Mechanical Properties Of Solids

Ans: 1
Length of the steel wire, L₁ = 4.7 m
Area of cross-section of the steel wire, A₁ = 3.0 × 10^–5m²
Length of the copper wire, L₂ = 3.5 m
Area of cross-section of the copper wire, A₂ = 4.0 × 10^–5m²
Change in length = ΔL₁= ΔL₂ = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:

… (i)
Young’s modulus of the copper wire:

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

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

Ans: 2
(a) It is clear from the given graph that for stress 150 × 10⁶ N/m², strain is 0.002.
∴Young’s modulus, Y

Hence, Young’s modulus for the given material is 7.5 ×10¹⁰N/m².
(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 300 × 10⁶ Nm/² or 3 × 10⁸ N/m².

Ans: 3
Answer: (a) A(b) A
(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
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.

Ans: 4
Answer: (a) False(b) True
(a) For a given stress, the strain in rubber is more than it is in steel.
Young’s modulus,
For a constant stress:
Hence, Young’s modulus for rubber is less than it is for steel.
(b) Shear modulus 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.

Ans: 5
Elongation of the steel wire = 1.49 × 10^–4 m
Elongation of the brass wire = 1.3 × 10^–4 m
Diameter of the wires,d = 0.25 m
Hence, the radius of the wires, = 0.125 cm
Length of the steel wire, L₁ = 1.5 m
Length of the brass wire, L₂ = 1.0 m
Total force exerted on the steel wire:
F₁ = (4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:

Where,
ΔL₁= Change in the length of the steel wire
A₁ = Area of cross-section of the steel wire
Young’s modulus of steel, Y₁ = 2.0 × 10¹¹ Pa

Total force on the brass wire:
F₂ = 6 × 9.8 = 58.8 N
Young’s modulus for brass:

Elongation of the steel wire = 1.49 × 10^–4 m
Elongation of the brass wire = 1.3 × 10^–4 m

Ans: 6
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 10⁹ Pa
Shear modulus, η
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m²
ΔL = Vertical deflection of the cube

= 3.92 × 10^–7 m
The vertical deflection of this face of the cube is 3.92 ×10^–7m.

Ans: 7
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 10¹¹ Pa
Total force exerted, F= Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = 122500 N
Young’s modulus,Y

Where,
Area, A = π(R² – r²) = π((0.6)² – (0.3)²)
= 7.22 × 10^–7
Hence, the compressional strain of each column is 7.22 × 10^–7.

Ans: 8
Length of the piece of copper, l = 19.1 mm = 19.1 × 10^–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10^–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10^–3 × 15.2 × 10^–3
= 2.9 × 10^–4 m²
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η= 42 × 10⁹ N/m²
Modulus of elasticity, η

= 3.65 × 10^–3


 Ans: 9
Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 10⁸ N m^–2
Maximum stress =
∴Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π(0.015)²
= 7.065 × 10⁴ N
Hence, the cable can support the maximum load of 7.065 × 10⁴ N.


Ans: 10

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

Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that
Young’s modulus for iron, Y₁ = 190 × 10⁹ Pa
Diameter of the iron wire = d₁
Young’s modulus for copper, Y₂ = 110 × 10⁹ Pa
Diameter of the copper wire = d₂
Therefore, the ratio of their diameters is given as:


Ans: 11
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω= 2 rev/s
Cross-sectional area of the wire, a = 0.065 cm²
Let δlbe 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:
F = mg +mlω²
= 14.5 × 9.8 + 14.5 × 1 × (2)²
= 200.1 N

Young’s modulus for steel = 2 × 10¹¹ Pa

Hence, the elongation of the wire is 1.539 × 10^–4 m.

Ans: 12
Initial volume, V₁ = 100.0l = 100.0 × 10 ^–3 m³
Final volume, V₂ = 100.5 l = 100.5 ×10 ^–3 m³
Increase in volume, ΔV = V₂ – V₁ = 0.5 × 10^–3 m³
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10⁵ Pa


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

Ans: 13
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10⁵ Pa
Density of water at the surface, ρ₁ = 1.03 × 10³ kg m^–3
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.



For equations (i) and (ii), we get:

Therefore, the density of water at the given depth (h) is 1.034 × 10³ kg m^–3.

Ans: 14
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 10⁵ Pa
Bulk modulus of glass,B = 37 × 10⁹ Nm^–2

Where,
= Fractional change in volume

Hence, the fractional change in the volume of the glass slab is 2.73 × 10^–5.

Ans: 15
Length of an edge of the solid copper cube, l = 10 cm = 0.1 m
Hydraulic pressure, p = 7.0 ×10⁶ Pa
Bulk modulus of copper, B = 140 × 10⁹ Pa

Where,
= Volumetric strain
ΔV = Change in volume
V = Original volume.

Original volume of the cube, V = l³

Therefore, the volume contraction of the solid copper cube is 5 × 10^–2 cm^–3.

Ans: 16
Volume of water, V = 1 L
It is given that water is to be compressed by 0.10%.


Therefore, the pressure on water should be 2.2 ×10⁶ Nm^–2.

Ans: 17
Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10^–3m

Compressional force,F = 50000 N
Pressure at the tip of the anvil:

Therefore, the pressure at the tip of the anvil is 2.55 × 10¹¹ Pa.

Ans: 18
Answer: (a) 0.7 m from the steel-wire end
(b) 0.432 m from the steel-wire end
Cross-sectional area of wire A, a₁ = 1.0 mm²= 1.0 × 10^–6 m²
Cross-sectional area of wire B, a₂ = 2.0 mm²= 2.0 × 10^–6 m²
Young’s modulus for steel, Y₁ = 2 × 10¹¹ Nm^–2
Young’s modulus for aluminium, Y₂ = 7.0 ×10¹⁰Nm^–2
(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.

If the two wires have equal stresses, then:

Where,
F₁= Force exerted on the steel wire
F₂= Force exerted on the aluminum wire

The situation is shown in the following figure.

Taking torque about the point of suspension, we have:

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

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)

If the strain in the two wires is equal, then:

Taking torque about the point where mass m, is suspended at a distance y₁ from the side where wire Aattached, we get:
F₁y₁= F₂ (1.05 – y₁)
…(iii)
Using equations (iii) and (iv), we get:

In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire Ais attached.

Ans: 19

Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 × 10^–2 cm² = 0.50 × 10^–6 m²
A mass 100 g is suspended from its midpoint.
m = 100 g = 0.1 kg
Hence, the wire dips, as shown in the given figure.

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO + OZ
Increase in the length of the wire:
Δl = (XO + OZ) – XZ
Where,
XO = OZ =


Let T be the tension in the wire.
∴mg = 2T cosθ
Using the figure, it can be written as:

Expanding the expression and eliminating the higher terms:



Young’s modulus of steel, Y =

Hence, the depression at the midpoint is 0.0106 m.

Ans: 20
Diameter of the metal strip, d = 6.0 mm = 6.0 × 10^–3 m

Maximum shearing stress = 6.9 × 10⁷ Pa

Maximum force = Maximum stress × Area
= 6.9 × 10⁷ × π× (r) ²
= 6.9 × 10⁷ × π× (3 ×10^–3)²
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N

Ans: 21
Water pressure at the bottom, p = 1.1 × 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.

Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10^–4 m³.