NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Physics
Chapter : 9
Chapter Name : Mechanical Properties Of Solids

Q9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m stretches by the same
2

amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m under a given
2

load. What is the ratio of the Young’s modulus of steel to that of copper?

Length of the steel wire, L1 = 4.7m

−5 2
Area of cross-section of the steel wire, A1 = 3.0 × 10 m

Length of the copper wire, L2 = 3.5m

−5 2
Area of cross-section of the copper wire, A2 = 4.0 × 10 m

Length of the copper wire, L2 = 3.5m

−5 2
Area of cross-section of the copper wire, A2 = 4.0 × 10 m

Change in length = ΔL1 = ΔL2 = ΔL

Force applied in both the cases = F

Young's modulus of the steel wire:
F1 L1
Y1 = ×
A1 ΔL

F ×4.7
=
−5
3.0×10 ×ΔL …(i)

Young's modulus of the copper wire:

F2 L2
Y2 = ×
A2 ΔL2

F ×3.5
=
−5
4.0×10 ×ΔL

Dividing (i) by (ii), we get:

−5
Y1 4.7×4.0×10
= −5
= 1.79 : 1
Y2 3.0×10 ×3.5

The ratio of Young's modulus of steel to that of copper is 1.79 : 1.

Page : 247 , Block Name : Exercise

Q9.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus
and (b) approximate yield strength for this material?

6 2
(a) It is clear from the given graph that for stress 150 × 10 N/m , strain is 0.002

Stress
=
Strain

∴ Young's modulus, Y

6
150×10 10 2
= = 7.5 × 10 N/m
0.002
10 2
Hence, Young's modulus for the given material is 7.5 × 10 N/m .

(b) The yield strength of a material is the maximum stress that the material can sustain

without crossing the elastic limit.

It is clear from the given graph that the approximate yield strength of this material is

6 2 8 2
300 × 10 Nm/ or 3 × 10 N/m

Page : 247 , Block Name : Exercise

Q9.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12.

The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?

Answer. (a) For a given strain, the stress for material A is more than it is for material B, as
shown in the two graphs.
Young's modulus

Stress
=
Strain

Young's modulus

For a given strain, if the stress for a material is more, then Young's modulus is also

greater for that material. Therefore, Young's modulus for material A is greater than it is

for material B.

(b) The amount of stress required for fracturing a material, corresponding to its fracture

point, gives the strength of that material. Fracture point is the extreme point in a stress-

strain curve. It can be observed that material A can withstand more strain than material

B. Hence, material A is stronger than material B.

Page : 247 , Block Name : Exercise

Q9.4 Read the following two statements below carefully and state, with reasons, if it is true or
false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.

Answer. (a) False (b) True
(a) For a given stress, the strain in rubber is more than it is in steel.
Young's modulus, Y = Stress

Strain

For a constant stress: Y ∝
1

Strain

Hence, Young's modulus for rubber is less than it is for steel.
(b) Shear modu us is the ratio of the applied stress to the change in the shape of a
body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is
involved in this process.

Page : 248 , Block Name : Exercise

Q9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as
shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.
Compute the elongations of the steel and the brass wires.

−4
Elongation of the steel wire = 1.49 × 10 m

−4
Elongation of the brass wire = 1.3 × 10 m

Diameter of the wires, d = 0.25m
d
r=
Hence, the radius of the wires, 2
= 0.125cm

Length of the steel wire, L1 = 1.5m

Length of the brass wire, L2 = 1.0m

Total force exerted on the steel wire:

F1 = (4 + 6)g = 10 × 9.8 = 98N

Young's modulus for steel:
F
1
( )
A
1
Y1 =
ΔL
1
( )
L
1

Where,

ΔL1 = Change in the length of the steel wire

2
A1 = Area of cross-section of the steel wire = πr
1

11
Young's modulus of steel, Y1 = 2.0 × 10 Pa
F1 ×L1 F1 ×L1
∴ ΔL1 = =
2
A1 ×Y1 πr ×Y1
1

98×1.5 −4
= = 1.49 × 10 m
2
−2 11
π(0.125×10 ) ×2×10

Total force on the brass wire:

F2 = 6 × 9.8 = 58.8N

Young's modulus for brass:
F
2
( )
A
2
Y2 =
ΔL
2
( )
L
2

Where,

ΔL2 = Change in length

A2 = Area of cross-section of the brass wire
F2 ×L2 F2 ×L2
∴ ΔL2 = =
2
A2 ×Y2 πr ×Y2
2

58.8×1.0 −4
= = 1.3 × 10 m
2
−2 11
π×(0.125×10 ) ×(0.91×10 )

−4
Elongation of the steel wire = 1.49 × 10 m

−4
Elongation of the brass wire = 1.3 × 10 m

Page : 248 , Block Name : Exercise

Q9.6 The edge of an aluminium cube is 10 cm long. One face of the cube is rmly xed to a
vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear
modulus of aluminium is 25 GPa. What is the vertical de ection of this face?

Edge of the aluminium cube, L = 10cm = 0.1m

The mass attached to the cube, m = 100kg

9
Shear modulus ( n) of aluminium = 25GPa = 25 × 10 Pa
F

Shear stress A
= = L
Shear strass
ΔL

Shear modulus, η

Where,

F = Applied force = mg = 100 × 9.8 = 980N

2
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01m

ΔL = vertical deflection of the cube
FL
∴ ΔL =
Aη

980×0.1
=
−2 9
10 ×(25×10 )

−7
= 3.92 × 10 m

−7
The vertical deflection of this face of the cube is 3.92 × 10 m .

Page : 248 , Block Name : Exercise

Q9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass
50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the
load distribution to be uniform, calculate the compressional strain of each column.

Mass of the big structure, M = 50, 000kg

Inner radius of the column, r = 30cm = 0.3m

Outer radius of the column, R = 60cm = 0.6m

11
Young's modulus of steel, Y = 2 × 10 Pa

Total force exerted, F = M g = 50000 × 9.8N
50000×9.8
Stress = Force exerted on a single column = = 122500N
4

Stress
Young's modulus, γ =
Strain

F
Strain =
A

Where,

2 2 2 2
Area, A = π (R − r ) = π ((0.6) − (0.3) )

122500 −7
Strain = = 7.22 × 10
2 2 11
π[(0.6) −(0.3) ]×2×10

−7
Hence, the compressional strain of each column is 7.22 × 10 .

Page : 248 , Block Name : Exercise

Q9.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in
tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

−3
Length of the piece of copper, l = 19.1mm = 19.1 × 10 m

−3
Breadth of the piece of copper, b = 15.2mm = 15.2 × 10 m

Area of the copper piece:

A = | × b

−3 −3
= 19.1 × 10 × 15.2 × 10

−4 2
= 2.9 × 10 m

Tension force applied on the piece of copper, F = 44500N

9 2
Modulus of elasticity of copper, n = 42 × 10 N/m
F

Stress A
Modulus of elasticity, n = =
Strain Strain

F
∴ Strain =
Aη

44500
=
−4 9
2.9×10 ×42×10

−3
= 3.65 × 10

Page : 248 , Block Name : Exercise

Q9.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress
is not to exceed 108 N m , what is the maximum load the cable can support ?
−2

Radius of the steel cable, r = 1.5cm = 0.015m

8 −2
Maximum allowable stress = 10 Nm

Maximum force

Maximum stress = Area of cross-section

∴ Maximum force = Maximum stress × Area of cross-section

8 2
= 10 × ∩(0.015)

4
= 7.065 × 10 N

4
Hence, the cable can support the maximum load of 7.065 × 10 N .

Page : 248 , Block Name : Exercise

Q9.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those
at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if
each is to have the same tension

The tension force acting on each wire is the same. Thus, the extension in each case i

the same. Since the wires are of the same length, the strain will also be the same.

The relation for Young's modulus is given as:
F 4F

2
Stress A πd
Y = = =
Strain Strain Strain

Where,

F = Tension force

A = Area of cross-section

d = Dlameter of the wire
1
Y∝
2
It can be inferred from equation (i) that d

9
Young's modulus for iron, Y1 = 190 × 10 Pa

Diameter of the iron wire = d1

9
Young's modulus for copper, Y2 = 110 × 10 Pa

Diameter of the copper wire = d2

Therefore, the ratio of their diameters is given as:

9
d2 Y1 190×10 19
= √ = √ 9
= √ = 1.31 : 1
d1 Y2 110×10 11

Page : 248 , Block Name : Exercise

Q9.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled
in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-
sectional area of the wire is 0.065 cm . Calculate the elongation of the wire when the mass is at
2

the lowest point of its path

Mass, m = 14.5kg

Length of the steel wire, / = 1.0m

Angular velocity, ω = 2rev/s

2
cross-sectional area of the wire, a = 0.065cm

Let δl be the elongation of the wire when the mass is at the lowest point of its path.

When the mass is placed at the position of the vertical circle, the total force on the mass

is:

Young's modulus = Stress

F

A F I
Y = =
Δl
A Δl
I

I
Fl
∴ Δl =
AY

11
Young's modulus for steel = 2 × 10 Pa

200.1 × 1
−3
∴ Δl = = 1539.23 × 10
−4 11
0.065 × 10 × 2 × 10
−4
= 1.539 × 10 m

Hence, the elongation Of the wire is 1.539 × 10
−4
m

Page : 248 , Block Name : Exercise

Q9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre,
Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the
bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the
ratio is so large.

−3 3
Initial volume, V1 = 100.01 = 100.0 × 10 m

−3 3
Final volume, V2 = 100.5 ∣
∣= 100.5 × 10 m

−3 3
Increase in volume, ΔV = V2 − V1 = 0.5 × 10 m

5
Increase in pressure, Δp = 100.0atm = 100 × 1.013 × 10 Pa

Δp Δp × V1
Bulk modulus = =
ΔV ΔV
5 −3
100 × 1.013 × 10 × 100 × 10
=
−3
0.5 × 10
9
= 2.026 × 10 Pa

5
Bulk modulus of air = 1.0 × 10 Pa
∘
Bulk modulus of water 2.026×10 4
∴ = 5
= 2.026 × 10
Bulk modulus of air 1.0×10

This ratio is very high because air is more compressible than water.

Page : 248 , Block Name : Exercise

Q9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at
the surface is 1.03 × 10 kg m ? 3 −3

Let the given depth be h.

5
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10 Pa

3 −3
Density of water at the surface, ρ1 = 1.03 × 10 kgm

Let ρ2 be the density of water at the depth h .

Let V1 be the volume of water of mass m at the surface.

Let V2 be the volume of water of mass m at the depth h

Let ΔV be the change in volume.

ΔV = V1 − V2

1 1
= m( − )
ρ1 ρ2

ΔV
∴ Volumetric strain =
V1

1 1 ρ1
= m( − ) ×
ρ1 ρ2 m

ΔV ρ1
∴ = 1 −
V1 ρ2

pV1
Bulk modulus, B =
ΔV

ΔV p
=
V1 B

1 −11 −1
Compressibity of water = = 45.8 × 10 Pa
B

ΔV 5 −11 −3
∴ = 80 × 1.013 × 10 × 45.8 × 10 = 3.71 × 10
V1

For equations (i) and (ii), we get:
ρ1 −3
1 − = 3.71 × 10
ρ2

3
1.03×10
ρ2 = −3
1−(3.71×10 )

3 −3
= 1.034 × 10 kgm

3 −3
Therefore, the density of water at the given depth (h) is 1.034 × 10 kgm .

Page : 248 , Block Name : Exercise

Q9.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic
pressure of 10 atm

Answer. Hydraulic pressure exerted on the glass slab,
p=10 atm = 10 × 1.013 × 10 Pa Bulk modulus of glass, B = 37 × 10 Nm
5 9 −2

p
Bulk modulus, B =
ΔV

V

Where,

ΔV
= Fractional change in volume
V

ΔV p
∴ =
V B
5
10 × 1.013 × 10
=
9
37 × 10
−5
= 2.73 × 10

Hence, the fractional change in the volume Of the glass slab is 2.73 x 10-5

Page : 248 , Block Name : Exercise

Q9.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when
subjected to a hydraulic pressure of 7.0 × 106 Pa

Length of an edge of the solid copper cube, I = 10cm = 0.1m

6
Hydraulic pressure, p = 7.0 × 10 Pa

9
Bulk modulus of copper, B = 140 × 10 Pa
p
Bulk modulus, B =
ΔV

Where,

ΔV
= Volumetric strain
V

ΔV = Change in volume

V = Original volume.

pV
ΔV =
B

3
Original volume of the cube, V = I
3
pl
∴ ΔV =
B
6 3
7 × 10 × (0.1)
=
9
140 × 10
−8 3
= 5 × 10 m
−8 −3
= 5 × 10 cm

Therefore, the volume contraction Of the solid copper cube is 5 × 10 −2
cm
−3

Page : 248 , Block Name : Exercise

Q9.16 How much should the pressure on a litre of water be changed to compress it by 0.10%?

Volume of water, V = 1L

It is given that water is to be compressed by 0.10%.

ΔV 0.1 −3
∴ Fractional change, = = 10
V 100×1
ρ
Bulk modulus, B =
ΔV

V

ΔV
p = B ×
V
9 −2
Bulk modulus of water, B = 2.2 × 10 Nm

9 −3
p = 2.2 × 10 × 10

6 −2
= 2.2 × 10 Nm

6 −2
Therefore, the pressure on water should be 2.2 × 10 Nm .

Page : 248 , Block Name : Exercise

Q9.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to
investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the
anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of
50,000 N. What is the pressure at the tip of the anvil?

−3
Diameter of the cones at the narrow ends, d = 0.50mm = 0.5 × 10 m

d −3
Radius, r = = 0.25 × 10 m
2

Compressional force, F = 50000N

Pressure at the tip of the anvil:

Therefore, the pressure at the tip of the anvil is 2.55 × 10 11

Force F
p = = 2
Areaπ r

50000
= 2
−3
π(0.25×10 )

11
= 2.55 × 10 Pa

Page : 249 , Block Name : Exercise

Q9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel
(wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas
of wires A and B are 1.0 mm and 2.0 mm , respectively. At what point along the rod should a
2 2

mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and
aluminium wires

(a) 0.7m from the steel-wire end

(b) 0.432m from the steel-wire end

2 −6 2
cross-sectional area of wire A, a1 = 1.0mm = 1.0 × 10 m

2 −6 2
cross-sectional area of wire B, a2 = 2.0mm = 2.0 × 10 m

11 −2
Young's modulus for steel, Y1 = 2 × 10 Nm

10 −2
Young's modulus for aluminium, Y2 = 7.0 × 10 Nm

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire

A is attached.

F F
Stress in the wire = =
Area a

If the two wires have equal stresses, then:
F1 F2
=
a1 a2

Where,

F1 = Force exerted on the steel wire

F2 = Force exerted on the aluminum wire
F1 a1 1
= =
F2 a2 2

The situation is shown in the following figure.

Taking torque about the point of suspension, we have:

F1 y = F2 (1.05 − y)

F1 (1.05−y)
=
F2 y

Using equations (i) and (ii), we can write:

(1.05−y) 1
=
y 2

2(1.05 − y) = y

2.1 − 2y = y

3y = 2.1

∴ y = 0.7m

In order to produce an equal stress in the two wires, the mass should be suspended at a
distance of 0.7 m from the end where wire A is attached.

(b)

Young's modulus = Stress
F

Stress G
Strain = =
Young's modulus Y

If the strain in the two wires is equal, then:
F1 F2
= a
a1 2

Y
2

11
F1 a1 Y1 1 2×10 10
= = × 10
=
F2 a2 Y2 2 7×10 7

Taking torque about the point where mass m, is suspended at a distance y1 from the

side where wire A attached, we get:

F1 y1 = F2 (1.05 − y1 )

F1 (1.05−y1 )
= … (iii)
F2 y1

Using equations (iii) and (iv), we get:

(1.05−y1 ) 10
=
y1 7

7 (1.05 − y1 ) = 10y1

17y1 = 7.35

∴ y1 = 0.432m

In order to produce an equal strain in the two wires, the mass should be suspended at a

distance of 0.432m from the end where wire A is attached.

Page : 249 , Block Name : Exercise

Q9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10 cm is stretched, −2 2

well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from
the mid-point of the wire. Calculate the depression at the midpoint.

Length of the steel wire = 1.0m

−2 2 −6 2
Area of cross-section, A = 0.50 × 10 cm = 0.50 × 10 m

A mass 100g is suspended from its midpoint.

m = 100g = 0.1kg

Hence, the wire dips, as shown in the given figure.

Original length = XZ

Depression = I

The length after mass m, is attached to the wire = ×0 + Oz

Increase in the length of the wire:

ΔI = (XO + OZ) − XZ

Where,
1
2 2 2
xO = Oz = [(0.5) + l ]

1
2 2 2
∴ Δl = 2[(0.5) + (l) ] − 1.0

1

2 2
l
= 2 × 0.5[1 + ( ) ] − 1.0
0.5

Expanding and neglecting higher terms, we get:
2
l
Δl =
0.5

Increase in length
Strain =
Original length

Let T be the tension in the wire.

∴ mg = 2T cos θ

Using the figure, it can be written as:
l
cos θ =
1
2 2
((0.5) +l ) 2

1
=
1
2 2
l
(0.5)(1+( ) )
0.5

Expanding the expression and eliminating the higher terms:

1
cos θ =
l2
(0.5)(1+ )
2
2(0.5)

2
l
(1 + ) = 1 for small l
0.5

t
∴ cos θ =
0.5

mg mg×0.5 mg
∴ T = = =
l 2l 4I
2( )
0.5

Tension mg
Stress = =
Area 4l×A

Young's modulus = Stress

mg×0.5
Y =
2
4l×A×l

3
mg×0.5
l = √
4Y A

lI
Young's modulus of steel, Y = 2 × 10 Pa

0.1×9.8×0.5
∴ l = √
11 −6
4×2×10 ×0.50×10

= 0.0106m

Hence, the depression at the midpoint is 0.0106m.

Page : 249 , Block Name : Exercise

Q9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0
mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress
on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one quarter of the
load

−3
Diameter of the metal strip, d = 6.0mm = 6.0 × 10 m

d −3
Radius, r = = 3.0 × 10 m
2

7
Maximum shearing stress = 6.9 × 10 Pa

Maximum load or force
Maximum stress =
Area

Maximum force = Maximum stress × Area

7 2
= 6.9 × 10 × π × (r)

2
7 −3
= 6.9 × 10 × π × (3 × 10 )

= 1949.94N

Each rivet carries one quarter of the load.

∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76N

Page : 249 , Block Name : Exercise

Q9.21 The Mariana trench is located in the Paci c Ocean, and at one place it is nearly eleven km
beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108
Pa. A steel ball of initial volume 0.32 m is dropped into the ocean and falls to the bottom of the
3

trench. What is the change in the volume of the ball when it reaches to the bottom?

8
Water pressure at the bottom, p = 1.1 × 10 Pa

3
Initial volume of the steel ball, V = 0.32m

11 −2
Bulk modulus of steel, B = 1.6 × 10 Nm

The ball falls at the bottom of the Pacific Ocean, which is 11km beneath the surface.

Let the change in the volume of the ball on reaching the bottom of the trench b ΔV
p
Bulk modulus, B =
ΔV

B
ΔV =
pV

8
1.1×10 ×0.32
= 11
1.6×10

−4 3
= 2.2 × 10 m

Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2

−4 3
×10 m .

Page : 249 , Block Name : Exercise