Chapter 14 : Some Mechanical Properties of Matter - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

JEE : Chapter 14 : Some Mechanical Properties of Matter - HC Verma Solution, Physics Class 11 Notes | EduRev

 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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Chapter 14 : Some Mechanical Properties of Matter - HC Verma Solution

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Physics Class 11 Notes | EduRev

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