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 Page 1


14.1
SOLUTIONS TO CONCEPTS
CHAPTER 14
1. F = mg
Stress = 
F
A
Strain = 
L
L
?
Y = 
FL L F
A L L YA
?
? ?
?
2. ? = stress = mg/A
e = strain = ?/Y
Compression ?L = eL ?
3. y = 
F L FL
L
A L AY
? ? ?
?
4. L
steel 
= L
cu
and A
steel 
= A
cu
a)
g
cu
cu g
A
F Stress of cu
Stress of st A F
? = 
cu
st
F
1
F
?
b) Strain = 
st st cu cu
st st cu cu
F L A Y Lst
lcu A Y F I
?
? ?
?
( ? L
cu
= I
st
; A
cu
= A
st
)
5.
st st
L F
L AY
? ? ?
?
? ?
? ?
cu cu
L F
L AY
? ? ?
?
? ?
? ?
cu cu
cu st
st st
AY Y strain steel wire F
( A A )
Strain om copper wire AY F Y
? ? ? ? ?
6. Stress in lower rod = 
1 1
1 1
T m g g
A A
? ?
? ? w = 14 kg
Stress in upper rod = 
2 2 1
u u
T m g m g wg
A A
? ?
? ? w = .18 kg
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break 
first.
1 1
1 1
T m g g
A A
? ?
? = 8 ? 10
8
? w = 14 kg
2 2 1
u u
T m g m g g
A A
? ? ?
? = 8 ? 10
8
? ?
0
= 2 kg
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. ?
7.
F L
Y
A L
?
?
8.
F L
Y
A L
?
?
?
YA L
F
L
?
?
9. m
2
g – T = m
2
a …(1)
and T – F = m
1
a …(2)
? a = 
2
1 2
m g F
m m
?
?
Page 2


14.1
SOLUTIONS TO CONCEPTS
CHAPTER 14
1. F = mg
Stress = 
F
A
Strain = 
L
L
?
Y = 
FL L F
A L L YA
?
? ?
?
2. ? = stress = mg/A
e = strain = ?/Y
Compression ?L = eL ?
3. y = 
F L FL
L
A L AY
? ? ?
?
4. L
steel 
= L
cu
and A
steel 
= A
cu
a)
g
cu
cu g
A
F Stress of cu
Stress of st A F
? = 
cu
st
F
1
F
?
b) Strain = 
st st cu cu
st st cu cu
F L A Y Lst
lcu A Y F I
?
? ?
?
( ? L
cu
= I
st
; A
cu
= A
st
)
5.
st st
L F
L AY
? ? ?
?
? ?
? ?
cu cu
L F
L AY
? ? ?
?
? ?
? ?
cu cu
cu st
st st
AY Y strain steel wire F
( A A )
Strain om copper wire AY F Y
? ? ? ? ?
6. Stress in lower rod = 
1 1
1 1
T m g g
A A
? ?
? ? w = 14 kg
Stress in upper rod = 
2 2 1
u u
T m g m g wg
A A
? ?
? ? w = .18 kg
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break 
first.
1 1
1 1
T m g g
A A
? ?
? = 8 ? 10
8
? w = 14 kg
2 2 1
u u
T m g m g g
A A
? ? ?
? = 8 ? 10
8
? ?
0
= 2 kg
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. ?
7.
F L
Y
A L
?
?
8.
F L
Y
A L
?
?
?
YA L
F
L
?
?
9. m
2
g – T = m
2
a …(1)
and T – F = m
1
a …(2)
? a = 
2
1 2
m g F
m m
?
?
Chapter-14
14.2
From equation (1) and (2), we get 
2
1 2
m g
2(m m ) ?
Again, T = F + m
1
a
?
2 2
1
1 2
m g m g
T m
2 2(m m )
? ?
?
?
2
2 1 2
1 2
m g 2m m g
2(m m )
?
?
Now Y = 
FL L F
A L L AY
?
? ?
?
?
2
2 1 2 2 2 1
1 2 1 2
(m 2m m )g m g(m 2m ) L
L 2(m m )AY 2AY(m m )
? ? ?
? ?
? ?
10. At equilibrium ? T = mg
When it moves to an angle ?, and released, the tension the T ? at lowest point is 
? T ? = mg + 
2
mv
r
The change in tension is due to centrifugal force ?T = 
2
mv
r
…(1)
? Again, by work energy principle,
?
2
1
mv
2
– 0 = mgr(1 – cos ?)
? v
2
= 2gr (1 – cos ?) …(2)
So, 
m[2gr(1 cos )]
T 2mg(1 cos )
r
? ?
? ? ? ? ?
? F = ?T
? F = 
YA L
L
?
= 2mg – 2mg cos ? ? 2mg cos ? = 2mg –
YA L
L
?
= cos ? = 1 –
YA L
L(2mg)
?
?
11. From figure cos ? = 
2 2
x
x l ?
= 
1/ 2
2
2
x x
1
l l
?
? ?
?
? ?
? ?
= x / l … (1)
Increase in length ?L = (AC + CB) – AB 
Here, AC = (l
2
+ x
2
)
1/2
So, ?L = 2(l
2
+ x
2
)
1/2
– 100 …(2)
Y = 
F l
A l ?
…(3)
From equation (1), (2) and (3) and the freebody diagram,
2l cos ? = mg. ?
12. Y = 
FL
A L ?
?
L F
L Ay
?
?
? = 
D/D
L /L
?
?
?
D L
D L
? ?
?
Again, 
A 2 r
A r
? ?
?
?
2 r
A
r
?
? ? ?
m 1
m 2
m 2g
a
T 
T 
a
F
? ?
B A
T
T
l l
T
x
mg
C
L ? L ?
Page 3


14.1
SOLUTIONS TO CONCEPTS
CHAPTER 14
1. F = mg
Stress = 
F
A
Strain = 
L
L
?
Y = 
FL L F
A L L YA
?
? ?
?
2. ? = stress = mg/A
e = strain = ?/Y
Compression ?L = eL ?
3. y = 
F L FL
L
A L AY
? ? ?
?
4. L
steel 
= L
cu
and A
steel 
= A
cu
a)
g
cu
cu g
A
F Stress of cu
Stress of st A F
? = 
cu
st
F
1
F
?
b) Strain = 
st st cu cu
st st cu cu
F L A Y Lst
lcu A Y F I
?
? ?
?
( ? L
cu
= I
st
; A
cu
= A
st
)
5.
st st
L F
L AY
? ? ?
?
? ?
? ?
cu cu
L F
L AY
? ? ?
?
? ?
? ?
cu cu
cu st
st st
AY Y strain steel wire F
( A A )
Strain om copper wire AY F Y
? ? ? ? ?
6. Stress in lower rod = 
1 1
1 1
T m g g
A A
? ?
? ? w = 14 kg
Stress in upper rod = 
2 2 1
u u
T m g m g wg
A A
? ?
? ? w = .18 kg
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break 
first.
1 1
1 1
T m g g
A A
? ?
? = 8 ? 10
8
? w = 14 kg
2 2 1
u u
T m g m g g
A A
? ? ?
? = 8 ? 10
8
? ?
0
= 2 kg
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. ?
7.
F L
Y
A L
?
?
8.
F L
Y
A L
?
?
?
YA L
F
L
?
?
9. m
2
g – T = m
2
a …(1)
and T – F = m
1
a …(2)
? a = 
2
1 2
m g F
m m
?
?
Chapter-14
14.2
From equation (1) and (2), we get 
2
1 2
m g
2(m m ) ?
Again, T = F + m
1
a
?
2 2
1
1 2
m g m g
T m
2 2(m m )
? ?
?
?
2
2 1 2
1 2
m g 2m m g
2(m m )
?
?
Now Y = 
FL L F
A L L AY
?
? ?
?
?
2
2 1 2 2 2 1
1 2 1 2
(m 2m m )g m g(m 2m ) L
L 2(m m )AY 2AY(m m )
? ? ?
? ?
? ?
10. At equilibrium ? T = mg
When it moves to an angle ?, and released, the tension the T ? at lowest point is 
? T ? = mg + 
2
mv
r
The change in tension is due to centrifugal force ?T = 
2
mv
r
…(1)
? Again, by work energy principle,
?
2
1
mv
2
– 0 = mgr(1 – cos ?)
? v
2
= 2gr (1 – cos ?) …(2)
So, 
m[2gr(1 cos )]
T 2mg(1 cos )
r
? ?
? ? ? ? ?
? F = ?T
? F = 
YA L
L
?
= 2mg – 2mg cos ? ? 2mg cos ? = 2mg –
YA L
L
?
= cos ? = 1 –
YA L
L(2mg)
?
?
11. From figure cos ? = 
2 2
x
x l ?
= 
1/ 2
2
2
x x
1
l l
?
? ?
?
? ?
? ?
= x / l … (1)
Increase in length ?L = (AC + CB) – AB 
Here, AC = (l
2
+ x
2
)
1/2
So, ?L = 2(l
2
+ x
2
)
1/2
– 100 …(2)
Y = 
F l
A l ?
…(3)
From equation (1), (2) and (3) and the freebody diagram,
2l cos ? = mg. ?
12. Y = 
FL
A L ?
?
L F
L Ay
?
?
? = 
D/D
L /L
?
?
?
D L
D L
? ?
?
Again, 
A 2 r
A r
? ?
?
?
2 r
A
r
?
? ? ?
m 1
m 2
m 2g
a
T 
T 
a
F
? ?
B A
T
T
l l
T
x
mg
C
L ? L ?
Chapter-14
14.3
13. B = 
Pv
v ?
? P = 
v
B
v
? ? ?
? ?
? ?
14.
0
0 d
m m
V V
? ? ?
so, 
d 0
0 d
V
V
?
?
?
…(1)
vol.strain = 
0 d
0
V V
V
?
B = 
0
0 d 0
gh
(V V )/ V
?
?
? 1 –
d
0
V
V
= 
0
gh
B
?
?
0
0
gh vD
1
v B
? ? ?
? ?
? ?
? ?
…(2)
Putting value of (2) in equation (1), we get
d
0 0
1
1 gh/B
?
?
? ? ?
?
d 0
0
1
(1 gh/B)
? ? ? ?
? ?
15.
F
A
? ?
?
Lateral displacement = l ?. ?
16. F = T l
17. a)
Hg
2T
P
r
? b) 
g
4T
P
r
? c) 
g
2T
P
r
?
18. a) F = P
0
A
b) Pressure = P
0
+ (2T/r)
F = P ?A = (P
0
+ (2T/r)A
c) P = 2T/r
F = PA = 
2T
A
r
19. a)
A
A
2Tcos
h
r g
?
?
? ?
b) 
B
B
2Tcos
h
r g
?
?
?
c) 
C
C
2Tcos
h
r g
?
?
?
20.
Hg Hg
Hg
Hg
2T cos
h
r g
?
?
?
2T cos
h
r g
? ?
?
?
?
?
?
where, the symbols have their usual meanings.
Hg
Hg Hg Hg
h T cos
h T cos
? ? ?
?
?
?
? ? ?
? ?
21.
2Tcos
h
r g
?
?
?
22. P = 
2T
r
P = F/r
23. A = ?r
2
24.
3 3
4 4
R r 8
3 3
? ? ? ?
? r = R/2 = 2
Increase in surface energy = TA ? – TA
Page 4


14.1
SOLUTIONS TO CONCEPTS
CHAPTER 14
1. F = mg
Stress = 
F
A
Strain = 
L
L
?
Y = 
FL L F
A L L YA
?
? ?
?
2. ? = stress = mg/A
e = strain = ?/Y
Compression ?L = eL ?
3. y = 
F L FL
L
A L AY
? ? ?
?
4. L
steel 
= L
cu
and A
steel 
= A
cu
a)
g
cu
cu g
A
F Stress of cu
Stress of st A F
? = 
cu
st
F
1
F
?
b) Strain = 
st st cu cu
st st cu cu
F L A Y Lst
lcu A Y F I
?
? ?
?
( ? L
cu
= I
st
; A
cu
= A
st
)
5.
st st
L F
L AY
? ? ?
?
? ?
? ?
cu cu
L F
L AY
? ? ?
?
? ?
? ?
cu cu
cu st
st st
AY Y strain steel wire F
( A A )
Strain om copper wire AY F Y
? ? ? ? ?
6. Stress in lower rod = 
1 1
1 1
T m g g
A A
? ?
? ? w = 14 kg
Stress in upper rod = 
2 2 1
u u
T m g m g wg
A A
? ?
? ? w = .18 kg
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break 
first.
1 1
1 1
T m g g
A A
? ?
? = 8 ? 10
8
? w = 14 kg
2 2 1
u u
T m g m g g
A A
? ? ?
? = 8 ? 10
8
? ?
0
= 2 kg
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. ?
7.
F L
Y
A L
?
?
8.
F L
Y
A L
?
?
?
YA L
F
L
?
?
9. m
2
g – T = m
2
a …(1)
and T – F = m
1
a …(2)
? a = 
2
1 2
m g F
m m
?
?
Chapter-14
14.2
From equation (1) and (2), we get 
2
1 2
m g
2(m m ) ?
Again, T = F + m
1
a
?
2 2
1
1 2
m g m g
T m
2 2(m m )
? ?
?
?
2
2 1 2
1 2
m g 2m m g
2(m m )
?
?
Now Y = 
FL L F
A L L AY
?
? ?
?
?
2
2 1 2 2 2 1
1 2 1 2
(m 2m m )g m g(m 2m ) L
L 2(m m )AY 2AY(m m )
? ? ?
? ?
? ?
10. At equilibrium ? T = mg
When it moves to an angle ?, and released, the tension the T ? at lowest point is 
? T ? = mg + 
2
mv
r
The change in tension is due to centrifugal force ?T = 
2
mv
r
…(1)
? Again, by work energy principle,
?
2
1
mv
2
– 0 = mgr(1 – cos ?)
? v
2
= 2gr (1 – cos ?) …(2)
So, 
m[2gr(1 cos )]
T 2mg(1 cos )
r
? ?
? ? ? ? ?
? F = ?T
? F = 
YA L
L
?
= 2mg – 2mg cos ? ? 2mg cos ? = 2mg –
YA L
L
?
= cos ? = 1 –
YA L
L(2mg)
?
?
11. From figure cos ? = 
2 2
x
x l ?
= 
1/ 2
2
2
x x
1
l l
?
? ?
?
? ?
? ?
= x / l … (1)
Increase in length ?L = (AC + CB) – AB 
Here, AC = (l
2
+ x
2
)
1/2
So, ?L = 2(l
2
+ x
2
)
1/2
– 100 …(2)
Y = 
F l
A l ?
…(3)
From equation (1), (2) and (3) and the freebody diagram,
2l cos ? = mg. ?
12. Y = 
FL
A L ?
?
L F
L Ay
?
?
? = 
D/D
L /L
?
?
?
D L
D L
? ?
?
Again, 
A 2 r
A r
? ?
?
?
2 r
A
r
?
? ? ?
m 1
m 2
m 2g
a
T 
T 
a
F
? ?
B A
T
T
l l
T
x
mg
C
L ? L ?
Chapter-14
14.3
13. B = 
Pv
v ?
? P = 
v
B
v
? ? ?
? ?
? ?
14.
0
0 d
m m
V V
? ? ?
so, 
d 0
0 d
V
V
?
?
?
…(1)
vol.strain = 
0 d
0
V V
V
?
B = 
0
0 d 0
gh
(V V )/ V
?
?
? 1 –
d
0
V
V
= 
0
gh
B
?
?
0
0
gh vD
1
v B
? ? ?
? ?
? ?
? ?
…(2)
Putting value of (2) in equation (1), we get
d
0 0
1
1 gh/B
?
?
? ? ?
?
d 0
0
1
(1 gh/B)
? ? ? ?
? ?
15.
F
A
? ?
?
Lateral displacement = l ?. ?
16. F = T l
17. a)
Hg
2T
P
r
? b) 
g
4T
P
r
? c) 
g
2T
P
r
?
18. a) F = P
0
A
b) Pressure = P
0
+ (2T/r)
F = P ?A = (P
0
+ (2T/r)A
c) P = 2T/r
F = PA = 
2T
A
r
19. a)
A
A
2Tcos
h
r g
?
?
? ?
b) 
B
B
2Tcos
h
r g
?
?
?
c) 
C
C
2Tcos
h
r g
?
?
?
20.
Hg Hg
Hg
Hg
2T cos
h
r g
?
?
?
2T cos
h
r g
? ?
?
?
?
?
?
where, the symbols have their usual meanings.
Hg
Hg Hg Hg
h T cos
h T cos
? ? ?
?
?
?
? ? ?
? ?
21.
2Tcos
h
r g
?
?
?
22. P = 
2T
r
P = F/r
23. A = ?r
2
24.
3 3
4 4
R r 8
3 3
? ? ? ?
? r = R/2 = 2
Increase in surface energy = TA ? – TA
Chapter-14
14.4
25. h = 
2Tcos
r g
?
?
, h ? = 
2Tcos
r g
?
?
? cos ? = 
h r g
2T
? ?
So, ? = cos
–1
(1/2) = 60°. ?
26. a) h = 
2Tcos
r g
?
?
b) T ? 2 ?r cos ? = ?r
2
h ? ? ? g
? ? cos ? = 
hr g
2T
?
?
27. T(2l) = [1 ? (10
–3
) ? h] ?g
28. Surface area = 4 ?r
2
29. The length of small element = r d ?
dF = T ? r d ?
considering symmetric elements, 
dF
y
= 2T rd ? . sin ? [dF
x
= 0]
so, F = 
/ 2
0
2Tr sin d
?
? ?
?
= 
/ 2
0
2Tr[cos ]
?
? = T ? 2 r
Tension ? 2T
1
= T ? 2r ? T
1
= Tr ?
30. a) Viscous force = 6 ??rv
b) Hydrostatic force = B = 
3
4
r g
3
? ?
? ?
? ?
? ?
c) 6 ?? rv + 
3
4
r g
3
? ?
? ?
? ?
? ?
= mg
v = 
2
2 r ( )g
9
? ? ?
?
?
3
2
m
g
(4 / 3) r 2
r
9 n
? ?
? ?
? ?
?
? ?
31. To find the terminal velocity of rain drops, the forces acting on the drop are,
i) The weight (4/3) ? r
3
?g downward.
ii) Force of buoyancy (4/3) ? r
3
?g upward.
iii) Force of viscosity 6 ??? r v upward.
Because, ? of air is very small, the force of buoyancy may be neglected.
Thus, 
6 ??? r v = 
2
4
r g
3
? ?
? ?
? ?
? ?
or v = 
2
2r g
9
?
?
?
32. v = 
R
D
?
?
? R = 
v D ?
?
? ? ? ?
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FAQs on HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter - Physics Class 11 - NEET

1. What are the mechanical properties of matter?
Ans. Mechanical properties of matter refer to the characteristics that describe how matter behaves under the action of external forces. These properties include elasticity, plasticity, strength, hardness, and toughness.
2. How is elasticity related to the mechanical properties of matter?
Ans. Elasticity is a mechanical property of matter that describes the ability of a material to return to its original shape and size after being deformed by an external force. Materials that exhibit high elasticity can undergo temporary deformation and then regain their original state when the force is removed.
3. What is the difference between strength and hardness in terms of mechanical properties of matter?
Ans. Strength and hardness are both mechanical properties of matter, but they refer to different aspects. Strength describes the ability of a material to withstand an applied force without breaking or deforming permanently, while hardness refers to the resistance of a material to indentation or scratching. Strength is related to the internal structure and bonding of the material, while hardness depends on the arrangement and density of atoms.
4. What is plasticity in the context of mechanical properties of matter?
Ans. Plasticity is a mechanical property of matter that describes the ability of a material to undergo permanent deformation without breaking. When a material is subjected to an external force beyond its elastic limit, it deforms permanently and does not regain its original shape and size. This property is important in processes like shaping and molding of materials.
5. How is toughness different from strength in terms of mechanical properties of matter?
Ans. Toughness and strength are both mechanical properties of matter, but they measure different aspects of a material's behavior under external forces. Strength refers to the ability of a material to withstand an applied force without breaking or deforming permanently. On the other hand, toughness describes the ability of a material to absorb energy and resist fracture when subjected to an impact or sudden load. Toughness takes into account both strength and the ability to deform plastically before breaking.
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Document Description: HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter covers topics like and HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 14 - Some Mechanical Properties of Matter.
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