base perimeter = 48cm
baseedge = \(\frac { 48 }{ 4 }\) = 12 cm, slant height = 10cm
height = \(\sqrt{(10)^{2}-(6)^{2}}=\sqrt{64}=8 \mathrm{cm}\)
Curved siuface area= 2 × 12 × 10 = 240 cm2
Volume = \(\frac { 1 }{ 3 }\) (12)2 × 8 = 384 cm3

Question 12.
The height of a square pyramid is 15cm, volume 1620cm. Calculate the total surface area.
Answer:
Volume of square pyramid =
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 49

Worksheet 4

Question 13.
The base area of a cone is 25 π cm, curved surface area 165 π. Calculate total surface area.
Answer:
Total surface area of cone = base area + curved surface area
= 25 π +165 π = 190 cm2

Question 14.
Base area of a cone is 81 π, height 12 cm. Calculate volume
Answer:
Volume of cone = \(\frac { 1 }{ 3 }\) × base perimeter ×
height = \(\frac { 1 }{ 3 }\) × 81 π × 12 = 324 π cm3

Question 15.
The height of a cone is 4cm, slant height 5cm. Calculate total surface area
Answer:
Height of cone = 4 cm
Slant height = 5 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 50
Surface area = π (3)2 + π x 3 x 5 = 9 π + 15 π =24 π cm2

Question 16.
Radius of a cone is 10cm, volume 3140 cubic centimeter. Calculate total surface area
Answer:
Radius of cone = 10 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 51

Question 17.
Calculate the surface area and volume of a sphere of radius 3 cm.
Answer:
Surface area of sphere = 4 π (3)2 = 36 π cm2
Volume = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \pi \times(3)^{3}=36 \pi \mathrm{cm}^{3}\)

Workshee 5

Question 18.
Calculate the volume of a sphere of surface area 144 π square centimetre.
Answer:
cKerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 51

Solids SCERT Questions & Answers

Question 19.
The measurements of the lateral surface of a square pyramid are shown in the figure. Calculate die base edge and slant height of die pyramid. [Score: 2, Time: 3 minute]
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 53
Answer:
Base edge = 10cm (1)
The given figure can be divided into n two right-angled triangles their angles are 30°, 60°, 90°, so ratio of their side will be 1: √3: 2.
2x = 10, x = 5
Slant height = 5 √3 cm (1)3

Question 20.
Is It possible to construct a pyramid of base edge 24 cm and lateral edge 13 cm? Justify [Score: 2, Time: 3 minute]
Answer:
Since slant height is 5 cm, such a pyramid can’t be constructed. (1)
Slant height should be greater than half of the base edge. \(\sqrt{13^{2}-12^{2}}=5 \mathrm{cm}\) (1)

Question 21.
Lateral surface of a square pyramid is shown in the Figure. All angles are equal
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 54
Find the total length of all edges of the square: pyramid. Find the slant height What is the |atio between height and slant height [Score: 4, Time: 5 minutes]
Answer:
Sum of edges = 8 x 8 = 64 cm (1)
Slant height = 4√3 cm (1)
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 55

Question 22.
Devikamade a square pyramid having base edge 40cii and height 15cm. Unfortunately, one lateral face got separated from the pyramid. Check which figure given below shows the isosceles triangle that got separated. [Score: 3, Time: 5 minute]
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 56
Square pyramid can’t be constructed since slant height 1 should be greater than half the base edge. (1)
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 57
If base edge = 40 and slant leight = 35, then height can’t be 15 (1)
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 58
Here height of pyramid is 15, so this is the isosceles triangle that got separated (1)

Question 23.
A tent constructed in the form of a square pyramid of base perimeter 80 metres and lateral edge 26 metres
a. Calculate the slant height of the tent
b. Calculate the area of tarpaulin sheet. required to cover the lateral faces of the tent. [Score: 3, Time: 5 minute]
Answer:
Base perimeter = 80 m, Base edge= 20 m Lateral edge = 26 m
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 59
= 960 sq.metre (2)

Question 24.
The triangle given in the figure is one lateral face of a square pyramid.
a. Calculate the slant height.
b. Find the lateral surface area of the pyramid, [Score: 3, Time: 4 minutes]
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 60
Answer:
Slant height \(=\sqrt{17^{2}-8^{2}}\)
\(=\sqrt{275}=15 \mathrm{cm}\) (2)

Question 25.
A square pyramid is made from a solid cube having edge 30cm. Calculate the surface area. [Score: 3, Time: 5 minute]
Answer:
Base edge of the square pyramid = 30 cm
Height = 30cm
Slant height = \(\sqrt{30^{2}+15^{2}}\)
\(=\sqrt{900+225}=\sqrt{1125}=15 \sqrt{5}\) (1)
Lateral surface area = \(=4 \times \frac{1}{2} \times 15 \sqrt{2} \times 30\)
= 60 x 15√5 = 900√5 sq.cm (1)
Total surface area = 900 + 900√5
= 900(1 + √5) sq.cm (1)

Question 26.
The lateral faces of a square pyramid are equilateral triangles Lateral, edge = 20 cm
a. Calculate the slant height
b. Find its surface area.
C. Find its volume. [Score: 5, Time: 6 minute]
Answer:
a. Slant height = 10 √3 cm (1)
b. Lateral surface area
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 61

Question 27.
Prove that the ratio between the base edge, slant height and height of a square pyrarmid having equal edges is 2: √3 : √2. [Score: 4, Time: 5 minute]
Answer:
Slant height = √3 a
Height = \(\begin{array}{l}{=\sqrt{(\sqrt{3} a)^{2}-a^{2}}} \\ {=\sqrt{2} a}\end{array}\)
Base edge: Slant height : Height
= 2a: √3a : √2 a = 2: √3 : √2 (1)

Question 28.
The ratio between the base edges of two square pyramids is 1: 2. The heights are also in the same ratio. If the volume of the first pyramid islO cubic centimeters, calculate the volume of the sec ond one. [Score: 3, Time: 4 minute]
Answer:
v1 : v2 = 1 : 8
v1 = \(\frac { 1 }{ 3 }\) a2 h
v2 = \(\frac{1}{3}(2 a)^{2} \times 2 h, v_{1}: v_{2}=1: 8\) (1)
Volume of the second pyramid = 800 cm2 (1)

Question 29.
Meera constructed a square pyramid of base edge 10cm and height 6cm Manu made a square pyramid having base edge 5cm and height 4cm. Find the volume of the pyramids and compare the measurements. [Score: 3, Time: 4 minutes]
Answer:
Volume of Meera’s pyramid = \(\frac { 1 }{ 3 }\) x Baste edge x height 3
= \(\frac { 1 }{ 3 }\) x 102 x 6 =200 cm3 (1)
Volume of Manu’s pyramid = \(\frac { 1 }{ 3 }\) x 52 x 24 = 200cm3
Volumes are equal (1)

Question 30.
The central angle of a sector is 288° If this sector is rolled up to make a cone, find the ratio between the radius and slant height of the cone. [Score :: 4, Time: 5 minutes]
Answer:
360 x \(\frac { 4 }{ 5 }\) = 288
∴ Radius of the cone 4 = \(\frac { 4 }{ 5 }\) x radius of the big circle (1)
∴ If r is the radius of the circle Radius of the
Radius of the cone = \(\frac { 4 }{ 5 }\) r (1)
But radius of the circle = slant height of cone
i.e., 1 = r (1)
∴Ratio between the radius of the cone and slant height
\(=\frac{4}{5} r: r=\frac{4}{5}: 1=4: 5\) (1)

Question 31.
The ratio between the radius and slant height of a cone is 2 : 3. Find the central angle of the sector to make the cone. [Score: 3, Time: 4 minutes]
Answer:
Ratio between the radius and slant height 2 : 3 (1)
Area length of the sector is equal to \(\frac { 2 }{ 3 }\) part of the circle perimeter. (1)
Central angle of the sector = 360 x \(\frac { 2 }{ 3 }\) = 240° (1)

Question 32.
The central angle of a circle is divided in the ratio 2 : 3 to form two sectors. Two cones are made by rolling up the two Rectors.
a. Find out the ratio between the base perimeters of the cones.
b. What is the ratio between the curved surface areas. [Score: 3, Time 6 minutes]
Answer:
a. Central angles are in the ratio 2 : 3 , so, let the perimeter of the two clones which made by rolling up this two sectors be \(\frac { 2 }{ 5 }\)
part and \(\frac { 3 }{ 5 }\) part of the perimeter of the circle (1)
That is perimeter of each sector be \(2 \pi r \times \frac{2}{5} \text { and } 2 \pi r \times \frac{3}{5}\)
Ratio between base perimeters of cone = \(2 \pi r \times \frac{2}{5}: 2 \pi r \times \frac{3}{5}=2: 3\)
b. Base perimeter of the cones will be \(\frac { 2 }{ 5 }\) and \(\frac { 3 }{ 5 }\) parts of the circumference of the 5 circle. (1)
Ratio between the perimeters of die cones = \(\pi r^{2} \times \frac{2}{5}: \pi r^{2} \times \frac{3}{5}=2 ; 3\) (1)

Question 33.
Find the ratio between the radius and slant height of a cone by roiling up a sector with central angle 120°. If the curved surface area is 108 π, find the radius and slant height of the cone. [Score:5,Time:7minutesv]
Answer:
Area of the sector with central angle 120° is one-third of the area of the circle. (1)
Curved surface area = 108 π
Ares of die; sector = 108 π,
which is one-third of the area of the circle Area of the circle = 108 π × 3
πr² = 324 π (1)
Radius of the circle r = 18 cm
Slant height = 18 cm (1)
Radius of the cone = 6 cm (1)

Question 34.
A wooden cone is has radius 30crn and height 40crn. Find its slant height. Calculate the cost to paint the face of 10 such cones at the rate of Rs.50/- per square metre. [Score: 5, Time: 7 minutes]
Answer:
Base Radius = 30 cm
Height =40 cm
Slant height = \(\sqrt{40^{2}+30^{2}}=50\) (1)
Surface area of the cone = πr² + πrl = π × 302 + π × 30 × 50 (1)
= 900π + 1500π = 2400π (2)
Total cost to paint 10 cones \(=\frac{2400 \pi \times 10 \times 50}{10000}\)
= \(=\frac{2400 \times 3.14 \times 10 \times 50}{10000}=377 \mathrm{Rs}\) (1)

Question 35.
Two cones are made using two sectors of central angles 60° and 120° of a circle. If the radius of the smaller cone is 5cm
a. Calculate the radius and base area of the smaller cone.
b. Find the surface area of the bigger cone. [Score: 5, Time: 8 minutes]
Answer:
a. Central angle of the small sector = 60°
\(\frac { 1 }{ 6 }\) part of the area of the circle Base radius of cone formed from above sector = 5 cm (1)
Radius of the circle = 5 × 6 = 30 (1)
Similarly, area of the sector of central angle 120° = \(\frac { 1 }{ 3 }\) of the area of the circle Base radius ofthe bigger cone = 30 × \(\frac { 1 }{ 3 }\) = 10 (1)
Base area of the bigger cone = π × 102 = 100× (1)
b. Curved surface area ofthe bigger cone = π × 10 × 30 = 300π (1)
Surface area = 100π + 300π = 400π cm2 (1)

Question 36.
Three solids a square pyramid, a cone and a sphere have been carved out from three solid cubes of the same size. Find the volume of each solid. [Score: 5, Time: 7 minutes]
Answer:
Volume of square pyramid = \(\frac { 1 }{ 3 }\) a3 (1)
Volume of cone
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 62

Question 37.
A metal sphere is melted and recasted into a cone. Both have same radii
a. Find the relationship between the height of the cone and the radius of the sphere.
b. Which solid has greater surface area? Justify. [Score: 5, Time: 7 minutes]
Answer:
a. If r is the radius of the sphere, then its volume = \(\frac { 4 }{ 3 }\) πr3 (1)
If h is the height of the cone, then its volume = \(\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi r^{3}\)
h = 4r (1)
Height of pyramid is four times the radius of sphere.
Surface area of the sphere = 4πr²
Slant height of the cone = \(\sqrt{(4 r)^{2}+r^{2}}=\sqrt{17 r^{2}}=\sqrt{17} r\) (1)
Surface area of the cone = \(\pi r^{2}+\pi r \sqrt{17} r=\pi r^{2}(1+\sqrt{17})\)
b. .Surface area of the cone is greater (1)

Question 38.
A hemisphere and a cone with same radii are attached to get a solid as given in the figure. Radius of the hemisphere is 9 cm. The height of the two solids together is 21 cm.
a. Find the height of the cone.
b. Find the volume of the cone
c. Find the volume of the solids [Score: 4, Time: 6 minutes]
Answer:
Height of the cone 21 – 9 = 12 cm (1)
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 63

Solids Exam Oriented Questi0ns& Answers

Short Answer Type Questions (Score 2)

Questi0n 39.
Total surface area of a solid hemisphere is 675 π sqcm. Find the curved surface area of the solid hemisphere.
Answer:
3 πr² = 675π cm2
r2 = 225
The CSA of the solid hemisphere,
CSA = 2πr² = 2π × 225 = 450π cm2

Questi0n 40.
The volume of a solid right circular cone is 4928 cm3. If its height is 24 cm, them find the radius of the cone.
Answer:
V = 4928 cm3 and h = 24 cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 64

Questi0n 41.
The figure given below has the total length 20cm height and common diameter 6cm. Find the volume of the figure.
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 65
Answer:
volume of solid
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 66

Questi0n 42.
The circular plate of radius 12cm is cut out into six sectors having same size. Calculate the slant height and radius of circular cone used to make one sector.
Answer:
slant height = 12cm
centre angle = \(\frac { 360 }{ 60 }\) = 60
radius of square pyramid = 12 × \(\frac { 60 }{ 360 }\) = 2 cm

Questi0n 43.
A circus tent is in the shape of a square pyramid. The area of the base is 1600m2 and its height is 375m then, how much canvas would be needed for this and also find its perimeter?
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 67

Questi0n 44.
Height of a cone is 40cm. Slant height is 41cm.
a. Find diameter of its base,
b. Find volume
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 68

Questi0n 45.
Radius and slant height of a solid right circular cone are 35cm and 37 cm respectively. Find the curved surface area and total surface area of the cone.
Answer:
r= 35 cm
l = 37 cm
CSA = πrI = π(35 x 37)=4070 cm2
TSA = πr (1 + r )
\(=\frac{22}{7} \times 35 \times(37+35)=7920 \mathrm{cm}^{2}\)

Questi0n 46.
Surface area of a wooden sphere is 40 cm2. It is cut into two identical hemispheres. Find
a. The area of the plane surface of one of the hemispheres,
b. Its surface area.
Answer:
a. Surface area of the sphere = 4 πr²
Here, 4 πr² = 40cm2
πr² = 10cm2
∴ Area of plane surface = πr² = 10cm2
b. Surface area of one piece (hemisphere)
= 3 πr² = 3 x 10 = 30cm2

Short Answer Type Questions (Score 3)

Questi0n 47.
A toy in the shape of a square pyramid has base edge 16 cm and slant height 10 cm. 500 of these are to be painted and the cost is 80 rupees per square meter. What would be the total cost?
Answer:
Surface area = Curved surface area + base area
4 × 1/2 × 16 × 10+ 162 = 320 + 256 = 576cm
Total surface area of 500 square prism is = 500 × 576 = 288000 cm2
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 69

Questi0n 48.
The base radius of a circular cone is 9cm and its height is 12cm. What is the radius and central angle of die sector used to make it?
Answer:
base radius r =9 cm
height h =12cm
slant height l = \(\begin{array}{l}{=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}} \\ {=\sqrt{225}=15 \mathrm{cm}}\end{array}\)
radius of the sector = 15cm
centre angle of sector = 360 × \(\frac { 9 }{ 15 }\) = 216°

Questi0n 49.
The central angle of a sector is 120°. What is the ratio of radius and slant height of a circular cone made by it? What is the radius and slant height of a cone if its curved surface area is 108 π cm2.
Answer:
The ratio of radius and slant height = \(\frac { 120 }{ 360 }\) = \(\frac { 1 }{ 3 }\) = 1 : 3
curved surface area (πrl) =108π
rl=108; r × 3 r = 108
3 × r2 = 108; r2 = 36
radius (r) = 6 cm; slant height (l) = 3r = 3 × 6 = 18cm

Questi0n 50.
For constructing a square pyramid, Rabiya cut of four triangles and a square. Figure given below shows the measures of these triangles and square. Can you make a square pyramid by using these measures? Explain the reason.
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 70
Answer:
Base edge = 42 cm
Slant edge = 29 cm
Slant height = \(\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}}=\sqrt{29^{2}-21^{2}}\)
\(\sqrt{841-441}=\sqrt{400}=20 \mathrm{cm}\)
Slant height is less than the half of the base edge 21cm so not possible for making a square pyramid

Long Answer Type Questions (Score 4)

Questi0n 51.
A sector shown in the figure is rolled up and made a cone. Find its
a. Slant height
b. Base radius
c. Volume
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 71
Answer:
a. Slant height = radius of the
sector = 30cm
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 72
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 73
\(\frac{2000 \sqrt{2 x}}{3} \mathrm{cm}^{3}\)

Questi0n 52.
Paddy is filled in a cylindrical shaped vessel. Then it has the following shape. How many litres of paddy does it contain?
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 74
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 75

Questi0n 53.
A petrol tank is in the shape of a cylinder with hemisphere of the same radius as the base of the cylinder attached to both ends. If the total length of the tank is 5 meters and the base radius of the cylinder is 1 metre, how many litres of petrol can it hold?
Answer:
Volume of cylinder = πr²h = 3πcm3
r=1, h=3
Volume of hemisphere = 2/3 πr3
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 76

Long Answer Type Questions (Score 5)

Questi0n 54.
A toy is in the shape of a hemisphere attached to one end of the cone Total height of the toy is 14.5cm, and common diameter is 7cm.
a. Draw a rough figure based on this fact,
b. Find the volume of the toy.
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 77

Questi0n 55.
Prove that:
i. ∠BAT = ∠BPA
ii. ∠BAS = ∠AQB
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 78
Answer:
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 79

Solids Memory Map

Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 80
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 81
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 82
Kerala Syllabus 10th Standard Maths Solutions Chapter 8 Solids - 83

Kerala Syllabus 9th Standard Biology Solutions Chapter 7 Division for Growth and Reproduction

You can Download Division for Growth and Reproduction Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 8 Division for Growth and Reproduction

Division for Growth and Reproduction Textual Questions and Answers

Division For Growth And Reproduction Class 9 Question 1.
The phase at which a cell prepares for division is called
Answer:
Interphase

9th Standard Biology Notes Kerala Syllabus Question 2.
………. takes place after karyokinesis
Answer:
Cytokinesis

Kerala Syllabus 9th Standard Biology Notes Question 3.
………… begins after interphase
Answer:
Cell division

Hss Live Guru 9th Biology Kerala Syllabus Question 4.
What are changes that take place during interphase?
Answer:
a) Division of nucleus (Karyokinesis)
b) Division of Cytoplasm (Cytokinesis)

Hss Live Biology Class 9 Kerala Syllabus Question 5.
State whether true or false
Cytokinesis takes place after karyokinesis
Answer:
True

Kerala Syllabus 9th Standard Biology Notes Pdf Question 6.
Main stages of cell division
Answer:
Interphase, Division of nucleus, Division of cytoplasm

Hss Live Guru Biology 9 Kerala Syllabus Question 7.
What are the important changes that take place during interphase?
Answer:

  • Number of cell organelles increase
  • Quantity of cytoplasm increases
  • Cell size increases
  • Genetic material duplicate

Cell cycle

A cell attains its complete growth during interphase. The fully grown cell undergoes division and becomes daughter cells. As the interphase and the division phase get repeated in a cyclic manner, they together constitute cell cycle.

UML full form, stands for, meaning, what is, description, example, explanation, acronym for, abbreviation, definitions, full name.

9th Biology Notes Kerala Syllabus Question 8.
…………..is brought about by cell division and cell growth
Answer:
Growth of the body

Kerala Syllabus Biology 9th Standard Question 9.
What are the two types of cell division?
Answer:
Mitosis and meiosis.

Biology Class 9 Kerala Syllabus Question 10.
What do you mean by mitosis?
Answer:
A parent cell divides to form two daughter cells are called mitosis.

Karyokinesis

Question 11.
Point out the phases taken place in the changes in nucleus?
Answer:
Prophase, metaphase, anaphase and telophase

Question 12.
In which phase does the chromatin reticulum become chromosomes?
Answer:
Prophase

Question 13.
What changes occurs in telophase?
Answer:
In telophase chromosomes that moved to the poles become chromatin reticulum and daughter nuclei are formed.

Question 14.
State whether true or false in prophase chromosomes become chromatin reticulum.
Answer:
False

Question 15.
Complete the table of stages of nuclear division
Answer:

Phases Changes
Prophase 1. Chromatin reticulum become chromosomes
2. Duplicated chromosomes.
3. Formation of spindle fibers
4. Nucleolus and nuclear membrane get disappeared
Metaphase Chromosomes have moved to the middle of the cell and, chromo­somes doubled.
Anaphase 1. Chromatids are starting to separate from each other.
2. Formation of two sets of daughter chromosomes
Telophase 1. Formation of daughter nuclei
2. Two daughter nuclei are formed.
3. There will be no change in chromosome number in each daughter nucleus

Cytokinesis (Division Of Cytoplasm)

Question 16.
The division of the cytoplasm is taken place in plant is entirely different. Give reason?
Answer:
Because it is due to the presence of the cell wall in plant cell.

Question 17.
What is the significance of mitosis?
Answer:
The significance of mitosis is that there is no change in the number of chromosomes.

Question 18.
Mitosis helps ………… & ……….
Answer:
For the repair of tissues and growth.

Question 19.
Which condition leads to cancer?
Answer:
Mitosis is a controlled process. A disruption in this controlled process leads to the excessive division of a cell and its proliferation. This condition leads to cancer.

Different Stages Of Growth

Question 20.
List out the different stages in the growth of human beings.
Answer:

  • Zygote
  • Infancy
  • Old age
  • Embryo
  • Childhood
  • Fetus
  • Adolescence
  • Youth

Question 21.
What are the physical peculiarities of old age?
Answer:
Rate of cell division decreases, Availability of oxygen to the cells decreases, Deterioration of cells increase Muscles shrink, Production of energy decrease

Question 22.
The elders should be cared. Do you agree with this statement? Why?
Answer:
Old age is inevitable in life. The aged who worked for the welfare of their family and society during their younger age deserve special consideration.

Question 23.
What are the differences between the growth in plants and animals? Draw a comparison and complete table
Answer:

Animals Plants
Animals grow only up to a certain stage Animals do not have localized centers of growth Plants can grow through­out their lives Growth in plants is localized only at certain parts

Question 24.
Plants grow due to the rapid division and differentiation of cells.
Answer:
Meristematic cells

Question 25.
What do you mean by meristematic cells?
Answer:
Meristematic cells are special types of cells that have the capacity for continuous division.

Question 26.
Plants can grow throughout their life due to the presence of
Answer:
Meristematic cells

Question 27.
……… helps to increase the length of root and stem.
Answer:
Apical meristem

Question 28.
……… helps to increase the girth of the stem.
Answer:
Lateral meristem

Question 29.
Name the meristematic cells which help to increase the length of the stem.
Answer:
Intercalary meristem

Question 30.
Where do you find intercalary meristem?
Answer:
It seen above the nodes of monocot plants.

Question 31.
The stem of monocots increases in length faster than dicots. Why?
Answer;
Because the intercalary meristem is visible only in the monocot plants.

Question 32.
Dicot plants: Lateral meristem
……………..: Intercalary meristem
Answer:
monocot plants

Question 33.
The stem of monocots does not increase its girth beyond an extent. Why?
Answer:
Because lateral meristem is absent in monocot plants.

Growth In Unicellular Organisms

Question 34.
Does cell division in unicellular organisms lead to growth or reproduction?
Kerala Syllabus 9th Std Biology Solutions
Answer:
Mitosis leads to reproduction in unicellular organisms.

Meiosis

Question 35.
What do you mean by meiosis? Explain.
Answer:
Meiosis is the mode of cell division in which gametes are formed. Meiosis occurs in the germinal cells of the reproductive organs. Human beings have 46 chromosomes. Germinal cells with 46 chromosomes divide continuously two times. These divisions in meiosis are known as meiosis I and meiosis II. Two daughter cells “with half the number of chromosomes (23 chromosomes) are formed in meiosis I. Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to. mitosis. As a result of meiosis, four daughter cells, each with 23 chromosomes, are formed from a germinal cell.

Question 36.
What do you mean by polar body?
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed.

Question 37.
Complete the illustration
Kerala Syllabus 9th Standard Biology Notes Malayalam Medium
Answer:
a = 46
b = 23
c= 23
d = 23
e = 23
f = 23
g = 23

Question 38.
What is the number of chromosomes in germinal cells?
Answer:
46

Question 39.
What is the number of chromosomes in the daughter cells formed after meiosis I?
Answer:
23

Question 40.
What is the peculiarity of meiosis II?
Answer:
Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to mitosis.

Question 41.
What are the kinds of cell division occur in sexually reproducing organisms?
Answer:
There are two kinds of cell division occur in sexually reproducing organisms that are mitosis and meiosis.

Question 42.
What are the different stages in the growth of human being?
Answer:
Infancy, childhood, adolescence, youth and old age.

Question 43.
Differentiate mitosis and Meiosis
9th Class Biology Notes Kerala Syllabus
Answer:

Differences Mitosis Meiosis
Type of reproduction Asexual Sexual
Genetically Chromosome Number Similar Remains same Different Reduced by half
Takes place in Somatic cells Germ cells
Number of daughters 2 diploid 4 haploid
cells produced cells cells

Let Us Assess

Question 1.
The stage of karyokinesis at which daughter nuclei are formed
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer:
Telophase

Question 2.
List the meristems in various parts of the plant and list their functions
Answer:

Meristem Function
Apical meristem Increase the length of root and stem
Lateral meristem Helps to increases the girth of stem
Intercalary meristem Helps to increase the length of the stem of monocot plants

Question 3.
In females, only a single ovum is formed from a germinal cell, whereas in males, more than one sperm is formed. Give reason.
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed. So only a single ovum is formed from a germinal cell. But in males, after meiosis, four sperms having 23 chromosomes are formed form one germinal cell.

Question 4.
Observe the figures
Biology Kerala Syllabus 9th Standard
a) Which stages of mitosis are indicated in the figures?
b) What are the changes that occur in the chromosomes during these stages?
Answer:
a) Metaphase
b) Chromosomes get aligned at the equator of the cell.

Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 Public Administration

You can Download Public Administration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Social Science Solutions Part 1  Chapter 3 Public Administration

Public Administration Text Book Questions and Answers

Public Administration Class 10 Kerala Syllabus Question 1.
Some of government institutions and their functions are given below. Expand the table by writing more.
Answer:

Institutions Functions
Primary health center Provide treatment facilities
Krishi bhavan Promotes agriculture
Police station Maintains law and order
Schools Provide learning opportunities
Post offices Provide communications
Courts Protecting liberties and rights

Sslc History Chapter 3 Notes Kerala Syllabus Question 2.
Discuss and list out the changes in the objectives of public administration in monarchy and democracy.
Answer:

Monarchy Democracy
Completely under king. Values on the human rights, liberty and democracy
Law is the king Ensures man’s liberty
King is last and cannot be questioned Humans are given complete control
Ruler may be from selected families All are given chances to vote and elect

State Syllabus Class 10 Social Science Notes Pdf  Question 3.
Whom do you wish to get the services of public administration? Discuss in your class and write your conclusion.
Answer:
The main aim of public administration is to ensure equality and justice to all is, especially, the deprived society. Government has taken measures for the social safety and enlistment. Crores are spent for the purpose. But corruption and political intervention are a hindrance.

  • Justice for the deserved.
  • Education, employment and treatment facilities for the deprived ones.
  • Control the cost of commodities.
  • Strict laws for the upliftment of women.
  • I believe these to be the main aims of public administration.

Public Administration Malayalam Notes Kerala Syllabus Question 4.
Visit a nearby Government. Office and prepare a report on the features of bureaucracy there.
Answer:
Bureaucracy is the main weapon of the government. They play various responsibilities. Employee may be highly experienced as they serve for long years. Ministers are the heads. Employees influence the decisions of the government. Though the ruling party changes, bureaucracy remains the same. Employees help the ministers in order to execute plans.

Kerala Syllabus 10th Standard Social Science Notes Malayalam Medium Question 5.
Discuss and prepare a note on the changes to be brought in the administrative system.
Answer:

  • Ensure justice and equal rights to all sections of the society.
  • Create corruption free administration and responsibility among the workers.
  • Constitution must be to save the victims and punish culprits.
  • Government service must be reached at fixed time.
  • Protect the rights of women and ensure their safety.
  • Ensure the welfare of all.

Use our free online relative standard deviation RSD calculator to know the standard deviation and %RSD for the given mean of data.

Social Science Class 10 Kerala Syllabus Question 6.How far the Right to Information Act make the general administration system efficient. Evaluate.
Answer:
In 2005, RTI Act was passed by the parliament. To get information is the fundamental right of all citizens. The main objectives of this Act are to prevent corruption, create responsibility and make the functioning of the government transparent. The citizens will get copies of public documents if they apply for them.

Sslc History Chapter 1 Notes Pdf Kerala Syllabus Question 7.
What are the situations in our society where the Right to Information Act can be positively used? Discuss and prepare a note.
Answer:
In 2005, RTI Act was passed by the parliament. To get information is the fundamental right of all citizens. RTI Act is helpful socially, for all the subjects, quickly in our society.
Example: Waiting for the results after University Exam if delayed, RTI commission could be approached :

  • To get information about certain legal affairs.
  • Any problems concerned with government.

Kerala Syllabus 10th Standard Social Science Notes Question 8.
Discuss whether the Right to Service Act is helpful for the people to get the service they should obtain from Government offices.
Answer:
The responsible employer would have to pay the penalty if he fails in his duty. So, the people can get the service from Government offices at the right time.

Hsslive Guru Social Science Kerala Syllabus Question 9.
What is the use of Ombudsman to the public? Prepare a note.
Answer:
Elected representatives and bureaucrats are part of administration. Complaints are filed against their corruptions to Ombudsman. People can directly approach Ombudsman with complaints.

Social Science 10th Kerala Syllabus Kerala Syllabus Question 10.
Can we make Government services transparent and corruption free through the above mentioned system? Conduct a debate.
Answer:
For:

  • Need not wait in Government offices for services.
  • Receives Govt, service with less expense and quickly.
  • Increases the efficiency and excellence of services.
  • Receives feedback quickly.

Against:

  • Is completely hidden in files.
  • Life is free of corruption.
  • Govt, services are obtained based on the financial status.

Public Administration Let Us Assess

Social 10th Class Notes State Syllabus Kerala Syllabus Question 1.
Explain the need of public administration in a country?
Answer:
Public administration is related to governmental administration. It is the effective way of utilizing men and materials for implementing the existing law policies and developmental projects. Primary Health Centers, Police station Krishi Bhavan, Village office, Corporation office, Election Commission, and other government organisations are all the parts of public administration.

The responsibilities of these institutions must be strictly executed for the public welfare.There must be an accurate method of public administration for all the sectors of people in our country. The following is a list of a few public administration.

Centers and their duties are given:
1. Vanitha Commission: Ensures the rights and welfare of women.

2. Election Commission: Relates with the election procedures.

3. Human Rights Commission: Protects the rights of men.

4. Village office: Ensures the primary needs of people.

10th Standard Social Science Notes Kerala Syllabus Question 2.
How are employees selected in a public administration system?
Answer:
Certain methods are adopted in the selection of employees to the public administrative system in India. The first step is the notification through Public Service Commission of India.Through this, the vacancies in various sections of governments are made known to the public. Later, through the exam conducted on the date notified or on the basis of interviews, candidates are selected and appointed.

Civil Service began during the British rule. It came to be the Civil Service of India after Independence. All the employees who work under the central and state governments and the employees under public sector undertakings are a part of India’s Civil Service. It is divided into All India Service, Central Service and State Service. The selection procedures of employees to these are given below.

All India Service:

  • Recruits at national level.
  • Appoints in Central Service or State Service.
    eg: IAS, IPS.

Central Service:

  •  Recruits at national level.
  •  Appoints in central government department only.
    eg: Indian Foreign Service, Indian Railway Service.

State Service:

  • Recruits at State level
  • Appoints in state government department
    eg: Sales Tax officer.

Candidates to All India Services and Central Services are recruited by the Union Public Service Commission (UPSC). The chairman and members of this commission are selected by the President of India. The UPSC has elaborate mechanisms for the selection based on qualifications.

In the state level, candidates are recruited by the Public Service Commission (PSC) of the state. The governor appoints the Chairman and the member of the State Public Service Commission. UPSC and state PSCs are on the basis of constitutional provisions and can be called constitutional institutions.

Std 10 Social Science Notes Kerala Syllabus Question 3.
What are the features of bureaucracy ?
Answer:
Hierarchical organisation:
Bureaucracy is organised in such a way that there is one employee at the top and the number increases, when it reaches the lower levels. This is known as hierarchical organisation.

Permanence:
Persons appointed will continue in service till the age of retirement.

Appointment based on qualification:
Employees are recruited and appointed on the basis of educational qualification.

Political Neutrality:
Bureaucrats are liable to implement the policies of which ever party comes to power. Party interests should not reflect in their work.

Professionalism:
Every government employee must be skilled in their work.

Sslc Social Science Notes Malayalam Medium Pdf 2021 Kerala Syllabus Question 4.
Classify the bureaucracy in India and explain ?
Answer:
Kerala Public Service Commission notifies for the recruitment of employees to the government service. Then candidates are selected on the basis of competitive examinations and interviews and are appointed in different government sectors. All those appointed in this way become a part of the civil service of India.

The aim of it is to bring welfare programmers speedily to all. Now there are specific services at the central and state levels. All the employees who work under public sector undertakings are the part of India’s civil service. There are All India services, central services and state services.

10th Social Science Notes Pdf State Syllabus Question 5.
What are the measures taken for the administrative reforms in India? Prepare a note.
Answer:
Government has taken a number of steps to increase the efficiency of services and to provide service to people within a time limit. This is called as administrative reforms. It makes the administration friendly and effective. Administrative reform commissions are made at national and state levels. Here are some steps taken for the reformation in our country.

E-Governance:
The single window system for Higher Secondary education, online applications for several scholarships etc are example of E-Governance. The use of electronic technology has helped to obtain the services of government effectively in a faster way.

Right to Information:
Every citizen has the right to collect the information from any government office about its working. This is under the Right to Information Act in 2005. The efforts of Mazdoor Kiran Shakthi Samghathan of Rajasthan led to this Act. The interventions of several organisations and social activities helped in passing this Act in 2005. The Right to Information ensures the right of all citizens of India to receive information.

Information Commission:
Files documents, circulars, memos, advice or orders, agreements statistics, reports, log books, press notes, samples, models, information in the form of electronic data, e-mail etc. related to government offices belong to public departments. Information commission helps to know everything.To perform the functions under the Right to information Act, Information Commissions are constituted at National and state levels.

A chief Information Commissioner and members not more than ten are in the Information Commission. If the information given is wrong and unsatisfactory we have the right to approach the Information Commission. If the commission is convinced a fine of Rs. 250 can be imposed on the employee.

Right to Service:
This ensures service to the people. This law determines the time limit for every service given by a government office. If the deserved service is not given within the time limit the responsible employee should pay fine. As per the Right to Service, an officer is appointed to give proper guidance and help to the applicants.

Lokpal and Lokayuktha:
These are the institutions to prevent corruption at administrative, bureaucratic and political levels. Lokpal has the power to register cases on issues of corruption against public workers and then suggest necessary actions. Lokayuktha hears the corruption cases at the state level. Both follow judicial procedures.

Central Vigilance Commission:
This is constituted at the national level to prevent corruption. It came in to force in 1964. It was formed to prevent corruption at central government offices.

Ombudsman:
Complaints of corruption against the elected representatives and bureaucrats who are a part of public administration are filed to ombudsman. A retired Judge of the High Court is appointed as Ombudsman. People can directly approach the Ombudsman with complaints.

He has the power to summon anyone on receiving the complaints, can order inquiry and recommend actions. Ombudsman began in the banking sector to hear the complaints of clients and rectify them.These actions recommended for the welfare of people as a part of administration maintains the public administration more trans parent.

Question 6.
How is E-Governance helpful to the public?
Answer:

  • Need not wait in the government offices for services.
  • Can receive service with the help of information technology.
  • It services offers fast and at less expense.
  • Enhances efficiency of offices and quality of the service.

Question 7.
Right to Information and Right to Service help to make public administration popular. Substantiate this statement.
Answer:
We can collect information from any government office about its working. People got this opportunity under the right to information act, 2005. This ensures the right of all citizens of India to receive information. The main objectives of this act are to prevent corruption, create responsibility and make the functioning of government transparent.

The citizens will get copies of public documents if they apply for them. Right to Service act is a law which ensures services to the public. As per the Right to Service Act, an officer is appointed in every government office to give guidance and proper help to the applicants. That is Right to Information and Right to Service helps to make public administration popular.

Question 8.
List out the steps taken to prevent corruption in India.
Answer:
1. Lokpal and Lokayuktha
2. Central vigilance commission
3. Ombudsman.

Public Administration Extended Activities

Question 1.
Prepare an application for getting information from an office under the Right to Information Act.
Answer:
From
Madhavan
Mahal Nivas
Ollavanna

To
Secretary
Grama Panchayath
Ollavanna

Based on RTI Act of 2005
Sir,
Sub: Amount spent for the electrification of houses for schedule tribes in the year 2015-16.
Could you kindly give details about how many applicants there were and how many money was spent in previous years. Expecting the reply within 15 days.
Ollavanna — Madhava
25/08/19  —  sd/-

Question 2.
Prepare a chart o the importance of All India Services.
Answer:
All the employees who work under the central and the state governments and the employees under the public sector under takings are part of India’s civil service. India’s civil service classification as All India Service, Central Service ans State Service.

All India Service.
Recruits at National level.
Appoints in the central or state service.
The UPSC has elaborate mechanisms for the recruitment of candidates based on qualification.
e.g., Indian Administrative Service.
Indian Foreign Service.
Indian Police Service.

Question 3.
Prepare a questionnaire to conduct an interview with the District Collector.
Answer:
Interview with kozhikode District Collector.

Sir, could you please share with me the inspiration behind your selection this position?

What were the changes you have thought to bring in Kozhikode before becoming the District Collector?

Sir, even in the presence of Lokpal, Lokayuktha, Central Vigilance Commission, Ombudsman being launched against corruption still corruption exists in Kerala. Don’t you long for a change ? How do you react as a district collector?

Sir, hope you have noted that vegetables from Tamil Nadu were sold less here in the Onam season. Do you expect the same for Vishu? As a District Collector, what all steps would you take to improve the agriculture sector of Kerala?

Though Kerala has developed in the educational field, the number of suicidal attempts have also increased. A change in our educational system is unavoidable. Sir, what all changes could be made possible?

Sir, M. C. Noufal was arrested here in Kozhikode for raping a woman from Bangladesh. Though many were arrested they were refused saved. Let me ask you sir with due respect, don’t you long for a change in the system of punishment here ?

Public Administration Orukkam Questions and Answers

Question 1.
List out the changes and objectives of public Administrative in monarchy and democracy.
Monarchy – The interest of the Monarch were the basis of public administration.
Democracy –
Answer:
Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 Public Administration 1
Democracy – Importance is given to the interests of the people.

Question 2.
Complete the diagram showing the importance of public administration
Answer:

  • Formulate government policies.
  • Ensure welfare of the people.
  • Find out solutions to public issues.
  • Provides goods and services.

Question 3.
Some features of public administration are given in column A. Find out the definitions of each one of them in column B.
i. Hierarchical Organisation – one employee at the top and the number increases when it reaches the lower levels.
ii. Permanence- i
iii.Appointment on the Basic of qualification- iii
iv. Political Neutrality – iii.
x Professionalism – iv
Answer:
i. Permanence – Persons appointed will continue in service till the age of retirement.
ii. Appointment on the basis of qualification- Employees are recruited and appointed on the basis of educational qualification.
iii. Political Neutrality – Bureaucrats are liable to implement the policies of which ever party comes to power. Party inters ts should not reflect in their work. They should act neutrally.
iv. Professionalism – Every government employee must be skilled in their work.

Question 4.
Find out the functions of the following constitutional institutions.
1. State public service commission.
2. Union Public Service Commission.
Answer:
1. State Public Service Commission – At the state level, candidates are recruited by the public service commission of the state.

2. Union Public Service Commission- Candidates to all India services and central services are recruited by the Indian Public service commission.

Question 5.
Complete the short showing the classification of India’s Civil Service.
All India Service
Central Service
State Service
Answer:
All India Service: Recruits at national level, Appoints in the central or state service.
Eg: Indian Administrative Service, Indian Police.

Central Service : Recruits at national level, Appoints in central government departments only.
Eg: Indian foreign service, Indian Railway Service.

State Service: Recruits at state level, Appoints in state government departments only.
Eg: Sales Tax Officer

Question 6.
Find out and list the benefits of E – Governance to Public.
Need not to wait in government offices for services.
Answer:

  • Can receive service with the help of information technology.
  • Government services offered speedily and with less expense.
  • Efficiency of the offices and quality of the service get enhanced.

Question 7.
Prepare a sample applications of Right to Information Act 2005
Answer:
From
Smitha Vijayan
DeviVihar
Arppookara

To
Secretary
Grama panchayath
Arpookara

Sir,
I request you to furnish information regarding the following questions under the right to information act 2005.

  • In the academic year 2015-2016 what is the amount of money taken from the fund for the constriction of houses for the backward castes in arpookara grama panchayath.
  • What was the amount of money spent in the past years for this purpose.
  • Expecting reply within 15 days.

Question 8.
Lokpal, and Lokayukta are institutions constituted to prevent corruption at administrative, bureacrative and political level, find out the features and complete the chart.
Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 Public Administration 2
Answer:

Lokpal Lokayukta
The institution constituted at national level to prevent corruption is called Lokpal. Lokayukta is the institution constituted at the state level to hear the corruption cases.
Lokpal has the power to register cases on issues of corruption against employees and pub lie workers and can suggest necessary actions. Follow judicial Procedures.

Question 9.
Right to service Act is a law which ensures services to the people. Find out and list the treasures of Right to service Act. This law determines the time limit for every service given by a government office.
Answer:

  • If the deserved service is not given within this time limit, the responsible employee should pay a fine.
  • An officer is appointed in every government office to give guidance and proper help to the applicants.

Question 10.
Compare and the list out central vigilance commission and state vigilance commission.
Answer:
Central Vigilance Commission :
The Central Vigilance Commission constituted at national level to prevent Corruption, Central Vigilance Commission, it came into effect into 1964, Formed to prevent corruption in central government offices, in every department there will be a chief vigilance officer.

State Vigilance Commission :
Constituted at State level to prevent corruption, Inquires in to corruption in the state government offices.

Question 11.
Complete the diagram of administrative reform measures adopted by the government.
Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 Public Administration 3
Answer:

  • Right to information.
  • Information commission.
  • Lokpal and lokayuktha.

12. Complete the concept map given below.
Kerala Syllabus 10th Standard Social Science Solutions Part 1 Chapter 3 Public Administration 4
Answer:

  • Create responsibility
  • The citizen will get copies of public documents if they apply for them.

Public Administration Evaluation Questions

Question 1.
‘Public Administration is about Governmental Administration’ on the basis of Gladden’s definition examine the features of public administration.
Answer:
From the definition of Gladden we can under stand that the important area of public administration is governmental administration. Public administration is the effective utilization of men and materials for the implementation of existing laws, governmental policies, programmer and developmental projects. Government try to find the solutions to various problems and ensure the welfare of the people through public administration.
Significance of Public administration.

  • Formulate government policies.
  • Provide goods and services.
  • Ensure welfare the people.
  • Find out solutions to public issues.

Question 2.
Explain Gandhiji’s vision on public administration.
Answer:
Gandhiji expected the protection of the interests of all through public administration. But there are a number of persons in our society who require special consideration and protection. Fie opined that public administration should consider them specially and protect them. Gandhiji’s concept of Grama Swaraj influenced India’s outlook of public administration to a great extent.

Question 3.
Define Bureaucracy. Examine the features of Bureaucracy?
Answer:
The employees who work under public administrative system are together known as Bureaucracy.

Features of Bureaucracy:
Hierarchical Organisation:
Bureaucracy is organised in such a way that there is one employee at the top and the number increases when it reaches the lower levels. This is known as Hierarchical organisation.

Permanance:
Persons appointed will continue in service till the age of retirement. Appointment on the basis of qualification Employees are recruited and appointed on the basis of educational qualification.

Political Neutrality:
Bureaucrats are liable to implement the policies of whichever party comes to power. Party interests should not reflect in their work.They should act neutrally.

Professionalism:
Every Government employee must be skilled in their work.

Question 4.
Name of the feature of bureaucracy which intercepts quick decisions.
a. Political Neutrality.
b.Professionalism.
c. Permanence.
d. Hierarchical organisation.
Answer:.
Hierarchical Organisation.

Question 5.
The Rajasthan based organisation paved the way for the legislation of right to Information Act.
a. Narmada Bachao Andolan.
b. Swabhimana Prasthanam
c. Mazdoor Kisan sakthi Sangathan.
d. Bharatiya Kissan Union.
Answer:
Mazdoor Kisan Shakti Sangathan.

Question 6.
Write a short note on the functions of central and state information commission.
Answer:
To perform the functions under the right to information act, Information Commissions are constituted at national and state levels.

Question 7.
State level organisation to prevent corruption at administrative, bureaucratize and political levels
a. Lokpal b. Lokayuktha
c. Central Vigilance Commission d Child rights commission
Answer:
Lokyukta

Question 8.
Match the items of column A with B

A B
1. All India Service i. Sales Tax Officer
2.Central Service ii. Indian Police Service
3.State Service iii.Indian Railway Service

Answer:
1-ii,
2-iii,
3-i.

Question 9.
The Institution constituted at the national level to prevent corruption in 1964
a. NitiAyog.
b. Administrative Tribunal.
c. Central Information Commission.
d. Central Vigilance Commission.
Answer:
Central Vigilance Commission.

Question 10.
UPSC and PSC are caused constitutional institutions why?
Answer:
UPSC( Union Public Service Commission) and state PSC are constituted on the basic of constitutional provisions.So they are called constitutional institutions.

Question 11.
Define E-Governance and write down two examples of E – Governance.
Answer:
E- governance is the use of electronic technology in administration . This help to obtain government services easily in a speedy manner. The single window system for admission to higher secondary courses, Online applications for various scholarships etc are exam pies for E- governance.

Question 12.
Explain different administrative reforms adopted by government for increasing the affiance of service.
Answer:
information Commission:
To perform the functions under the right to information act, In formation Commissions are constituted at the national and state levels. There will be a chief Information commission and not more than ten members in the Information Com mission.

Central Vigilance Commission:
The Central Vigilance Commission is the institution constituted at the national level to prevent corruption. It came into effect in 1964. It is formed to prevent corruption in the central government offices.The Central Vigilance Commissioner is the head of the Central Vigilance Commission.

Lokpal and Lokayuktha:
Lokpal and lokayuktha are institutions constituted to prevent corruption at administrative, bureaucratic and political levels, the institution constituted at the national level to prevent corruption is lokpal.

Lokpal has the power to register cases on issues of corruption against employees and public workers and can suggest necessary actions. Lokayukta is the institution constituted at the state level to hear the corruption cases.

Ombudsman:
Elected representatives and bureaucrats are part of public administration. Complaints can be filed against their corruption,nepotism or financial misappropriation or negligence of duties. Ombudsman is constituted for this purpose.

Public Administration SCERT Questions and Answers

Question 1.
Pick out any two public administration institutions and write about their functions.
Answer:

Institution Functions
a. Krishi Bhavan Promotes agriculture
b. Police Station Maintains law and order
c. Primary health center Provides treatment for illness

Question 2.
What are the differences found in the public administration under monarchy and democracy?
Answer:

  • In monarchy, the interest of the monarch is the basis of public administration,
  • In democracy importance is given to the interests of the people.

Question 3.
Explain the importance of public administration.
Answer:

  • Ensures welfare of people
  • Formulates government policies
  • Provides goods and services
  • Finds out solution for public grievances

Question 4.
Explain the importance of bureaucracy in public administration.
Answer:

  • Makes the public administration system dynamic,
  • Services of the government made accessible to the people ,
  • Performs the day- to-day administration of the country.

Question 5.
What is meant by hierarchical organisation and permanence of bureaucracy.
Answer:

  • Organization made up of one employee at the top and more towards the bottom.
  • Persons once appointed as employee will continue in service till the age of retirement. This is permanency.

Question 6.
Prepare a note on civil service in India.
Answer:

  • All India service
  • Central services
  • Stateservice. Explain hints

Question 7.
Why PSC and UPSC are called as constitutional institutions?
Answer:
UPSC and PSC are constituted on the basis of constitutional provisions. So they are cal led as constitutional institutions.

Question 8.
Find out the factors that adversely affect the efficiency of public administration.
Answer:

  • Inefficiency of bureaucracy,
  • Corruption
  • Shortage of employees

Question 9.
Write a brief note on e-Governance implemented as part of administrative reforms?
Answer:

  • e-Governance is the use of electronic technology in administration. This helps people to obtain government services quickly and easily .
  • Information technology is used in the field of public service.
  • Government service is made available less expensively expensively

Question 10.
A road constructed before six months in your place is damaged now. You became aware that there is some corruption, Prepare an application under Right to Information Act addressing panchayat secretary to get the details of it.
Answer:
To Prepare an application under the RTI Act.

Question 11.
What are the benefits attained by society as a result of formulation of the Right to Information Act.
Answer:

  • Controlled corruption,
  • Increased the responsibility of bureaucrats,
  • Functioning of government became transparent.

Question 12.
Explain the structure of Information Commission?
Answer:

  • Central Information Commission,
  • Chief Information Commissioner and not more than ten members,
  • State Information Commission.

Question 13.
How the Right to Service Act is helpful to people?
Answer:

  • Ensures government services to people.
  • Gets service within time limit.
  • Employee should pay fine in case of delay.
  • Appoints an officer in charge as per Right to Service Act in all offices to give service.

Question 14.
Differentiate the functioning of Lokpal and Lokayuktha?
Answer:
Lokpal:

  • Institution constituted to prevent corruption charges at national level
  • Has the power to register cases of corruption against employees and public workers.

Lokayuktha :

  • Constituted at state level to hear corruption cases.
  • Follow Judicial procedures

Question 15.
From the following select the statement appropriate to central vigilance commission and state vigilance commission?
a. Esquires about corruption in state government offices.
b. Institution to prevent corruption at national level.
Answer:
a. State vigilance commission .
b. Central vigilance commission

Question 16.
How the functioning of Ombudsman helps the public to prevent corruption.
Answer:

  • Complaints against corruption among elected representatives and bureaucrats can be filed in Ombudsman,
  • People can directly approach Ombudsman to give complaints.
  • Ombudsman can enquirer into such complaints and recommend actions.

Question 17.
Compare and list the different levels of civil services in India.
Answer:

  • All India Service
  • Central Service
  • State Service

Question 18.
What is considered as corruption by modern society?
Answer:

  • Delayed service, Denial of right to service is corruption.
  • Making service as a favor.

Public Administration Exam Oriented Questions and Answers

Question 1.
The chairman and the members of this commission are appointed by
Answer:
The president of India

NSUI full form stands for National Students’ Union of India that is a student wing of famous politician Indian National Congress party.

Question 2.
What is the full form of U. P. S.C ?
Answer:
Union Public Service Commission.

Question 3.
In which year Central Vigilance Commission come into effect ?
Answer:
1964

Question 4.
Write a note on Akshaya Center and E-literacy.
Answer:
For the benefit of people, Akshaya centers have been constituted to make use of Government service delivered through E – governance. It also aims at making people E-literate. E – literacy is the awareness about basic information about technology.

Question 5.
Prepare a seminar report on the importance of public administration.
Answer:
Without public administration, the government cannot operate and manage activities effectively and efficiently. The administration plays a vital role for delivering and distributing the public services to all comers of the country.

Administration spreads all over the country for supplying the governmental and public goods and services up to the villages and door to door. The administration is not operating and managing the activities properly and smoothly in developing countries. The scopes of administration shows the importance of public administration.

Following are the importance of Public Administration:
Management of Public Service, Distribution Social Change, Disaster Management, Population Control, Preservation of Human Right, Management of Industrial Relationship, Internal and External attached Economic Development. These points show that the administration is used all over the sections of the country.

Question 6.
Write a note on Administrative Tribunal.
Answer:
Actions are taken by various government departments against the government officials. The Administrative Tribunal is the institution where the employees can lodge their complaints against such actions. ”

Question 7.
The famous western administrative thinker Gladden says “ Public administration is concerned with an administration of the government”. On the basis of this definition, explain the relationship between Public administration and the government.
Answer:
From this definition we can understand that the important area of public administration is governmental administration. Public administration is the effective utilization of men and materials for the implementation of existing laws, governmental policies, programmers and developmental projects and the government has constituted a number of institutions for this purpose. All these governmental institutions are part of public administration. They function for the welfare of the people.

An administrative system is needed for governments to exist and function. The history of public administration begins with the formation of state. Based on differences in the form of government we can find differences in public administration also. In a monarchy, the interests of the monarch was the basis of public administration. But in a democratic system, importance is given to the interests of the people. Democratic administration becomes more effective and efficient through public administration.

Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion

You can Download Proportion Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 12 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 12 Proportion

Proportion Textual Questions and Answers

Textbook Page No. 182

Hss Live Guru 9th Maths Kerala Syllabus Question 1.
A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.
i. Are the interests proportional to the investments?
ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.
iii. What is the annual rate of interest in the first scheme? And in the second?
Answer:
i. The ratio between the amounts invested = 10000:15000 = 2 : 3
The ratio between the interests = 900 : 1200 = 3 : 4
Since the ratios are different. So, interests will be not proportional to the investments.

ii. The ratio between the amount invested and interest in scheme 1
= 10000 : 900= 100 : 9
The ratio between the amount invested and interest in scheme 2
= 15000 : 1200 = 25 : 2

iii. Rate of interest in the first scheme
= 900/10000 × 100 = 9%
Rate of interest in the second scheme =
1200/15000 × 100 = 8%

Make use of this free RSD calculator online that is specifically designed to calculate relative deviation of a data set.

Kerala State Class 9 Maths Solutions Question 2.
The area of A0 paper is one square metre. Calculate the lengths of the sides of A4 paper correct to a millimetre, using a calculator.
Answer:
The area of A1 paper is half of the area of A0 paper. The area of A2 paper is half of the area of A1 paper. Let con-sider the area of the A0 paper be 1 square metre, the area of A1 paper is
1/2 square metre , the area of A2 paper is 1/4 square metre , the area of A3 paper is 1/8 square metre , and the area of A4 paper is 1/16 square metre ,
If the length of A4 paper is √2 x and breadth x
length of A4 paper : breadth = √2 : 1
Area of A4 paper = length x breadth
√2x × x = √2x2
This is 1/16 m2, therefore √2 x2 = 1/16
Hss Live Guru 9th Maths Kerala Syllabus

Hsslive Guru 9th Maths Kerala Syllabus Question 3.
In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?
Answer:
The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12 The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7
Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

Textbook Page No. 185

Hss Live 9 Maths Kerala Syllabus Question 1.
For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.
i. Perimeter and radius of circles.
ii. Area and radius of circles.
iii. The distance travelled and the number of rotations of a circular ring moving along a line.
iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.
v. The volume of water poured into a hollow prism and the height of the water level.
Answer:
i. The perimeter of a circle is n times its diameter. That is 2n times the radius.
∴Perimeter = 2πr
∴ The perimeter and radius are proportional.
The constant of proportionality is 2π

ii. The area of a circle is π times the square of the radius.
∴ Area = πr2
∴ Area and radius are not proportional.

iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring.

iv. If the amount deposited is P and the rate of interest is R.
Annual interest = I = PNR,
I = P × R (N = 1)
The amount and interest are propor-tional.
Constant of proportionality is rate of interest, R.

v. Volume of a hollow prism
= base × height
Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

Kerala Syllabus 9th Standard Maths Notes Question 2.
During rainfall, the volume of water falling in each square metre may be considered equal.
i. Prove that the volume of water falling in a region is proportional to the area of the region.
ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.
Answer:
i. The volume of water falling in each square metre are equal.
Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k
The volume of the rain falling on the 3 square metre = 3k
The volume of the rain falling on the x square metre y = kx Here x and y are proportional.

ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh.
y/x =h
The volume of water collected is different in hollow prisms having different base area.

But \(\frac { volume }{ Area }\) is always equal to the height of water level. So the height at which rainwater collected is same.

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
When a weight is suspended by a spring, the extension is proportional to the weight. Explain how this can be used to mark weights on a spring balance.
Kerala State Class 9 Maths Solutions
Answer:
Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e.,
2, 4, 6 and 8.
These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

Hss Live Guru 9 Maths Kerala Syllabus Question 4.
In the angle shown below, for different points on the slanted line, as the distance from the vertex of the angle changes, the height from the horizontal line also changes.
Hsslive Guru 9th Maths Kerala Syllabus
i. In the picture, perpendicular to the horizontal lines are drawn from points on the slanted line.
Δ ABP, Δ ACQ, Δ ADRand Δ AES are similar triangles. (Because these are a right-angled triangle with is common to all)
∴ Ratio of their sides are equal. That is sides are proportional.
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 4
Let k be the constant of proportionality
PB = k × AB, QC = k × AC
RD = k × AD, SE = k × AE ,
i.e., the change in height is proportional to the distance.

Hss Live 9 Maths Kerala Syllabus
Kerala Syllabus 9th Standard Maths Notes

An online geometric sequence calculator helps you to find geometric Sequence, first term, common ratio calculator, and the number of terms.

Textbook Page No. 189

Class 9 Maths Chapter 12 Kerala Syllabus Question 1.
i. Prove that for equilateral triangles, area is proportional to the square of the length of aside. What is the constant of proportionality?
ii For squares, is area proportional to square of the length of a side? If so, what is the constant of proportionality?
Answer:
i. Δ ABC is an equilateral triangle
Consider their sides are ‘a’,
Draw a perpendicular line CD to AB
from C. In right-angled triangle ADC, according to the Pythagoras theorem AD2 + DC2 = AC2
DC2 = AC2 – AD2
Hsslive Guru Maths 9th Kerala Syllabus

ii. If x be the one side of a square then its area y = x2
Area of the square is proportional to the square of their sides.
Constant of proportionality = 1

Hsslive Guru Class 9 Maths Kerala Syllabus Question 2.
In rectangles of area one square metre, as the length of one side changes, so does the length of the other side. Write the relation between the lengths as an algebraic equation. How do we say this in the language of proportions?
Answer:
Area of the rectangle is the sum product of length and breadth.
Let × be the length and y be the breadth, then
x × y = 1 m2
Hss Live Guru 9 Maths Kerala Syllabus
x = 1/y = m
Example when y = 3 m
x = 1/3 meter
when y = 4 m
x = 1/4 meter
The algebraic equation is x = 1/y
i.e., length of rectangle is proportional to the reciprocal of the breadth, i.e., length of rectangle is proportional to its breadth.

Chapter 12 Maths Class 9 Kerala Syllabus Question 3.
In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Answer:
Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 9
Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Hss Live Guru Class 9 Maths Kerala Syllabus Question 4.
In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?
Answer:
The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { Sum of exterior angles }}{\text { No. of sides }}\)
One outer angle = 360/n
(n = number of sides)
If the measure of an exterior angle is ‘x’.
x= \(\frac{1}{n} \times 360^{0}\)
One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

Hsslive Guru Maths Kerala Syllabus 9th Question 5.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i. The rate of water flow and the height of the water level.
ii The rate of water flow and the time taken to fill the tank.
Answer:
i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then
x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by
C = V × t
\(V=\frac{C}{t}=C \times \frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

Proportion Exam oriented Questions and Answers

Hsslive 9th Maths Kerala Syllabus Question 1.
the weight of 6 spheres of same size made of the same metal is 14 kg. When 9 more spheres are added the weight is 35 kg. Check whether the number of spheres and their weights are proportional.
Answer:
Weight of 6 spheres = 14kg
Ratio of number and weight = 6 : 14 = 3 : 7
Total number of sphers are added = 15
Total weight = 35kg
Ration of number and weight = 15 : 35 = 3 : 7
Since the ratios are equal, the number of sphers and their weights are propor-tional.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5
Ratio of profit divided
1800 : 3000 = 18 : 30 = 3 : 5
Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

Question 3.
The two sides of a triangle having perimeter 10 m are 2\(\frac { 1 }{ 2 }\) m and 3 \(\frac { 1 }{ 2 }\) m.
What is the ratio of the length of the three sides of triangles?
Answer:
Perimeter of triangle = 10 m
First side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 5 }{ 2 }\) m
Second side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 7 }{ 2 }\) m
Third side = 4 m Ratio of three sides of a triangle
\(=\frac{5}{2}: \frac{7}{2}: 4 = 5: 7: 8\)

Question 4.
150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?
Answer:
Quantity of water flowing in 6 minutes = 150 litres
Ratio between the quantity of water and timeofflow= 150: 6 = 25: 1
Quantity of water flowing in 8 minutes = 200 litres
Ratio between quantity of water and times of flow = 200 : 8 = 25: 1
Since the radios are equal, the amount of water flowing and the time of flow are proportional.

Question 5.
Unniyappam was made using 1 kg rice, 250 g plantain and 750g jaggery. Find the ratio between the ingredients.
Answer:
Rice = 1
kg= 1000g
Plantain = 250 g, Jaggery = 750 g.
Ratio between the ingredients
= 1000 : 250 : 750 = 4 : 1 : 3

Question 6.
Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4\(\frac { 1 }{ 2 }\) hours. Are the wages obtained proportional to the working time ?
Answer:
Ratio of working hours = 6 : 4 \(\frac { 1 }{ 2 }\)
= 12 : 9 = 4 : 3
Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3
Since the ratios are equal, the working hours are proportional to the wages.

Question 7.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
for a square the perimeter is 20 cm. so,
length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
x — y
9 — 1
8 — 2
7 — 3
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
hence length and breadth are not in inversely proportional.

Question 8.
The weight of an object having mass 5 kg is 49 Newton. The weight of another object having mass 15 kg is 147 Newton. Check whether the mass and weight are proportional ? What is the constant of proportionality? What is the weight of an object having mass 8 kg ?
Answer:
Weight of 5 kg object = 49 N
Ratio between mass and weight = 5 : 49
Weight of 15 kg massed object = 147N
Ratio between mass and weight = 15:147 = 5:49
Mass and weight are proportional.
Weight of 8 kg massed object = 8 × 9.8 = 78.4N
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 10
Constant of proportionality = 9.8

Question 9.
The perimeter of a triangle is 60 cm. Sides are in the ratio 3: 4: 5. Then j find the length of the sides.
Answer:
Ratio of sides = 3 : 4 : 5
Therefore, the sides are the \(\frac { 3 }{ 12 }\), \(\frac { 4 }{ 12 }\) and \(\frac { 5 }{ 12 }\) part of the perimeter .
The perimeter is 60 cm, then the length of sides are
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 11
\(60 \times \frac{5}{12}=25 \mathrm{cm}\)

Question 10.
Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?
Answer:
In the first dye , blue colour: white colour = 10 : 15 = 2 : 3
In the second dye, blue colour: white colour = 12: 17
The ratios are not same. Hence the both will not have same colour.

Question 11.
The face perimeter of some vessels in the shape of square prisms are equal. 12 litres of water can be filled in the vessel having height 15 cm. 16 litres of water can be filled in the vessel having height 20 cm. Check whether the volumes of the vessel and height are proportional. What is the constant of proportionality?
Answer:
Since the face perimeters are equal,
Volume = face perimeter × height
Volume of the vessel having height 15 cm = 12 litres.
Ratio of height to volume =15 : 12 = 5 : 4 Volume of the vessel having height 20 cm = 16 litres
Ratio of height to volume =20 : 16 = 5 : 4 Since the ratios are equal, height and volume are proportional.
\(\frac { volume }{ height }\) = \(\frac { 4 }{ 5 }\) constant of proportionality
Volume of the vessel having height 35 cm
= 35 × \(\frac { 4 }{ 5 }\) = 28 liters.

Question 12.
The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 12

Question 13.
Raju got Rs. 400 after working for 8 hours. Damu worked for 6 hours and got Rs. 300. Are the wages obtained proportional to the work time?
Answer:
Ramu working hours = 8 hour
Wages Raju obtained = Rs. 400
Damu working hours = 6 hour
Wages obtained by Damu = Rs. 300
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 13
∴ Wages obtained are proportional to the work time.

Question 14.
150 L of water flows through a pipe for 6 minutes. 200 L of water flows for 8 minutes through the same pipe. Are the time and amount of water flowing proportional?
Answer:
Ratio of quantity of water
= 150 : 200 = 15 : 20 = 3 : 4
Ratio oftime =6 : 8 = 3 : 4
The ratios are same , hence they are proportional.

Question 15.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 14
The quantity of petrol needed to travel 180 km= 12 litre

Question 16.
The angles of a triangle are in the ratio 1 : 3: 5. How much is each angle of the triangle?
Answer:
Sum of the angles of a triangle = 180°
The angles of the traingle are \(\frac { 1 }{ 9 }\), \(\frac { 3 }{ 9 }\) and \(\frac { 5 }{ 9 }\)
parts of 180° , hence the angles are
Kerala Syllabus 9th Standard Maths Solutions Chapter 12 Proportion 15

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting

Students can Download Chapter 2 National Income Accounting Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting

Plus Two Economics National Income Accounting One Mark Questions and Answers

Question 1.
GNP – depreciation is called
(a) GDP
(b) NNP
(c) PCI
(d) PI
Answer:
(b) NNP

Question 2.
The GDP deflator is equal to
i) Real GDP-Nominal GDP
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img1
Answer:
iii) \(\frac{{ No minal GDP }}{\text { Real GDP }} \times 100\)

Question 3.
NFIA is included in:
(a) NNPFC
(b) NDPFC
(c) GDPFC
(d) All the above
Answer:
(a) NNPFC

Question 4.
Which among the following in a flow concept?
(a) export
(b) wealth
(c) capital
(d) foreign exchange reserve
Answer:
(a) export

HSSLive.Guru

Question 5.
When does net factor income from abroad become negative?
(a) NDP < NNP
(b) NNP < NDP
(c) NDP = NNP
(d) none of the above
Answer:
(b) NNP < NDP

Question 6.
When does GDP and GNP of an economy become equal?
(a) When net factor income from abroad is positive
(b) When net factor income from abroad is negative
(c) When net factor income from abroad is zero
(d) None ofthe above.
Answer:
(c) When net factor income from abroad is zero

Plus Two Economics National Income Accounting Two Mark Questions and Answers

Question 1.
Same job is done by a servant and housewife, whose service is included in the national income calculation? Why?
Answer:
Service of a servant is included in the national income calculation, whereas, the service of housewife is not included in the national income. This is because the housewife is not paid for the service she does.

Question 2.
From the following, classify the material into final goods and intermediary goods. Wheat, Bench, Bread, Wood, Rubber, Tyre.
Answer:

Final Goods Intermediary goods
Bench Wheat
Bread Wood
Tyre Rubber

Question 3.
Distinguish between real flow and money flow?
Answer:
Flow of goods and services from firms to households is called real flow. Factors of production receive reward for their services in the form of money. Households use this money to buy goods and services produced by firms. This flow of money from firms to households and back to firms is called money flow.

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Question 4.
Some variables are given below. Classify them into Stock and Flow

  1. Wealth
  2. Income of a household
  3. Consumption
  4. Capital
  5. Money Supply
  6. Capital formation
  7. Inventories
  8. Saving of a household

Answer:
a. Stock

  • Wealth
  • Inventories
  • Capital
  • Money supply

b. Flow

  • Income of a household
  • Consumption
  • Capital formation
  • Saving of a household

Question 5.
GDP = C + I + G + (X – M) = C + S + T Derive the Budget Deficit and Trade Deficit equations from the above identity.
Answer:
GDP = C + I + G + (X – M) = C + S + T
Budget deficit = G – T
Trade deficit = M – X

Plus Two Economics National Income Accounting Three Mark Questions and Answers

Question 1.
“Transfer payments are not included in the national income calculation”. Do you agree? Justify your answer.
Answer:
Yes. Transfer payments like pension, old age pension, etc. are not included in the national income. This is because they are transfer earnings not generated by any economic activity. These payments are usually made by the government out of tax revenue collected from the public. Since these generated incomes are already included in national income calculation there is no need to include transfer payment in the national income calculation again.

Question 2.
State whether the following are included or excluded in the national income.

  1. purchase of second hand goods
  2. operating surplus
  3. production for self-consumption
  4. interest
  5. windfall gains and loses

Answer:

  1. Purchase of second hand goods – excluded
  2. operating surplus – included
  3. old age pension – excluded
  4. Production for self consumption – excluded
  5. interest – included
  6. windfall gains and loses – excluded

Question 3.
Provide appropriate term.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img2
Answer:

  1. Value added
  2. GNP
  3. NNP
  4. NNPFC

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Question 4.
Point out any 3 uses of national income accounting.
Answer:
The uses of national income accounting are given below.

  1. It shows the distribution of national income among the various factors of production.
  2. National income statistics indicate the contribution of different sectors in the economy.
  3. Structural changes in the economy can be assessed by the national income accounting.

Uniform Distribution Calculator is an online tool that helps to calculate the probability distribution for the given values.

Question 5.
Classify the following under proper heads.
Flow of teacher services, Flow of subsidies and taxes, Flow of factor rewards, flow of finished goods, Flow of consumption expenditure, Flow of import goods.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img3
Answer:

Real Flow Money Flow
Flow of teacher services Flow of subsidies and taxes
Flow of finished goods Flow of factor rewards
Flow of import goods Flow of consumption

Question 6.

  • Does not includes prices of imported goods
  • Weights are different
  • It includes all goods and services
  • Includes prices of imported goods
  • Weights are constant
  • Does not include all goods and services

Answer:
a. Consumer price index

  • Includes prices of imported goods
  • Weights are constant
  • Does not include all goods and services

b. GDP deflator

  • Does not include prices of imported goods
  • Weights are different
  • It includes all goods and services

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Question 7.
Assume that there are three goods produced in an economy and they are sold at different prices in dif-ferent years. Calculate GDP Deflator.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img4
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img5

Question 8.
Calculate Depreciation, Net Indirect Tax and NNPFC from the below data.
GDPMP = 11300
NDPMP = 10300
NDPFC = 10000
NFIA = 1500
Answer:
1. Depreciation = GDPMP – NDPMP
= 11300 – 10300
= 1000

2. Net Indirect tax = NDPMP – NDPFC
= 10300 – 10000 = 300

3. NNPFC = NDPFC + NFIA
= 10000 + 1500
= 11500

Plus Two Economics National Income Accounting Five Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

  1. GNP, NNP, CSO, GDP
  2. Salary, bonus, GPF, free housing, saving
  3. Smuggling, production of wheat, sale of second-hand goods, services of housewives
  4. Services of teacher, services of engineer, services of lawyer, services of housewife
  5. Unemployment allowances, scholarships, old age pension, support price.

Answer:

  1. C.S.O. Others are national income concepts.
  2. Saving. Others come under compensation to employees
  3. Production of wheat. Others are excluded from national income
  4. Services of housewife. Others are included in the national income calculation.
  5. Support price. Others are transfer payments.

Question 2.
Match the following.

A B
NNP GDP – net factor income from abroad
GNP Personal income – direct taxes
Value added GNP-depreciation
GDP at market prices value of output – intermediate consumption
Disposable income GDP at factor cost – net indirect tax

Answer:

A B
NNP GNP – depreciation
GNP GDP – net factor income from abroad
Value added Value of output- intermediate consumption
GDP at market prices GDP at factor cost – net indirect tax
Disposable income Personal income – direct taxes

Question 3.
Categorize the following into stocks and flows, wealth, salary, food grain stock, foreign exchange reserves, export, gross domestic saving, capital, change in money supply, quantity of money, capital formation.
Answer:

Stock Flow
Wealth Export
Foreign exchange reserves Salary
Food grain stock Gross domestic saving
Capital Change in money supply
Quantity of money Capital formation

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Question 4.
The phase of circular flow of income in a two sector economy is given below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img6

  1. Complete the diagram.
  2. Explain the process of circular flow

Answer:

1.

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img7
2. Circular flow of income:
The concept that the aggregate value of goods and services produced in an economy is going around in a circular way. Either as factor payments, or as expenditures on goods and services, or as the value of aggregate production.

Question 5.
Suppose that in a two sector economy the value of finished goods is equal to ₹100 crore and the income generated as factor rewards is also equal to ₹100 crore. The households spend only ₹80 crore.

  1. What will happen to the circular flow?
  2. Which system can be introduced to correct the circular flow?
  3. Name the leakages and injections.

Answer:

  1. There will be a mismatch between the real flow and money flow in the circular flow. In other words, the flow will be broken.
  2. As a corrective measure, the financial system can be introduced.
  3. The leakages is the difference between the income generates and household spending.

This is saving. The injection are the savings that the households, firms and the government take from the financial institutions as borrowings.

HSSLive.Guru

Question 6.
1. Estimate the NI of India and Pakistan from the data given below.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img8
2. Which method is used here?
3. What are the other methods of measuring national income?
Answer:

  1. National income of India = ₹2885 crore
    National income of Pakistan = ₹1860 crore
  2. The method used here is the product method or value added method.
  3. Income method and expenditure method are the other two method of measuring national income.

Question 7.
What do you mean by GDP deflator? How far GDP deflator differs from Consumer Price Index?
Answer:
The ratio of nominal to real GDP is a well known index of prices. This is called GDP Deflator. GDP deflator differs from Consumer Price Index. The major points of difference are given below.

1. The goods purchased by consumers do not represent all the goods which are produced in a country. GDP deflator takes into account all such goods and services.

2. CPI includes prices of goods consumed by the representative consumer; hence it includes prices of imported goods. GDP deflator does not include prices of imported goods.

3. The weights are constant in CPI – but they differ according to production level of each good in GDP deflator.

Question 8.
Write down some of the limitations of using GDP as an index of welfare of a country.
Answer:
GDP is the sum total of value of goods and services created within the geographical boundary of a country in a particular year. It gets distributed among the people as incomes. So we may be tempted to treat higher level of GDP of a country as an index of greater well-being of the people of that country. But there are at least three reasons why this may not be correct. They are discussed below.

1. Distribution of GDP – how uniform is it:
If the GDP of the country is rising, the welfare may not rise as a consequence. This is because the rise in GDP may be concentrated in the hands of very few individuals or firms. For the rest, the income may, in fact, have fallen.

In such a case the welfare of the entire country cannot be said to have increased. If we relate welfare improvement in the country to the percentage of people who are better off, then surely GDP is not a good index.

2. Non-monetary exchanges:
Many activities in an economy are not evaluated in monetary terms. For example, the domestic services women perform at home are not paid for. The exchanges which take place in the informal sector without the help of money are called barter exchanges.

This is a case of underestimation of GDP. Hence GDP calculated in the standard manner may not give us a clear indication of the productive activity and well-being of a country.

3. Externalities:
Externalities refer to the benefits (or harms) a firm or an individual causes to another for which they are not paid (or penalized). Externalities do not have any market in which they can be bought and sold. Therefore, if we take GDP as a measure of welfare of the economy we shall be overestimating the actual welfare.

This was an example of negative externality. There can be cases of positive externalities as well. In such cases, GDP will underestimate the actual welfare of the economy.

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Question 9.
Assume that GDP in the year 2007 was ₹1,200 which rose to ₹1,800 in 2008. Calculate GDP deflator.
Answer:
GDP deflator = Current year GDP / Base year GDP x 100
= 1800/1200 × 100
= 1.5 × 100
= 1.5 (in percentage terms 150)

Question 10.
Relate and complete the identities/equations in column A with column B.
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img9
Answer:
Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img10

Question 11.
Estimate the Gross National Product at market price and GNP at factor cost through the expenditure method.

Item Amount (in Crores)
Inventory investment 15
Net factor income from abroad 10
Personal consumption expenditure 475
Gross residential construction investment 48
Exports 25
Government purchase of goods and services 175
Gross public investment 15
Gross business fixed investment 38
Imports 12
Net indirect tax 8

Answer:
GNPMP = private consumption expenditure + govt, final consumption expenditure( gross fixed capital formation + change in stock or inventory investment) + net export + net factor income from abroad
= 475 + 175 + 101 (i.e., 48 + 15 + 38) + 15 + 13
= ₹779 crores.
GNPC = GNPUD – net indirect taxes
= 779 – 8 = ₹771 crores

Question 12.
Suppose that in a two sector economy, the value o finished goods is equal to ₹200 crore and the income generated as factor rewards is equal to ₹200 crore. The households spend only ₹180 crore. The remaing 20 crore economy saved then.

  1. Is ₹20 (saving) included in the circular flow?
  2. Which system can be introduced to correct the circular flow?
  3. Is saving leakage or injection.

Answer:

  1. No, saving (₹20) is excluded in the circular flow.
  2. Financial system can be introduced to correct the circular flow.
  3. Yes, saving is a leakage.

Question 13.
Fill in the blanks

  1. GNPMP – ……….. = NNPMP
  2. NNPMP – ………… = NNPFC
  3. GDPFC+ – ………… = GDPMP
  4. GDP + -………….. = GNP

Answer:

  1. GNPMP – depreciation = NNPMP
  2. NNPMP – net indirect tax = NNPFC
  3. GDPFC + net indirect tax = GDPMP
  4. GDP + net factor income from aborad = GNP

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Question 14.
Write down the 3 identities of calculating the GDP of a country by the 3 methods. Also briefly explain why each of those should give us the same value of GDP.
Answer:
Gross National Product (GNP) equals Gross National Income equals Gross National Expenditure, i.e.
GNP = GNI = GNE
These are equal because national income is a circular flow of income. Aggregate expenditure is equal to aggregate output which in turn, is equal to aggregate income. However each method has some different items, yet they show exactly identical results.

Their identity can be shown in the following manner:
Reconciling Three Methods of Measuring Gross

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img11

Question 15.
The economic recession of 2008 affected the market economics in general and the US in particular. Thou-sands of Indians working abroad lost their job especially in IT and banking sectors and they returned to India. Evaluate its consequences on Indian economy with regard to the following macro variables.

  1. The value of GNP
  2. Gneral unemployment level
  3. Foreign exchange rate

Answer:

  1. The value of GNP decreases due to reduction in NFIA.
  2. General unemployment level increases.
  3. Foreign exchange rate increases.

Plus Two Economics National Income Accounting Eight Mark Questions and Answers

Question 1.
Given below some macro economic indicators. Derive the equations of the following terms:

  1. GNP
  2. NNP
  3. NNP at factor cost
  4. Personal income
  5. Personal disposable income
  6. Private Income
  7. National Disposable Income

Answer:
1. GNP = GDP + Factor income earned by the domestic factors of production employed in the rest of the world – Factor income earned by the factors of production of the rest of the world employed in the domestic economy

2. NNP = GNP – Depreciation

3. NNP at factor cost = National Income (NI) = NNP at market prices – (Indirect taxes – Subsidies)

4. Personal income (PI) = NI – Undistributed profits – Net interest payments made by households – Corporate tax + Transfer payments to the households from the government and firms.

5. Personal Disposable Income (PDI) = PI – Personal tax payments – Non-tax payments.

6. Private Income = Factor income from net domestic product accruing to the private sector + National debt interest + Net factor income from abroad + Current transfers from government + Other net transfers from the rest of the world

7. National Disposable Income = Net National Product at market prices + other current transfers from the rest of the world

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Question 2.
Prepare a seminar report on the topic ‘Measurement of National Income’.
Answer:
Measurement of National Income Respected teachers and dear friends,
The topic of my seminar paper is ‘measurement of national income or the methods of measuring national income’. The concept of national income occupies an important place in economic theory.

National income is the aggregate money value of all goods and services produced in a country during an accounting year. In this seminar paper, I would like to present various methods of measuring national income.

Content:
National income can be measured in different ways. Generally there are three methods for measuring national income. They are

  1. Value-added method
  2. Expenditure method
  3. Income method

1. Value-added method:
The term that is used to denote the net contribution made by a firm is called its value-added. We have seen that the raw materials that a firm buys from another firm which are completely used up in the process of production are called ‘intermediate goods’.

Therefore the value-added of a firm is the value of production of the firm – value of intermediate goods used by the firm. The value-added of a firm is distributed among its four factors of production, namely, labor, capital, entrepreneurship, and land.

Therefore wages, interest, profits, and rents paid out by the firm must add up to the value-added of the firm. Value-added is a flow variable.

2. Expenditure Method:
An alternative way to calculate the GDP is by looking at the demand side of the products. This method is referred to as the expenditure method. The aggregate value of the output in the economy by expenditure method will be calculated.

In this method we add the final expenditures that each firm makes. Final expenditure is that part of expenditure which is undertaken not for intermediate purposes.

3. Income Method:
As we mentioned in the beginning, the sum of final expenditures in the economy must be equal to the incomes received by all the factors of production taken together (final expenditure is the spending on final goods, it does not include spending on intermediate goods).

This follows from the simple idea that the revenues earned by all the firms put together must be distributed among the factors of production as salaries, wages, profits, interest earnings, and rents.
That is GDP = W + P + In + R

Conclusion:
Thus it can be concluded that there are three methods for measuring national income. These methods are value-added method, income method and expenditure method. Usually in estimating national income, different methods are employed for different sectors and sub sectors.

HSSLive.Guru

Question 3.
From the following data, calculate personal income and personal disposable income (₹in Crores).

  1. NDPFC – 8,000
  2. net factor income from abroad – 200
  3. Undistributed profit – 1,000
  4. Corporate tax – 500
  5. Interest received by households – 1,500
  6. Interest paid by households – 1,200
  7. Transfer income – 300
  8. Personal Tax – 500

Answer:
Personal income = NDPfc + Net factor income from abroad – undistributed profits – corporate taxes + transfer payments + net interest received from households.
= 8000 + 200-1000 – 500 + 300 (1500 -1200)
= 7,300 crores
Personal disposable income = Personal income – personal tax
= 7,300 – 500 = 6,800 crores

Question 4.
Production generates income. Prove this statement with the help of a simple two sector model of circular flow of income.
Answer:
circular flow of income:
It is a pictorial representation of interdependence or interrelationship between the various sectors of the economy. It is a concept associated with income earning and spending. The circular flow of income in a simple economy works on the basis of certain assumptions.
They are as follows:

  1. Households and firms are the only two sectors in an economy (2 sector model)
  2. Households supply factor services to firms.
  3. Firms hire factor services households
  4. Household spends their entire income on consumption and thereby no savings are left with them.
  5. Firms sell their entire products to the households
  6. There is no government in the economy.
  7. The economy is not related to any other economies or the economy is a ‘closed’ system. As a result, there is no export or imports from the economy.

In such an economy, there would be two types of markets.
They are:

  1. product-market for goods and services
  2. factor markets for buying and selling various factor services.

The relationship between the sectors of an economy can be explained with the help of a diagram.

Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting img12

The households own the factors of production such as land, labour, capital, and organization. The households sell these factors of production to the firms for producing goods and services are known as real flow. The rewards for factors of production are rent to land, interest to capital, wage to the labour and profit to the entrepreneur is known as the money flow.