Mechanical Properties of Solids

Given,

• Length of the steel wire, L_S = 4.7 m
• Area of cross-section of the steel wire, A_S = 3.0 × 10^-5 m²
• Length of the copper wire, L_C = 3.5 m
• Area of cross-section of the copper wire, A_C = 4.0 × 10^-5 m²

If a wire of length L and area of cross-section A stretches in length by ΔL under a load Mg, then the Young's modulus of the material of the wire is

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} = \frac{\text{Mg}/\text A}{\Delta \text L/\text L} = \frac{\text{Mg}\text L}{\text A\Delta \text L}$

The load Mg and the stretching ΔL are the same for the two wires. Hence, for the steel and the copper wires,

$\frac{\text Y_S}{\text Y_C} = \frac{\text L_S}{\text A_S} \times \frac{\text A_C}{\text L_C}$

Substituting the given values,

$$\begin{aligned}\frac{\text Y_S}{\text Y_C} = \frac{4.7 \times (4.0 \times 10^{-5})}{(3.0 \times 10^{-5}) \times 3.5} \\[1em] = \frac{18.8}{10.5} = 1.79\end{aligned}$$

Hence, the ratio of the Young's modulus of steel to that of copper is 1.79.

(a) Young's modulus : In the initial straight portion of the graph, Hooke's law is obeyed and the ratio of stress to strain is constant. From the graph, a stress of 150 × 10⁶ N m^-2 corresponds to a strain of 0.002. Therefore,

$$\begin{aligned}\text Y = \frac{\text{stress}}{\text{strain}} = \frac{150 \times 10^{6}}{0.002} \\[1em] = 7.5 \times 10^{10}\ \text{N m}^{-2}\end{aligned}$$

Hence, the Young's modulus of the material is 7.5 × 10¹⁰ N m^-2.

(b) Yield strength : The yield point of the given graph is the point B, at which the stress is roughly 300 × 10⁶ N m^-2. The stress at the yield point is the yield strength.

Hence, the approximate yield strength of the material is 3.0 × 10⁸ N m^-2.

(a) Material A has the greater Young's modulus.

The Young's modulus is the ratio of stress to strain, which is the slope of the linear portion of the stress-strain curve. The linear portion of the curve of material A is steeper than that of material B, that is, its slope is greater. Hence material A has the greater Young's modulus.

(b) Material A is the stronger material.

The strength of a material is decided by the maximum stress it can bear before it breaks, that is, by the stress at its fracture point. Material A withstands a greater stress before breaking than material B, and is therefore stronger.

(a) False.

The Young's modulus measures the stiffness of a material, that is, its resistance to a change in length under a longitudinal stress. Rubber is highly elastic in the everyday sense — it produces a large elongation for a small stress — so it is much less stiff. Steel, on the other hand, is very stiff, as a large stress produces only a small elongation in it. Since

$\text Y = \frac{\text{stress}}{\text{strain}}$

a smaller strain for the same stress means a larger Y. Therefore Y of steel is very much greater than Y of rubber.

(b) True.

When a coil (spring) is stretched, the wire of the coil is not stretched along its length; it is twisted. This twisting is a shearing deformation, and the restoring force arises from the shear deformation of the material of the wire. Hence the extension of a coil is governed by the modulus of rigidity η of its material and not by the Young's modulus.

Given,

• Diameter of each wire, d = 0.25 cm, so radius r = $\frac{0.25}{2}$ cm = 0.125 × 10^-2 m
• Unloaded length of the steel wire, L_S = 1.5 m
• Unloaded length of the brass wire, L_B = 1.0 m
• Y_S = 2.0 × 10¹¹ N m^-2, Y_B = 0.91 × 10¹¹ N m^-2
• g = 9.8 N/kg

If a wire of length L and radius r increases in length by ΔL under a suspended load Mg, then

$\text Y = \frac{\text{Mg}\text L}{(\pi \text r^2)\Delta \text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{Mg}\text L}{(\pi \text r^2)\text Y}$

Elongation of the steel wire : The steel wire carries both the loads hanging below it, so the suspended load is (4.0 + 6.0) kg,

$(\text{Mg})_S = 10.0 \times 9.8 = 98\ \text N$

$$\begin{aligned}\Delta \text L_S = \frac{98 \times 1.5}{3.14 \times (0.125 \times 10^{-2})^2 \times (2.0 \times 10^{11})} \\[1em] = 1.5 \times 10^{-4}\ \text m\end{aligned}$$

Elongation of the brass wire : The brass wire carries only the lower load of 6.0 kg,

$(\text{Mg})_B = 6.0 \times 9.8 = 58.8\ \text N$

$$\begin{aligned}\Delta \text L_B = \frac{58.8 \times 1.0}{3.14 \times (0.125 \times 10^{-2})^2 \times (0.91 \times 10^{11})} \\[1em] = 1.3 \times 10^{-4}\ \text m\end{aligned}$$

Hence, the elongation of the steel wire is 1.5 × 10^-4 m and that of the brass wire is 1.3 × 10^-4 m.

Given,

• Edge of the cube, L = 10 cm = 0.10 m
• Mass attached, M = 100 kg
• Shear modulus of aluminium, η = 25 GPa = 25 × 10⁹ N m^-2
• g = 10 N/kg

One face of the cube is fixed to the vertical wall and the mass is attached to the opposite face, so a tangential force F = Mg acts vertically downwards along that face. Let x be the vertical deflection of the face.

The area of the face is A = L², and the shearing strain is $\theta = \frac{\text x}{\text L}$. The modulus of rigidity is

$\eta = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{\text F/\text A}{\theta} = \frac{\text{Mg}/\text L^2}{\text x/\text L}$

$\Rightarrow \quad \text x = \frac{\text{Mg}}{\eta\text L}$

Substituting the given values,

$$\begin{aligned}\text x = \frac{100 \times 10}{(25 \times 10^{9}) \times 0.10} \\[1em] = \frac{1000}{2.5 \times 10^{9}} = 4.0 \times 10^{-7}\ \text m\end{aligned}$$

Hence, the vertical deflection of the face is 4.0 × 10^-7 m.

Given,

• Mass of the structure, M = 50,000 kg
• Inner radius of each column, r₁ = 30 cm = 0.30 m
• Outer radius of each column, r₂ = 40 cm = 0.40 m
• Young's modulus of steel, Y = 2.0 × 10¹¹ N m^-2
• g = 9.8 N/kg

The cross-sectional area of steel in each hollow column is

$$\begin{aligned}\text A = \pi \text r_2^2 - \pi \text r_1^2 = \pi(\text r_2^2 - \text r_1^2) \\[1em] = \frac{22}{7}\left[(0.40)^2 - (0.30)^2\right] = \frac{22}{7} \times 0.07 = 0.22\ \text m^2\end{aligned}$$

The weight of the structure is

$\text W = \text{Mg} = 50{,}000 \times 9.8 = 4.9 \times 10^{5}\ \text N$

Since the load is distributed uniformly among the four columns, the compressional force on each column is

$\text F = \frac{4.9 \times 10^{5}}{4} = 1.225 \times 10^{5}\ \text N$

The Young's modulus of steel is

$\text Y = \frac{\text{compressional stress}}{\text{compressional strain}} = \frac{\text F/\text A}{\text{strain}}$

$$\begin{aligned}\Rightarrow \quad \text{compressional strain} = \frac{\text F}{\text{AY}} = \frac{1.225 \times 10^{5}}{0.22 \times (2.0 \times 10^{11})} \\[1em] = 2.78 \times 10^{-6}\end{aligned}$$

Hence, the compressional strain of each column is 2.78 × 10^-6.

Given,

• Cross-section of the copper piece = 15.2 mm × 19.1 mm
• Force applied, F = 44,500 N
• Modulus of rigidity of copper, η = 4.20 × 10¹⁰ N m^-2

The area of cross-section of the body is

$\text A = 15.2 \times 19.1 = 290\ \text{mm}^2 = 2.90 \times 10^{-4}\ \text m^2$

The modulus of rigidity of the material of the body is

$\eta = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{\text F/\text A}{\theta}$

where F is the tangential force applied. Therefore the strain produced is

$\theta = \frac{\text F}{\text A\eta}$

Substituting the given values,

$$\begin{aligned}\theta = \frac{44500}{(2.90 \times 10^{-4}) \times (4.20 \times 10^{10})} \\[1em] = \frac{44500}{1.218 \times 10^{7}} = 3.65 \times 10^{-3}\end{aligned}$$

Hence, the resulting strain is 3.65 × 10^-3.

Given,

• Radius of the steel cable, r = 1.5 cm = 1.5 × 10^-2 m
• Maximum (breaking) stress = 10⁸ N m^-2

The stress in the cable is the load per unit area of cross-section, that is,

$\text{stress} = \frac{\text{load}}{\text{area of cross-section}}$

Therefore the maximum load the cable can support is

$\text{maximum load} = \text{maximum stress} \times \pi \text r^2$

Substituting the given values,

$$\begin{aligned}\text{maximum load} = 10^{8} \times \lbrace3.14 \times (1.5 \times 10^{-2})^2\rbrace \\[1em] = 10^{8} \times (7.065 \times 10^{-4}) = 7.065 \times 10^{4}\ \text N\end{aligned}$$

Hence, the maximum load the cable can support is 7.065 × 10⁴ N.

Given,

• Mass of the rigid bar, M = 15 kg
• Length of each wire, L = 2.0 m
• Y_copper = 1.20 × 10¹¹ N m^-2, Y_iron = 1.90 × 10¹¹ N m^-2
• g = 9.8 N/kg

Since the bar is supported symmetrically by three wires and each wire is to have the same tension,

$\text T = \frac{1}{3}\text{Mg} = \frac{15 \times 9.8}{3} = 49\ \text N$

For a wire of radius r, the Young's modulus of its material is

$\text Y = \frac{\text{Mg}/\pi \text r^2}{\Delta \text L/\text L} = \frac{\text{Mg}\text L}{\pi \text r^2\Delta \text L}$

If D be the diameter of the wire, then $\text r = \frac{\text D}{2}$, and therefore

$\text Y = \frac{4\text{Mg}\text L}{\pi \text D^2\Delta \text L}$

In the given case the length L, the elongation ΔL and the tension $\text T = \frac{1}{3}\text{Mg}$ are the same for all the three wires. Hence

$\text Y \propto \frac{1}{\text D^2} \quad \text{or} \quad \text D \propto \frac{1}{\sqrt{\text Y}}$

Therefore,

$$\begin{aligned}\frac{\text D_{copper}}{\text D_{iron}} = \sqrt{\frac{\text Y_{iron}}{\text Y_{copper}}} = \sqrt{\frac{1.90 \times 10^{11}}{1.20 \times 10^{11}}} \\[1em] = \sqrt{\frac{19}{12}} = 1.258\end{aligned}$$

Hence, the ratio of the diameter of the copper wire to that of the iron wire is 1.258.

Given,

• Mass fastened to the wire, M = 14.5 kg
• Unstretched length of the wire, L = 1.0 m
• Angular velocity, ω = 2 rev/s
• Area of cross-section, A = 0.065 cm² = 0.065 × 10^-4 m²
• Young's modulus of steel, Y = 2.0 × 10¹¹ N m^-2
• g = 9.8 m s^-2

The angular velocity must be expressed in radian per second. Since one revolution corresponds to 2π radian,

$$\begin{aligned}\omega = 2\ \text{rev s}^{-1} = 2 \times 2\pi\ \text{rad s}^{-1} = 4\pi\ \text{rad s}^{-1} \\[1em] \omega^2 = (4\pi)^2 = 157.9\ \text{rad}^2\text s^{-2}\end{aligned}$$

At the lowest point of the vertical circle, the effective force acting on the mass is its weight plus the centrifugal force. Thus,

$\text F = \text{Mg} + \text{ML}\omega^2 = \text M(\text g + \text L\omega^2)$

Substituting the given values,

$$\begin{aligned}\text F = 14.5\left[9.8 + 1.0 \times 157.9\right] \\[1em] = 14.5 \times 167.7 = 2431.8\ \text N\end{aligned}$$

By Hooke's law, the Young's modulus of steel is

$\text Y = \frac{\text{stress}}{\text{strain}} = \frac{\text F/\text A}{\Delta \text L/\text L} = \frac{\text F\text L}{\text A\Delta \text L}$

$\Rightarrow \quad \Delta \text L = \frac{\text F\text L}{\text A\text Y}$

Substituting the values,

$$\begin{aligned}\Delta \text L = \frac{2431.8 \times 1.0}{(0.065 \times 10^{-4}) \times (2.0 \times 10^{11})} \\[1em] = \frac{2431.8}{1.3 \times 10^{6}} = 1.87 \times 10^{-3}\ \text m\end{aligned}$$

Hence, the elongation of the wire at the lowest point is 1.87 × 10^-3 m, that is, 1.87 mm.

Note: The textbook solution substitutes ω = 2 directly into Lω², which treats 2 rev/s as 2 rad s^-1 and gives 1.539 × 10^-4 m. Since 2 rev/s = 4π rad s^-1 and ω occurs squared, the centripetal term is understated by a factor of 4π². The correct elongation is 1.87 × 10^-3 m.

Given,

• Initial volume of water, V = 100.0 litre
• Final volume of water = 100.5 litre
• Increase in pressure, p = 100.0 atm
• 1 atm = 1.013 × 10⁵ Pa

The increase in pressure is

$\text p = 100.0 \times (1.013 \times 10^{5}) = 1.013 \times 10^{7}\ \text{Pa}$

The change in volume is

$\Delta \text V = 100.5 - 100.0 = 0.5\ \text{litre}$

The bulk modulus of a substance is the ratio of the normal stress to the volume strain,

$\text K = \frac{\text{normal stress}}{\text{volume strain}} = \frac{\text p}{\Delta \text V/\text V}$

Substituting the values,

$$\begin{aligned}\text K_{water} = \frac{1.013 \times 10^{7}}{0.5/100.0} \\[1em] = \frac{1.013 \times 10^{7}}{5 \times 10^{-3}} = 2.026 \times 10^{9}\ \text{Pa}\end{aligned}$$

The bulk modulus of air at S.T.P. is K_air = 1.0 × 10^-4 Pa. Therefore,

$\frac{\text K_{water}}{\text K_{air}} = \frac{2.026 \times 10^{9}}{1.0 \times 10^{-4}} = 2.026 \times 10^{13}$

Hence, the bulk modulus of water is 2.026 × 10⁹ Pa and it is about 2.026 × 10¹³ times that of air.

The ratio is very large because the intermolecular forces in air are negligible in comparison with those in water. The molecules of a gas are far apart and hardly attract one another, so a gas can be compressed easily and a small pressure produces a large volume strain in it. In water the molecules are closely packed and strongly attract one another, so the same pressure produces a very small volume strain. A small volume strain for a given normal stress means a large bulk modulus.

Note: The textbook gives the bulk modulus of water as 2.036 × 10⁹ Pa. However, the correct calculation gives 2.026 × 10⁹ Pa. Therefore, the textbook value has a small calculation error.

Given,

• Density of ocean water at the surface, ρ = 1.03 × 10³ kg m^-3
• Pressure at the depth, p = 80.0 atm
• Compressibility of water, B = 45.8 × 10^-11 Pa^-1
• 1 atm = 1.013 × 10⁵ Pa

Compressibility is the reciprocal of the bulk modulus. Hence the bulk modulus of water is

$\text K = \frac{1}{45.8 \times 10^{-11}} = 2.18 \times 10^{9}\ \text{Pa}$

The pressure at the given depth is

$\text p = 80.0 \times (1.013 \times 10^{5}) = 81.04 \times 10^{5}\ \text{Pa}$

Let V and V′ be the volumes of a certain mass M of ocean water at the surface and at the depth respectively, and ρ and ρ′ the corresponding densities. Then

$\text V = \frac{\text M}{\rho} \quad \text{and} \quad \text V' = \frac{\text M}{\rho'}$

The decrease in volume at the depth is

$-\Delta \text V = \text V - \text V' = \text M\left(\frac{1}{\rho} - \frac{1}{\rho'}\right)$

Therefore the volume strain is

$-\frac{\Delta \text V}{\text V} = \frac{\text M}{\text V}\left(\frac{1}{\rho} - \frac{1}{\rho'}\right) = \rho\left(\frac{1}{\rho} - \frac{1}{\rho'}\right) = 1 - \frac{\rho}{\rho'} \qquad \ldots(\text i)$

By definition of the bulk modulus,

$\text K = \frac{\text p}{-\Delta \text V/\text V} \quad \Rightarrow \quad -\frac{\Delta \text V}{\text V} = \frac{\text p}{\text K} \qquad \ldots(\text{ii})$

From equations (i) and (ii),

$1 - \frac{\rho}{\rho'} = \frac{\text p}{\text K} \quad \Rightarrow \quad \rho' = \frac{\rho}{1 - \frac{\text p}{\text K}}$

Substituting the values,

$$\begin{aligned}\rho' = \frac{1.03 \times 10^{3}}{1 - \frac{81.04 \times 10^{5}}{2.18 \times 10^{9}}} = \frac{1.03 \times 10^{3}}{1 - 0.003717} \\[1em] = \frac{1.03 \times 10^{3}}{0.996} = 1.034 \times 10^{3}\ \text{kg m}^{-3}\end{aligned}$$

Hence, the density of ocean water at that depth is 1.034 × 10³ kg m^-3.

Given,

• Hydraulic pressure applied, p = 10 atm = 10 × 1.013 × 10⁵ Pa
• Bulk modulus of glass, K = 37 × 10⁹ Pa

The bulk modulus of a substance is

$\text K = -\frac{\text p}{\Delta \text V/\text V}$

Therefore the fractional change in volume is

$\frac{\Delta \text V}{\text V} = -\frac{\text p}{\text K}$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text V}{\text V} = -\frac{10 \times (1.013 \times 10^{5})}{37 \times 10^{9}} \\[1em] = -\frac{1.013 \times 10^{6}}{3.7 \times 10^{10}} = -2.74 \times 10^{-5}\end{aligned}$$

Hence, the fractional change in volume of the glass slab is − 2.74 × 10^-5.

The negative sign shows that the volume of the slab decreases under the applied pressure.

Given,

• Edge of the copper cube = 10 cm
• Hydraulic pressure applied, p = 7.0 × 10⁶ Pa
• Bulk modulus of copper, K = 1.40 × 10¹¹ Pa

The volume strain produced in the cube is

$$\begin{aligned}\frac{\Delta \text V}{\text V} = -\frac{\text p}{\text K} = -\frac{7.0 \times 10^{6}}{1.40 \times 10^{11}} \\[1em] = -5.0 \times 10^{-5}\end{aligned}$$

The volume of the cube is

$\text V = (10)^3 = 1000\ \text{cm}^3$

Therefore the volume contraction is

$$\begin{aligned}\Delta \text V = -\frac{\text p}{\text K} \times \text V = (-5.0 \times 10^{-5}) \times 1000 \\[1em] = -0.050\ \text{cm}^3\end{aligned}$$

Hence, the volume of the copper cube contracts by 0.050 cm³.

Given,

• Volume of water, V = 1 litre
• Compression required = 0.10%
• Bulk modulus of water, K = 2.2 × 10⁹ Pa

The volume strain is

$\frac{\Delta \text V}{\text V} = -\frac{0.10}{100} = -1.0 \times 10^{-3}$

The bulk modulus of water is

$\text K = -\frac{\text p}{\Delta \text V/\text V} \quad \Rightarrow \quad \text p = \text K\left(-\frac{\Delta \text V}{\text V}\right)$

Substituting the values,

$$\begin{aligned}\text p = (2.2 \times 10^{9}) \times (1.0 \times 10^{-3}) \\[1em] = 2.2 \times 10^{6}\ \text{Pa}\end{aligned}$$

Hence, the pressure on the water should be increased by 2.2 × 10⁶ Pa.

Given,

• Coefficient of linear thermal expansion of steel, α = 10^-5/°C
• Young's modulus of steel, Y = 10¹¹ N/m²
• Radius of the wire, r = 1 mm = 10^-3 m
• Fall in temperature, Δt = 40 − 30 = 10°C
• g = 10 m/s²

When the mass m is hung, the wire is stretched by ΔL. On cooling through Δt, the wire contracts by the same amount and regains its original length L. Hence the longitudinal strain produced by the load is equal to the contraction strain produced by cooling,

$\frac{\Delta \text L}{\text L} = \alpha\Delta \text t$

By Hooke's law, the Young's modulus of the material of the wire is

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \frac{\text{mg}}{\text A} = \text Y(\alpha\Delta \text t)$

Therefore,

$\text m = \frac{\text{AY}(\alpha\Delta \text t)}{\text g} = \frac{\pi \text r^2\text Y(\alpha\Delta \text t)}{\text g}$

Substituting the given values,

$$\begin{aligned}\text m = \frac{\pi (10^{-3})^2 \times 10^{11} \times (10^{-5} \times 10)}{10} \\[1em] = \frac{\pi \times 10^{-6} \times 10^{11} \times 10^{-4}}{10} = \pi \approx 3\end{aligned}$$

Hence, the value of m is about 3 kg.

Given,

• Mass suspended, m = 0.1 kg
• Length of the wire, L = 1 m
• Area of cross-section, A = 4.9 × 10^-7 m²
• Angular frequency, ω = 140 rad s^-1
• Young's modulus, Y = n × 10⁹ N m^-2

The Young's modulus of the material of the wire is

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} = \frac{\text F\text L}{\text A\Delta \text L} \quad \Rightarrow \quad \text F = \left(\frac{\text{YA}}{\text L}\right)\Delta \text L$

When the mass is pulled down through ΔL and released, the restoring force developed in the wire is

$\text F = \text k\Delta \text L$

Comparing the two expressions, the force constant of the wire is

$\text k = \frac{\text{YA}}{\text L}$

The angular frequency of the resulting simple harmonic motion is

$\omega = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{\text{YA}}{\text{mL}}}$

Substituting the given values,

$$\begin{aligned}140 = \sqrt{\frac{(\text n \times 10^{9}) \times (4.9 \times 10^{-7})}{0.1 \times 1}} \\[1em] 140 = \sqrt{4900\text n} = 70\sqrt{\text n}\end{aligned}$$

$\sqrt{\text n} = 2 \quad \Rightarrow \quad \text n = 4$

Hence, the value of n is 4.

2.00

When a force F is applied at the two ends of the composite wire, the same force is transmitted longitudinally to each part. However, the two parts have different areas of cross-section, so the stresses induced in them are different and hence their elongations are different.

Let l₁ be the elongation of the thin wire (length L, radius R) and l₂ that of the thick wire (length 2L, radius 2R). From the relation

$\text Y = \frac{\text F\text L}{\text Al} \quad \Rightarrow \quad l = \frac{\text{FL}}{\text{AY}}$

For the thin wire, A₁ = πR² and length = L,

$l_1 = \frac{\text F\text L}{\pi \text R^2\text Y}$

For the thick wire, A₂ = π(2R)² = 4πR² and length = 2L,

$l_2 = \frac{\text F(2\text L)}{4\pi \text R^2\text Y} = \frac{\text F\text L}{2\pi \text R^2\text Y}$

Therefore,

$\frac{l_1}{l_2} = \frac{\text{FL}}{\pi \text R^2 \text Y} \times \frac{2\pi \text R^2 \text Y}{\text{FL}} = 2$

Hence, the ratio of the elongation in the thin wire to that in the thick wire is 2.00.

2 A Y α ΔT

When the ring cools down, it contracts and presses the two semicircular parts together. Let T be the tension developed in the ring. The free body diagram of one half of the wheel shows that the ring pulls it inward at two places, so the force applied by one part on the other part is 2T.

Let l be the change in the length of the ring when it is heated through ΔT. By the definition of the coefficient of linear expansion,

$\alpha = \frac{\text{change in length}}{\text{initial length} \times \text{rise in temperature}} = \frac{l}{\text L\Delta \text T}$

Therefore the longitudinal strain produced in the ring is

$\frac{l}{\text L} = \alpha\Delta \text T$

The thermal stress developed in the ring is

$\text P_T = \text Y \times \text{strain} = \text Y \alpha\Delta \text T$

But the stress is also the tension per unit area of cross-section, that is, $\text P_T = \frac{\text T}{\text A}$. Hence

$\frac{\text T}{\text A} = \text Y \alpha\Delta \text T \quad \Rightarrow \quad \text T = \text{YA}\alpha\Delta \text T$

Therefore the force applied by one part of the wheel on the other part is

$\text F = \text T + \text T = 2\text T = 2\text{AY}\alpha\Delta \text T$

Hence, the required force is 2 A Y α ΔT.

2880 J/m³

Given,

• Rise in temperature, ΔT = 20 − 0 = 20°C
• Coefficient of linear expansion, α = 12 × 10^-6 per °C
• Young's modulus, Y = 10¹¹ N/m²

The elastic potential energy stored per unit volume of a stretched body is

$\text U_E = \frac{1}{2} \times \text{Young's modulus} \times \text{strain}^2$

Since the rod is prevented from expanding freely, the strain produced in it is

$\frac{\Delta \text L}{\text L} = \alpha \times \Delta \text T = \alpha \times 20$

Substituting the values,

$$\begin{aligned}\text U_E = \frac{1}{2} \times 10^{11} \times (12 \times 10^{-6} \times 20)^2 \\[1em] = \frac{1}{2} \times 10^{11} \times (2.4 \times 10^{-4})^2\end{aligned}$$

$= \frac{1}{2} \times 10^{11} \times (5.76 \times 10^{-8}) = 2880\ \text{J m}^{-3}$

Hence, the energy stored per unit volume of the rod is 2880 J/m³.

1.55 × 10⁵ Pa

Given,

• Initial pressure = 1.01 × 10⁵ Pa
• Final pressure = 1.165 × 10⁵ Pa
• Volume strain, $\frac{\text{dV}}{\text V}$ = − 10% = − 0.1

The bulk modulus of a medium is the ratio of the change in pressure to the volume strain,

$\text B = -\frac{\text{dP}}{\text{dV}/\text V}$

Substituting the values,

$$\begin{aligned}\text B = -\frac{(1.165 - 1.01) \times 10^{5}}{-0.1} \\[1em] = \frac{0.155 \times 10^{5}}{0.1} = 1.55 \times 10^{5}\ \text{Pa}\end{aligned}$$

Hence, the bulk modulus of the medium is 1.55 × 10⁵ Pa.

Given,

• Diameter of the flat face at the narrow end = 0.50 mm, so radius r = 0.25 × 10^-3 m
• Compressional force, F = 50,000 N

The compressional force at each section of the anvil is the same. The pressure at the top of the anvil is the compressional stress there, that is, the force per unit area of the narrow flat face,

$\text{pressure} = \text{compressional stress} = \frac{\text F}{\pi \text r^2}$

Substituting the values,

$$\begin{aligned}\text{pressure} = \frac{50{,}000}{3.14 \times (0.25 \times 10^{-3})^2} \\[1em] = \frac{50{,}000}{1.9625 \times 10^{-7}} = 2.55 \times 10^{11}\ \text{N m}^{-2}\end{aligned}$$

Hence, the pressure at the top of the anvil is 2.55 × 10¹¹ N m^-2.

Given,

• Length of the wire, L = 1.0 m
• Area of cross-section, A = 0.50 × 10^-2 cm² = 5 × 10^-7 m²
• Mass suspended, m = 100 g = 0.1 kg
• Young's modulus of steel, Y = 2 × 10¹¹ N m^-2
• g = 9.8 m s^-2

Let the wire AB be stretched between the pillars and let the mass be suspended at its mid-point C. Let y be the depression at the mid-point, so that the wire takes the shape ACB. Here

$\text{AD} = \text{BD} = \frac{\text L}{2} = 0.50\ \text m$

From the geometry of the figure,

$\text{AC} = \sqrt{(\text{AD})^2 + (\text{CD})^2} = \sqrt{\left(\frac{\text L}{2}\right)^2 + \text y^2} = \frac{\text L}{2}\left[1 + \frac{\text y^2}{(\text L/2)^2}\right]^{\frac{1}{2}}$

Since $\text y \ll \frac{\text L}{2}$, by the binomial theorem

$\text{AC} = \frac{\text L}{2}\left[1 + \frac{\text y^2}{2 \times (\text L/2)^2}\right]$

Therefore the stretched length of the wire is

$\text{ACB} = 2 \times \frac{\text L}{2}\left[1 + \frac{2\text y^2}{\text L^2}\right] = \text L + \frac{2\text y^2}{\text L}$

The increase in length is

$\Delta \text L = \text{ACB} - \text{ADB} = \frac{2\text y^2}{\text L}$

so that the longitudinal strain is

$\frac{\Delta \text L}{\text L} = \frac{2\text y^2}{\text L^2} \qquad \ldots(\text i)$

For the vertical equilibrium of the point C, if T be the tension in each half of the wire and θ the angle each half makes with the vertical,

$2\text T\cos \theta = \text{mg} \quad \Rightarrow \quad \text T = \frac{\text{mg}}{2\cos \theta} \qquad \ldots(\text{ii})$

Thus the stress in the wire is

$\frac{\text T}{\text A} = \frac{\text{mg}}{2\text A\cos \theta}$

From the figure,

$\cos \theta = \frac{\text{CD}}{\text{AC}} = \frac{\text y}{\text L/2} = \frac{2\text y}{\text L}$

Hence the Young's modulus of the wire is

$\text Y = \frac{\text{stress}}{\text{strain}} = \frac{\text{mg}/2\text A\cos \theta}{2\text y^2/\text L^2} = \frac{\text{mg}\text L^2}{4\text A\left(\frac{2\text y}{\text L}\right)\text y^2} = \frac{\text{mg}\text L^3}{8\text{Ay}^3}$

$\Rightarrow \quad \text y^3 = \frac{\text{mg}\text L^3}{8\text A\text Y}$

Substituting the given values,

$$\begin{aligned}\text y^3 = \frac{0.10 \times 9.8 \times (1)^3}{8 \times (5 \times 10^{-7}) \times (2 \times 10^{11})} \\[1em] = \frac{0.98}{8 \times 10^{5}} = 1.225 \times 10^{-6}\end{aligned}$$

$\text y = (1.225)^{1/3} \times 10^{-2} = 1.07 \times 10^{-2}\ \text m$

Hence, the depression at the mid-point of the wire is 1.07 × 10^-2 m.

Given,

• Length of the rod = 1.05 m
• Area of cross-section of the steel wire A, A_A = 1.0 mm²
• Area of cross-section of the aluminium wire B, A_B = 2.0 mm²
• Y_A (steel) = 2 × 10¹¹ N m^-2, Y_B (aluminium) = 7 × 10¹⁰ N m^-2

Let the tensions in the wires A and B be T_A and T_B respectively, and let the mass m be suspended at a point P at a distance x from the wire A.

For the rotational equilibrium of the rod, the moment of the forces about the point P must be zero,

$\text T_A \times \text x = \text T_B \times (1.05 - \text x) \qquad \ldots(\text i)$

(a) For equal stresses in the two wires :

$\frac{\text T_A}{\text A_A} = \frac{\text T_B}{\text A_B} \quad \Rightarrow \quad \frac{\text T_A}{1.0} = \frac{\text T_B}{2.0} \quad \Rightarrow \quad \text T_B = 2\text T_A \qquad \ldots(\text{ii})$

Substituting equation (ii) in equation (i),

$$\begin{aligned}\text T_A \times \text x = 2\text T_A(1.05 - \text x) \\[1em] \text x = 2.10 - 2\text x \quad \Rightarrow \quad 3\text x = 2.10\end{aligned}$$

$\text x = 0.70\ \text m$

Hence, for equal stresses the mass should be suspended at a distance of 0.70 m from the steel wire A.

(b) For equal strains in the two wires :

$(\text{strain})_A = (\text{strain})_B \quad \Rightarrow \quad \frac{(\text{stress})_A}{\text Y_A} = \frac{(\text{stress})_B}{\text Y_B}$

$\frac{(\text{stress})_A}{(\text{stress})_B} = \frac{\text Y_A}{\text Y_B} \quad \Rightarrow \quad \frac{\text T_A/1.0}{\text T_B/2.0} = \frac{2 \times 10^{11}}{7 \times 10^{11}}$

$\Rightarrow \quad \text T_B = 0.70\text T_A$

Substituting this in equation (i),

$$\begin{aligned}\text T_A \times \text x = 0.70\text T_A(1.05 - \text x) \\[1em] \text x = 0.735 - 0.70\text x \quad \Rightarrow \quad 1.70\text x = 0.735\end{aligned}$$

$\text x = \frac{0.70 \times 1.05}{1.70} = 0.43\ \text m$

Hence, for equal strains the mass should be suspended at a distance of 0.43 m from the steel wire A.

Given,

• Number of rivets = 4
• Diameter of each rivet = 6.0 mm, so radius r = 3.0 × 10^-3 m
• Maximum shearing stress = 6.9 × 10⁷ Pa

The maximum tension which each rivet can bear is the maximum shearing stress multiplied by its area of cross-section,

$\text{tension per rivet} = \text{maximum shearing stress} \times \pi \text r^2$

Substituting the values,

$$\begin{aligned}\text{tension per rivet} = (6.9 \times 10^{7}) \times 3.14 \times (3.0 \times 10^{-3})^2 \\[1em] = (6.9 \times 10^{7}) \times (2.826 \times 10^{-5}) = 1.95 \times 10^{3}\ \text N\end{aligned}$$

Since each of the four rivets carries one-quarter of the load, the maximum tension that can be exerted by the riveted strip is

$\text T = 4 \times (1.95 \times 10^{3}) = 7.8 \times 10^{3}\ \text N$

Hence, the maximum tension that can be exerted by the riveted strip is 7.8 × 10³ N.

Given,

• Initial volume of the steel ball, V = 0.32 m³
• Water pressure at the bottom of the trench, p = 1.1 × 10⁸ Pa
• Bulk modulus of steel, K = 160 GPa = 160 × 10⁹ Pa

The bulk modulus of steel is

$\text K = \frac{\text{change in pressure }(\text p)}{\text{volume strain }(-\Delta \text V/\text V)}$

Therefore the volume strain is

$$\begin{aligned}-\frac{\Delta \text V}{\text V} = \frac{\text p}{\text K} = \frac{1.1 \times 10^{8}}{160 \times 10^{9}} \\[1em] = 6.875 \times 10^{-4}\end{aligned}$$

The change in the volume of the ball is

$$\begin{aligned}\Delta \text V = -(6.875 \times 10^{-4}) \times \text V = -(6.875 \times 10^{-4}) \times 0.32 \\[1em] = -2.2 \times 10^{-4}\ \text m^3\end{aligned}$$

Hence, the volume of the ball decreases by 2.2 × 10^-4 m³ when it reaches the bottom.

N/m²

Reason — The restoring force acting per unit area within a body is termed as stress, that is, $\text{stress} = \frac{\text F}{\text A}$. In the S.I. system force is measured in newton (N) and area in metre² (m²), so the unit of stress is N/m², which is also called the pascal (Pa). The newton is the unit of force, N/m of force per unit length and N-m of work.

Young's modulus

Reason — Young's modulus deals with longitudinal stress and longitudinal strain. When a wire or a rod is stretched or compressed along one direction only its length changes, and within the elastic limit the ratio of the longitudinal stress to the longitudinal strain is the Young's modulus of the material. The bulk modulus refers to a change in volume under uniform pressure and the shear (rigidity) modulus to a change in shape under tangential stress.

within elastic limit

Reason — Hooke's law states that within the elastic limit of a material, stress is directly proportional to strain. It is valid only up to the proportionality limit of the material, that is, in the linear part of the stress-strain curve. Beyond the elastic limit permanent deformation occurs and the stress is no longer proportional to the strain, so the law is no longer applicable.

volumetric stress to volumetric strain

Reason — When a uniform pressure is applied all over the surface of a body, the volume of the body changes but its shape remains unchanged. Within the elastic limit, the ratio of the normal stress (pressure) to the volume strain is called the bulk modulus,

$\text K = \frac{\text{normal stress}}{\text{volume strain}} = -\frac{\text{pV}}{\Delta \text V}$

Thus bulk modulus deals with a uniform pressure change causing a change in volume.

ratio of lateral strain to longitudinal strain

Reason — When a material elongates along a certain direction under a tensile stress, it contracts along the perpendicular directions. The fractional change in the direction along which the forces have been applied is the longitudinal strain, and the fractional change in a perpendicular direction is the lateral strain. The ratio of the lateral strain to the longitudinal strain is called the Poisson's ratio σ, which is a constant for the material of the body. Its value theoretically lies between − 1 and 0.5, and for most solid materials it lies between 0.25 and 0.35.

Steel has higher Young's modulus

Reason — Elasticity depends on the modulus of elasticity of a material and not on its softness or hardness. If a steel wire and a rubber wire of the same length and the same area of cross-section are taken, a much larger force has to be applied to the steel wire to produce the same elongation. Hence Y_steel > Y_rubber, and due to its higher Young's modulus steel resists deformation more strongly, that is, steel is more elastic than rubber.

depends on both stress and original length

Reason — The longitudinal strain produced in a stretched wire is

$\text{strain} = \frac{\text{change in length}}{\text{initial length}} = \frac{\Delta \text L}{\text L}$

The elongation ΔL is itself produced by the stress applied, while L is the original length of the wire. Hence the strain depends on both the stress applied and the original length of the wire.

its bulk modulus is very large

Reason — The bulk modulus K quantifies a material's resistance to uniform compression, $\text K = -\frac{\text p}{\Delta \text V/\text V}$. A large bulk modulus means that even under a high pressure the change in volume is very small, which is the defining trait of an incompressible fluid. Liquids typically have a high bulk modulus compared to gases, which is why they are often treated as incompressible. A very small bulk modulus would mean the liquid compresses easily, and the shear modulus relates to a change in shape, not to compressibility.

It relates shear stress and shear strain

Reason — Within the elastic limit, the ratio of the shearing stress to the shearing strain is called the modulus of rigidity of the material of the body,

$\eta = \frac{\text F/\text A}{\theta}$

Thus the modulus of rigidity quantifies a material's resistance to deformation under a shear stress, as when its layers are twisted or made to slide over one another. Option 1 relates to the Young's modulus, option 3 to the bulk modulus, and option 4 to thermal stress.

A is more elastic

Reason — The slope of the linear portion of the stress-strain curve is the ratio of stress to strain, which is the Young's modulus of the material. A steeper slope therefore means a higher Young's modulus. Since a higher modulus of elasticity indicates a higher resistance to elastic deformation, the material A is more elastic than B.

permanent deformation begins

Reason — Up to the elastic limit the body regains its original length when the load is removed, and the curve is retraced back to the origin. Beyond the elastic limit the plastic region starts, in which the strain increases much more rapidly than the stress and the wire does not return to its original length, leaving a permanent set. It is this beginning of permanent deformation that makes the curve non-linear.

one-fourth

Reason — Stress is inversely proportional to the area of cross-section, since $\text{stress} = \frac{\text F}{\text A}$ and F is the same for the two wires, so $\text{stress} \propto \frac{1}{\text A}$.

The area of cross-section is proportional to the square of the diameter. Since the diameter of A is twice that of B, its area is four times that of B. Hence the stress in wire A will be one-fourth of that in wire B.

longitudinal and shear

Reason — The load applied to the spring causes it to extend, leading to an increase in its length. The strain associated with this increase in length is called longitudinal strain. At the same time, as the spring stretches, the individual coils undergo a slight deformation due to the forces acting tangentially to the coils, which leads to a shear strain. Hence both longitudinal and shear strains are produced in a stretched spring.

To allow thermal expansion of material

Reason — The material of a bridge expands when heated and contracts when cooled. If no gap were left, this expansion would be prevented and a large thermal stress, F = Y A α Δt, would develop in the structure, which could cause the bridge to buckle or crack. Gaps are therefore provided so that the material can expand thermally without developing dangerous stresses.

Elastic: temporary deformation; Plastic: permanent deformation

Reason — Elastic deformation is reversible and temporary — when the deforming force is removed the body returns to its original shape and size. Plastic deformation, which occurs beyond the elastic limit, is irreversible — the body retains its modified form even after the load is removed and a permanent set is left in it.

Yes, because steel has higher modulus of elasticity

Reason — In physics, elasticity is quantified by the Young's modulus, that is, the ratio of stress to strain. Steel has a much higher Young's modulus than rubber, so it resists deformation more and returns to its original shape more precisely. Rubber stretches a great deal for a small stress, but in the technical sense this makes it less elastic, since it deforms much more under the same stress. The common belief that stretchiness equals elasticity is a misconception.

To increase load capacity using high tensile strength of steel

Reason — Reinforced concrete combines the compressive strength of concrete with the tensile strength of steel. Tall buildings face large vertical and lateral loads arising from gravity, wind and seismic forces, and steel rods help resist the tensile stresses that concrete alone cannot handle. Thus the reinforcement increases the load-carrying capacity of the structure.

Plastic region

Reason — The plastic region represents permanent deformation — once a material enters this zone, it will not return to its original shape even after the load is removed. In engineering design, materials are used within the elastic region, especially the proportional and Hooke's law regions, where stress and strain are linearly related and the deformation is reversible.

1.0 mm

Reason — Given,

• Length of the wire, L = 2 m
• Area of cross-section, A = 2 mm² = 2 × 10^-6 m²
• Force applied, F = 200 N
• Young's modulus, Y = 2 × 10¹¹ N/m²

The elongation produced in a stretched wire is

$\Delta \text L = \frac{\text{FL}}{\text{AY}}$

Substituting the values,

$$\begin{aligned}\Delta \text L = \frac{200 \times 2}{(2 \times 10^{-6}) \times (2 \times 10^{11})} \\[1em] = \frac{400}{4 \times 10^{5}} = 1.0 \times 10^{-3}\ \text m = 1.0\ \text{mm}\end{aligned}$$

Note: The textbook answer key gives 0.1 mm. However, the correct calculation gives 1.0 × 10⁻³ m, or 1 mm. Therefore, the answer key has a small calculation error.

40 N

Reason — For a wire of given length, area of cross-section and material,

$\text Y = \frac{\text{WL}}{\text A\Delta \text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{WL}}{\text{AY}}$

For the same wire, L, A and Y are constant, so ΔL ∝ W, that is, the extension is directly proportional to the load. Doubling the extension from 1 mm to 2 mm therefore requires the load to be doubled, from 20 N to 40 N.

1 × 10^-3

Reason — Given,

• Original length of the wire, L = 2 m
• Extension at breaking, ΔL = 0.2 cm = 0.002 m

The longitudinal strain is the ratio of the change in length to the original length,

$$\begin{aligned}\text{strain} = \frac{\Delta \text L}{\text L} = \frac{0.002}{2} \\[1em] = 1 \times 10^{-3}\end{aligned}$$

Note: The textbook answer key gives 1 × 10⁻⁴. However, 0.002 ÷ 2 = 1 × 10⁻³. Therefore, the answer key has a small calculation error.

1.2 mm

Reason — Given,

• Length of the rod, L = 2 m
• Rise in temperature, ΔT = 50°C
• Coefficient of linear expansion, α = 1.2 × 10^-5/°C

The increase in length on heating is

$$\begin{aligned}\Delta \text L = \alpha\text L\Delta \text T = (1.2 \times 10^{-5}) \times 2 \times 50 \\[1em] = 0.0012\ \text m = 1.2\ \text{mm}\end{aligned}$$

Copper

Reason — The elongation of a loaded wire is $\Delta \text L = \frac{\text{FL}}{\text{AY}}$. In the given condition F, L and A are the same for both wires, so $\Delta \text L \propto \frac{1}{\text Y}$. Copper has a smaller Young's modulus than steel, and therefore copper elongates more.

Thinner rod

Reason — The elongation of a stretched rod is $\Delta \text L = \frac{\text{FL}}{\text{AY}}$. Here F, L and Y are the same for both rods, so $\Delta \text L \propto \frac{1}{\text A}$, that is, a smaller area of cross-section produces a larger elongation. Hence the thinner rod elongates more.

Moderate elasticity + high breaking stress

Reason — In mountain climbing, the rope is designed with moderate elasticity so that it can stretch a little and absorb the shock if a climber falls, preventing a sudden jerk that could cause injury. At the same time it must have a high breaking stress so that it can bear very large forces without breaking. Together, these two properties make the rope both safe and reliable.

increase three-fold

Reason — Stress is the force per unit area. When the wire is stretched to three times its original length, the volume of the wire remains the same, so its area of cross-section becomes one-third of the original.

For the two states,

$(\text{stress})_1 = \frac{\text F}{\text A_1}, \qquad (\text{stress})_2 = \frac{\text F}{\text A_2} \quad \Rightarrow \quad \frac{(\text{stress})_2}{(\text{stress})_1} = \frac{\text A_1}{\text A_2}$

Since the volume is conserved, A₁l₁ = A₂l₂, so

$\frac{\text A_1}{\text A_2} = \frac{l_2}{l_1} = \frac{3l_1}{l_1} = 3$

Therefore (stress)₂ = 3 × (stress)₁, that is, the stress increases three-fold.

Note: The question is incomplete because the information that the wire is stretched to three times its original length is missing. The solution above follows the textbook’s explanation and gives a three-fold increase. However, the printed answer key gives a six-fold increase.

Bulk modulus reflects resistance to uniform compression, which dominates in deep-sea environments, whereas Young's modulus relates to uniaxial stress.

Reason — A submarine deep underwater is subjected to hydrostatic pressure acting equally in all directions, which is a volumetric (bulk) stress. The bulk modulus measures how incompressible a material is when pressure is applied uniformly from all directions, and is therefore the key property in such conditions. The Young's modulus, while important for structural stiffness under tension or compression along one axis, does not describe the volumetric compressibility that dominates in submerged environments.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The maximum deforming force upto which a body retains its property of elasticity is called the limit of elasticity. Upto this limit the body returns to its original state when the deforming force is removed, so it is indeed the maximum extent to which a solid can be stretched without permanent deformation.

Reason (R) is also correct: Beyond the elastic limit the material will not return to its original shape and Hooke's law is no longer applicable, since the strain increases much more rapidly than the stress.

The failure of Hooke's law beyond this point is exactly what marks the onset of permanent deformation, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Young's modulus is the ratio of the longitudinal stress to the longitudinal strain within the elastic limit, and a high value of Y means a small elastic strain for a given stress. It is therefore a measure of the stiffness of the material.

Reason (R) is false: A material with a higher Young's modulus is stiffer and deforms less under a given load, not more. For example, steel has a much higher Y than rubber and stretches far less under the same stress.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: This is the statement of Hooke's law — within the elastic limit of a material, stress is directly proportional to strain.

Reason (R) is false: The constant of proportionality between stress and strain within the elastic limit is called the modulus of elasticity (for a stretched wire, the Young's modulus), not the Poisson's ratio. The Poisson's ratio is the ratio of the lateral strain to the longitudinal strain.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a uniform pressure is applied all over the surface of a body, its volume changes while its shape remains unchanged. The bulk modulus is the ratio of the normal stress to the volume strain, and so measures the resistance of the material to uniform compression.

Reason (R) is also correct: The reciprocal of the bulk modulus of the material of a body is called the compressibility of that material, $\text B = \frac{1}{\text K}$.

Since compressibility is the reciprocal of K, a large bulk modulus means a small compressibility, that is, a high resistance to uniform compression. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Within the elastic limit, the ratio of the shearing stress to the shearing strain is called the modulus of rigidity or shear modulus of the material of the body, $\eta = \frac{\text F/\text A}{\theta}$.

Reason (R) is false: Shear stress involves the force applied parallel (tangential) to a surface area, not a change in length per unit area. The tangential force acting per unit area of the surface is the shearing stress, and it produces a change in shape without any change in volume.

Therefore, assertion is true but reason is false.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Hooke's law is obeyed only so long as stress remains directly proportional to strain, that is, only in the straight-line portion OA of the stress-strain curve (Fig. 3). The point A is the limit of proportionality and it lies below the elastic limit B. In the region AB the wire is still elastic, since it regains its original length completely when the load is removed, but the graph is already curved and stress is no longer proportional to strain. Hooke's law therefore fails inside the elastic region itself, and the elastic limit is not the limit up to which the law is valid.

Reason (R) is correct: Beyond the elastic limit the deformation becomes permanent. The material behaves plastically and does not return to its original shape when the deforming load is removed.

Therefore, assertion is false but reason is true.

Note: The printed answer key gives option (a). Option (a) requires the limit of proportionality and the elastic limit to be the same point, whereas the textbook's own Fig. 3 marks A as the Proportional Limit and B as the Elastic Limit, and the Hot Info box on the same page states that Hooke's law is valid up to the proportionality limit, that is, in the linear part of the stress-strain curve.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: High elasticity does not necessarily mean a high breaking stress; these are different properties. Elasticity is measured by the modulus of elasticity, which describes how stiffly a material resists deformation, whereas the breaking stress describes how much stress it can bear before it fractures. Glass, for example, has a high Young's modulus but breaks easily.

Reason (R) is correct: The breaking stress is indeed the maximum stress that a material can withstand before breaking, and corresponds to the fracture point of the stress-strain curve.

Therefore, assertion is false but reason is true.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: For a given material the Young's modulus is greater than the modulus of rigidity, since a material offers greater resistance to a change in length than to a change in shape.

Reason (R) is false: The Young's modulus and the shear modulus are not independent of each other; they are related through the Poisson's ratio of the material.

Therefore, assertion is true but reason is false.

Note: The textbook's printed answer key marks option 2 for this question, but its own Hints section states that the Young's modulus and the shear modulus are related by the Poisson's ratio, which makes the Reason false. The answer above follows the textbook's explanation.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The modulus of elasticity is the ratio of stress to the corresponding strain within the elastic limit, and therefore describes how a material deforms elastically under a given stress.

Reason (R) is false: A higher modulus of elasticity indicates a higher resistance to elastic deformation, not a lower one, since a larger stress is then needed to produce a given strain.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Ductile materials such as copper, aluminium, silver and iron have a large plastic range of extension, so in a tensile test their yield point and fracture point are well separated and a large plastic deformation occurs before fracture.

Reason (R) is false: Ductility is related to the ability of a material to undergo plastic deformation, and not necessarily to its modulus of elasticity. A material may have a high modulus of elasticity and yet be brittle.

Therefore, assertion is true but reason is false.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Stress depends on the force applied and on the area of cross-section of the wire, and not on the length of the wire. Hence the stress is not inversely proportional to the length.

Reason (R) is correct: Stress is indeed defined as the restoring force acting per unit area, $\text{stress} = \frac{\text F}{\text A}$.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The thicker rod has a larger area of cross-section, leading to less elongation under the same applied force.

Reason (R) is also correct: The elongation of a loaded rod is $\Delta \text L = \frac{\text{FL}}{\text{AY}}$, so for the same F, L and Y the elongation is inversely proportional to the area of cross-section.

Since a greater thickness means a greater area of cross-section, and the elongation varies inversely as the area, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a spring is stretched, work has to be done against the restoring force. This work is stored in the spring as its strain (elastic potential) energy, so the two are equal.

Reason (R) is also correct: The strain energy stored in a spring stretched through x is indeed $\text U = \frac{1}{2}\text{kx}^2$, where k is the spring constant.

Since the work done in stretching the spring through x works out to exactly $\frac{1}{2}\text{kx}^2$, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: A material with a high Young's modulus is stiff, and therefore requires a large force for a significant stretching, not a small one.

Reason (R) is correct: The Young's modulus is directly related to the stiffness of a material — the greater the value of Y, the stiffer the material.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a given material the modulus of rigidity is indeed less than the Young's modulus, since a material resists a change in shape less strongly than a change in length.

Reason (R) is also correct: The modulus of rigidity and the Young's modulus of a material are related to each other through its Poisson's ratio.

It is this relation, involving the Poisson's ratio, that fixes the modulus of rigidity at a value smaller than the Young's modulus. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The ratio of the lateral strain to the longitudinal strain is called the Poisson's ratio of the material of the body, and it is a constant for that material.

Reason (R) is also correct: Poisson's ratio defines how much a material changes laterally when it is stretched longitudinally, so a high value of σ means a larger lateral change for the same longitudinal strain.

Since σ is defined as the ratio of the two strains, a large σ directly implies a larger lateral change accompanying a given longitudinal strain. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The coefficient of linear expansion is defined as the change in length per unit original length per degree rise in temperature, and therefore measures how much a material expands when heated.

Reason (R) is also correct: Materials expand when heated and contract when cooled, and in both cases the amount of expansion or contraction depends on the thermal expansion coefficient. A material with a higher coefficient contracts more when cooled.

Since the same coefficient governs both the expansion and the contraction, the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: The elastic region extends from the origin O up to the elastic limit B (Fig. 3). Only the portion OA, up to the limit of proportionality, is a straight line. Between A and B the material is still elastic, since it recovers fully when the load is removed, but the graph is curved. The slope there changes from point to point and does not represent the Young's modulus. Hence the slope represents the Young's modulus only in the linear portion OA, not throughout the elastic region.

Reason (R) is correct: The Young's modulus is defined as the ratio of the longitudinal stress to the corresponding longitudinal strain within the elastic limit.

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}$

Therefore, assertion is false but reason is true.

Note: The printed answer key gives option (a). Option (a) would require the whole elastic region to be a straight line, which contradicts Fig. 3, where the curve departs from the straight line at A while remaining elastic up to B.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In a brittle material such as cast iron, glass or ceramic, the yield point and the fracture point are very close together, so it breaks suddenly with little or no noticeable plastic deformation.

Reason (R) is false: Brittle materials fracture with little plastic deformation, but this is not necessarily due to a low breaking stress. Many brittle materials can withstand a high stress before breaking; what they lack is the plastic range, not the strength.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For most metals, increasing the temperature decreases the modulus of elasticity, which is why metals become softer on heating and are easier to forge and shape.

Reason (R) is also correct: At higher temperatures the atoms vibrate more vigorously, that is, their internal energy increases and they can move more freely.

Since the more vigorous atomic vibrations make the interatomic bonds less effective in resisting deformation, the material offers less resistance and its elastic modulus falls. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

When a body is acted upon by an external (deforming) force, the intermolecular separation is affected and internal forces arise which tend to restore the body to its original shape and size. These forces are called restoring forces.

The restoring force acting per unit area within a body is termed as stress.

$\text{stress} = \frac{\text F}{\text A}$

Its S.I. unit is the pascal (Pa) and its dimensional formula is [M L^-1 T^-2].

The maximum stress that a material can withstand before it breaks is called its breaking stress.

It corresponds to the fracture point of the stress-strain curve, and is also known as the ultimate tensile strength of the material. If the stress in a wire exceeds this value, the wire snaps.

Hooke's law states that within the elastic limit of a material, stress is directly proportional to strain.

$\text{stress} \propto \text{strain} \quad \Rightarrow \quad \text{stress} = \text E \times \text{strain}$

where E is a constant of proportionality called the modulus of elasticity of the material. Hooke's law is valid only upto the proportionality limit, that is, in the linear part of the stress-strain curve.

The work done in stretching a wire through a length x, in terms of the force-constant k, is

$\text W = \frac{1}{2}\text{kx}^2$

This work remains stored in the wire in the form of elastic potential energy.

Shearing stress : When two equal and opposite forces act along the tangents to the surfaces of two opposite faces of a body, the tangential force acting per unit area of the surface is called the shearing stress.

$\text{shearing stress} = \frac{\text{tangential force }(\text F)}{\text{area }(\text A)}$

Shearing strain : The angular deformation produced when a body is subjected to a tangential stress, that is, the angle θ through which a face of the body originally perpendicular to the fixed face is turned, is called the shearing strain.

$\theta = \frac{\text{displacement of the upper surface}}{\text{distance of the upper surface from the fixed surface}}$

Within the elastic limit upto which Hooke's law is applicable, the ratio of the shearing stress to the shearing strain is called the modulus of rigidity of the material of the body. It is denoted by η.

$\eta = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{\text F/\text A}{\theta} = \frac{\text F}{\text A\theta}$

Its S.I. unit is newton/metre² (N m^-2) or pascal (Pa).

Within the elastic limit upto which Hooke's law is applicable, the ratio of the normal stress to the volume strain is called the bulk modulus of the material of the body. It is denoted by K.

$\text K = \frac{\text{normal stress}}{\text{volume strain}} = \frac{\text p}{-\Delta \text V/\text V} = -\frac{\text{pV}}{\Delta \text V}$

Its S.I. unit is newton/metre² (N m^-2) or pascal (Pa).

The Young's modulus of elasticity is

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} = \frac{\text F/\text A}{\Delta \text L/\text L}$

The longitudinal strain is a ratio of two lengths and is therefore a pure number having no unit. Hence the unit of Y is the same as the unit of stress, that is, the unit of force divided by the unit of area,

$\text{unit of Y} = \frac{\text{newton}}{\text{metre}^2}$

Hence, the S.I. unit of Young's modulus is newton/metre² (N m^-2) or pascal (Pa).

In stretching a metal wire the interatomic separation is increased, so work has to be done against the inter-atomic forces of attraction between the atoms of the wire.

This work is not lost. It remains stored in the stretched wire in the form of elastic potential energy, which is recovered when the deforming force is removed and the wire returns to its original length.

The elastic potential energy stored in a stretched wire is equal to the work done in stretching it. Since the stretching force increases uniformly from zero to F as the extension increases from zero to x, the average force is $\frac{\text F}{2}$, and therefore

$\text U = \text{average force} \times \text{extension} = \frac{1}{2}\text F\text x$

Hence, the potential energy of the stretched wire is $\frac{1}{2}$ F x.

The reciprocal of the bulk modulus of the material of a body is called the compressibility of that material.

$\text B = \frac{1}{\text K}$

Its S.I. unit is metre²/newton.

Gas is the most compressible.

The compressibility of gases is large, while that of liquids and solids is comparatively very small. This is because the molecules of a gas are far apart and the intermolecular forces between them are negligible, so a given pressure produces a large volume strain in a gas.

A steel spring is stretched to a smaller extent than a copper spring under the same deforming force, because the Young's modulus of steel is greater than that of copper. Moreover, steel recovers its original state quicker than copper after the deforming force is removed.

Hence, springs are made of steel and not of copper.

The Young's modulus of a perfectly rigid body is infinite.

In a perfectly rigid body there is no change in shape or size on applying an external force, so the strain produced is zero. Since $\text Y = \frac{\text{stress}}{\text{strain}}$, a zero strain makes Y infinitely large.

The Young's modulus of a material decreases with a rise in temperature.

At a higher temperature the atoms vibrate more vigorously and the interatomic bonds become less effective in resisting deformation. Consequently the material becomes easier to stretch or compress, that is, its modulus of elasticity falls. This is why metals become softer on heating.

The ratio stress/strain will decrease.

When the deforming force is applied beyond the elastic limit, the amount of strain produced is more than that observed within the elastic limit for the same stress. Since the strain in the denominator becomes larger, the ratio of stress to strain decreases.

The breaking force is the breaking stress multiplied by the area of cross-section, so for a given material the breaking force is directly proportional to the area of cross-section.

(i) Two parallel wires of this size : The total area of cross-section becomes twice that of a single wire, so the breaking force is 2 F.

(ii) A single wire of double thickness : If the thickness (diameter) is doubled, the area of cross-section becomes four times, so the breaking force is 4 F.

Given,

• $\text L_1 = \frac{\text L_2}{2}$ and D₁ = 2D₂
• Equal loads Mg are applied on both the wires and the metal is the same, so Y is the same

The increase in length of a loaded wire is

$\Delta \text L = \frac{\text{Mg}\text L}{\text A\text Y} = \frac{4\text{Mg}\text L}{\pi \text D^2\text Y}$

Since Mg and Y are the same for both wires,

$$\begin{aligned}\frac{\Delta \text L_1}{\Delta \text L_2} = \frac{\text L_1}{\text L_2} \times \frac{\text D_2^2}{\text D_1^2} = \frac{1}{2} \times \frac{\text D_2^2}{(2\text D_2)^2} \\[1em] = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\end{aligned}$$

Hence, the ratio of the increase in their lengths is 1 : 8.

For wires of the same area of cross-section producing the same longitudinal strain,

$\text Y = \frac{\text F/\text A}{\text{strain}} \quad \Rightarrow \quad \text Y \propto \text F$

Hence the material requiring a greater force is more elastic. Arranging the given forces in increasing order — 690 N, 900 N, 1100 N and 2000 N — the corresponding materials are aluminium, brass, copper and steel.

Hence, in the order of increasing elasticity the materials are aluminium, brass, copper, steel.

Given,

• Y₁ (steel) = 2.0 × 10¹¹ N/m²
• Y₂ (copper) = 1.2 × 10¹¹ N/m²

The increase in length of a loaded wire is $\Delta \text L = \frac{\text{Mg}\text L}{\text A\text Y}$.

(i) Here Mg, L and A are the same for both wires, so $\Delta \text L \propto \frac{1}{\text Y}$,

$\frac{\Delta \text L_1}{\Delta \text L_2} = \frac{\text Y_2}{\text Y_1} = \frac{1.2 \times 10^{11}}{2.0 \times 10^{11}} = \frac{3}{5}$

Hence, the increase in lengths of steel and copper wires are in the ratio 3 : 5.

(ii) Here Mg, A and ΔL are the same for both wires, so L ∝ Y,

$\frac{\text L_1}{\text L_2} = \frac{\text Y_1}{\text Y_2} = \frac{2.0 \times 10^{11}}{1.2 \times 10^{11}} = \frac{5}{3}$

Hence, the initial lengths of steel and copper wires are in the ratio 5 : 3.

The increase in length of a loaded wire is $\Delta \text L = \frac{\text{Mg}\text L}{\text A\text Y}$, and the maximum load it can bear is (breaking stress × A).

(i) For a given load, ΔL ∝ L. When the length is cut to half, the increase in length is also halved.

Hence, the increase in length will be reduced to half.

(ii) The maximum load depends on the breaking stress of the material and on the area of cross-section, neither of which changes on cutting the wire.

Hence, there will be no effect on the maximum load which the wire can bear.

The part CDE of the graph indicates the extension beyond the elastic limit of the wire.

Upto the point C the graph is a straight line, showing that Hooke's law is obeyed and the extension is proportional to the load. Beyond C the graph becomes curved, which shows that the extension is no longer proportional to the load, that is, Hooke's law is not obeyed. This curvature of CDE is the evidence that the wire has been extended beyond its elastic limit and has entered the plastic region.

Point A — limit of proportionality : Upto the point A the graph is a straight line, so the stress is directly proportional to the strain and Hooke's law is obeyed. A is therefore called the limit of proportionality.

Point B — elastic limit : Between A and B the curve deviates from the straight line, so the stress is no longer exactly proportional to the strain, but the wire still regains its original length when the load is removed. B is therefore the elastic limit of the wire.

Point C — yield point : Beyond B the wire enters the plastic region. On removing the load at C, the wire does not return to its original length and a permanent extension is left in it. C is therefore called the yield point.

The work done in stretching a wire through x is

$\text W = \frac{1}{2}\frac{\text{YA}}{\text L}\text x^2$

The springs are identical, so A and L are equal, and they are equally stretched, so x is the same. Hence W ∝ Y.

Since the Young's modulus of steel is greater than that of copper, more work has to be done on the steel spring.

Hence, more work will have to be done on the steel spring.

Given,

• Young's modulus of the material, Y = 2.5 × 10¹² N/m²
• Original length of the wire, L = 1 m
• Final length = 2 m, so the increase in length ΔL = 1 m

To double the length of the wire, the longitudinal strain required is

$\frac{\Delta \text L}{\text L} = \frac{1}{1} = 1$

From the definition of the Young's modulus,

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \text F = \text{YA}\left(\frac{\Delta \text L}{\text L}\right) = \text{YA}$

Hence, the force required is (2.5 × 10¹²) × A newton, where A is the area of cross-section of the wire.

Given,

• Young's modulus of the material, Y = 20 × 10¹² N/m²
• Side of the cube, L = 1 m, so the area of cross-section A = L² = 1 m²
• To double the length, ΔL = 1 m

The longitudinal strain required is

$\frac{\Delta \text L}{\text L} = \frac{1}{1} = 1$

Therefore the force required is

$$\begin{aligned}\text F = \text{YA}\left(\frac{\Delta \text L}{\text L}\right) = (20 \times 10^{12}) \times 1 \times 1 \\[1em] = 20 \times 10^{12}\ \text N\end{aligned}$$

Hence, the force required is 20 × 10¹² N.

Given,

• Increase in the length of the wire = 1% of its original length

The linear (longitudinal) strain is the ratio of the change in length to the original length,

$$\begin{aligned}\text{linear strain} = \frac{\Delta \text L}{\text L} = \frac{1}{100} \\[1em] = 0.01\end{aligned}$$

Hence, the linear strain produced in the wire is 0.01.

Given,

• Increase in length with the original wire, ΔL₁ = 8 mm
• The radius is doubled, so r₂ = 2r₁

The increase in length of a loaded wire is

$\Delta \text L = \frac{\text{Mg}\text L}{\pi \text r^2\text Y}$

Since Mg, L and Y remain unchanged, $\Delta \text L \propto \frac{1}{\text r^2}$. Therefore

$\frac{\Delta \text L_2}{\Delta \text L_1} = \frac{\text r_1^2}{\text r_2^2} = \frac{\text r_1^2}{(2\text r_1)^2} = \frac{1}{4}$

$\Delta \text L_2 = \frac{8}{4} = 2\ \text{mm}$

Hence, the change in length of the wire will be 2 mm.

The ratio of the tensions developed in the two wires is 1 : 1.

The tension developed in a stretched wire is equal to the force applied along its length. Since equal forces F, F are applied on both the wires, the tensions developed in them are also equal, irrespective of their lengths.

Given,

• First wire : length L, radius 2r
• Second wire : length 2L, radius r
• Equal weights Mg are applied and the material is the same, so Y is the same

The Young's modulus of the material of a wire is

$\text Y = \frac{\text{Mg}\text L}{\pi \text r^2\Delta \text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{Mg}\text L}{\pi \text r^2\text Y}$

Therefore, for the two wires,

$$\begin{aligned}\frac{\Delta \text L_1}{\Delta \text L_2} = \frac{\text L/(2\text r)^2}{2\text L/\text r^2} = \frac{\text L}{4\text r^2} \times \frac{\text r^2}{2\text L} \\[1em] = \frac{1}{8}\end{aligned}$$

Hence, the ratio of the elongations produced in the two wires is 1 : 8.

Given,

• $\text Y_{brass} = \frac{1}{2}\text Y_{iron}$
• The lengths of the two wires are equal and equal stress is applied on both

For a given stress and a given length,

$\text Y = \frac{\text{stress}}{\Delta \text L/\text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{stress} \times \text L}{\text Y}$

that is, $\Delta \text L \propto \frac{1}{\text Y}$. Therefore

$\frac{\Delta \text L_{brass}}{\Delta \text L_{iron}} = \frac{\text Y_{iron}}{\text Y_{brass}} = \frac{\text Y_{iron}}{\frac{1}{2}\text Y_{iron}} = \frac{2}{1}$

Hence, the ratio of the increase in lengths of the brass and the iron wires is 2 : 1.

Given,

• Y₁ (steel) = 2.0 × 10¹¹ N/m²
• Y₂ (copper) = 1.2 × 10¹¹ N/m²

The increase in length of a loaded wire is $\Delta \text L = \frac{\text{Mg}\text L}{\text A\text Y}$.

(i) Here Mg, L and A are the same, so $\Delta \text L \propto \frac{1}{\text Y}$,

$\frac{\Delta \text L_1}{\Delta \text L_2} = \frac{\text Y_2}{\text Y_1} = \frac{1.2 \times 10^{11}}{2.0 \times 10^{11}} = \frac{3}{5}$

Hence, the increase in lengths of the steel and the copper wires are in the ratio 3 : 5.

(ii) Here Mg, A and ΔL are the same, so L ∝ Y,

$\frac{\text L_1}{\text L_2} = \frac{\text Y_1}{\text Y_2} = \frac{2.0 \times 10^{11}}{1.2 \times 10^{11}} = \frac{5}{3}$

Hence, the initial lengths of the steel and the copper wires are in the ratio 5 : 3.

The atoms of a solid are held together in a regular array by electric forces, in a way as if they were connected by springs. Under these forces the solid remains in its natural equilibrium state, in which the distribution of positive and negative atomic charges is such that there is no net force between the atoms. The intermolecular separation corresponding to this state is called the equilibrium distance r₀.

When the solid is compressed : The distance between the atoms decreases and the distribution of charges changes in such a way that a net force of repulsion begins to act between them. This is called the interatomic force. When the external force is removed, this force pushes the atoms back to their initial positions so that the solid returns to its original size.

When the solid is stretched : The interatomic space increases and the distribution of charges changes so that there is a net force of attraction between the atoms. When the external force is removed, this force of attraction brings the atoms close to each other, back to their initial positions.

Thus the property of elasticity arises from the interatomic forces which tend to restore the equilibrium distance between the atoms. If the applied force is very large, the atoms move so far apart that the force of attraction between them becomes negligible; there is then a permanent dislocation of the atoms and, on removing the force, the solid does not return to its original size.

Let A be the area of cross-section of the rope. The weight of the rope is

$\text F = \text V \times \text{density} \times \text g = \text{AL}\rho\text g$

This weight acts at the centre of gravity of the rope, which lies at a distance $\frac{\text L}{2}$ from the rigid support. Hence the effective length over which the extension is produced is $\frac{\text L}{2}$.

The Young's modulus of the material of the rope is

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} = \frac{\text F/\text A}{\Delta \text L\Big/\frac{\text L}{2}}$

Substituting F = ALρg,

$\text Y = \frac{\text{AL}\rho \text g/\text A}{\Delta \text L\Big/\frac{\text L}{2}} = \frac{\text L\rho \text g\text L}{2\Delta \text L}$

$\Rightarrow \quad \Delta \text L = \frac{\text L^2\rho \text g}{2\text Y}$

Hence, the increase in length of the rope due to its own weight is $\frac{\text L^2\rho \text g}{2\text Y}$.

The force F acting on the free surface makes an angle θ with the vertical. It is resolved into two components — a component F cos θ normal to the free surface and a component F sin θ along (tangential to) the free surface.

(i) Longitudinal stress : This is produced by the normal component of the force,

$\text{longitudinal stress} = \frac{\text F_\perp}{\text A} = \frac{\text F\cos \theta}{\text A}$

(ii) Shear stress : This is produced by the tangential component of the force,

$\text{shear stress} = \frac{\text F_\parallel}{\text A} = \frac{\text F\sin \theta}{\text A}$

The work done in stretching a spring through x is

$\text W = \text U = \frac{1}{2}\text{kx}^2 \quad \text{and} \quad \text f = \text{kx}\ \text{(magnitude only)}$

(i) Stretched by equal force : Putting $\text x = \frac{\text f}{\text k}$,

$\text U = \frac{1}{2}\text k\left(\frac{\text f}{\text k}\right)^2 = \frac{\text f^2}{2\text k}$

For an equal value of the force, $\text W \propto \frac{1}{\text k}$.

Hence, for equal forces a greater amount of work will be done in stretching the spring of smaller force constant.

(ii) Stretched by equal distance : For an equal value of x, W ∝ k.

Hence, for equal extensions more work will be done in stretching the spring of larger force constant.

In the decreasing order of their modulus of elasticity the materials are : steel, copper, glass, rubber.

Steel has the highest Young's modulus of the four and therefore resists deformation most strongly, while rubber has the lowest and stretches a great deal even under a small stress.

(i) For a given load, the increase in length of a wire is $\Delta \text L = \frac{\text{Mg}\text L}{\text A\text Y}$, that is, ΔL ∝ L. On cutting the wire to half its length, the increase in length is also halved.

Hence, the increase in length will be reduced to half.

(ii) The maximum load a wire can bear is (breaking stress × area of cross-section). Cutting the wire changes neither the breaking stress of the material nor the area of cross-section.

Hence, there will be no effect on the maximum load which the wire can bear.

(i) On suspending the same load on wires of the same length and the same material,

$\text A\Delta \text L = \text{constant}$

If the diameter is doubled, the radius becomes 2r and

$\pi (2\text r)^2 \times \Delta \text L' = \pi \text r^2 \times \Delta \text L \quad \Rightarrow \quad \Delta \text L' = \frac{\Delta \text L}{4}$

Hence, the increase in length will be reduced to one-fourth.

(ii) The maximum load is proportional to the area of cross-section. Since the diameter is doubled, the area becomes four times,

$\text Y = \frac{\text{WL}}{(\pi \text r^2) \times \Delta \text L} = \frac{\text W'\text L}{\pi (2\text r)^2 \times \Delta \text L} \quad \Rightarrow \quad \text W' = 4\text W$

Hence, the maximum bearable load will become four times.

When a material elongates along a certain direction under a tensile stress, it contracts along the perpendicular directions. The fractional change in the direction along which the forces have been applied is called the longitudinal strain, while the fractional change in a perpendicular direction is called the lateral strain.

The ratio of the lateral strain to the longitudinal strain is called the Poisson's ratio. It is a constant for the material of the body and is denoted by σ.

For a wire of original length L and diameter D,

$\sigma = \frac{\Delta \text D/\text D}{\Delta \text L/\text L} = \frac{\Delta \text D}{\Delta \text L} \times \frac{\text L}{\text D}$

Poisson's ratio has no unit and no dimensions. Theoretically its value lies between − 1 and 0.5, but for most solid materials it lies between 0.25 and 0.35.

(a) The curves of the two rubbers differ from the corresponding curve of a metal in the following respects :

(i) Hooke's law is not obeyed even for small stresses, so there is no initial straight portion in the curve.

(ii) There is no permanent strain even for large stresses, that is, the rubber returns to its original length.

(iii) The same curve is not retraced during unloading, so a closed loop is formed. This is the presence of elastic hysteresis.

(b) Rubber B should be used as the shock-absorber sheet. The area of the hysteresis loop for rubber B is larger than that for A. This means that rubber B dissipates a larger amount of energy as heat in each cycle of loading and unloading, and so it will dissipate a larger amount of the vibrational energy of the machine.

(c) Rubber A should be used for making a car tyre. Since the area of its hysteresis loop is smaller, it will dissipate a smaller amount of kinetic energy as heat and will not get heated too much. This will decrease the wear and tear of the tyre.

The work done in stretching the wire is equal to the elastic potential energy stored in it,

$\text W = \frac{1}{2} \times \text Y \times (\text{strain})^2 \times \text{volume}$

Here the strain is $\frac{\text x}{\text L}$ and the volume of the wire is AL. Substituting these,

$$\begin{aligned}\text W = \frac{1}{2} \times \text Y \times \left(\frac{\text x}{\text L}\right)^2 \times (\text{AL}) \\[1em] = \frac{\text{YA}\text x^2}{2\text L}\end{aligned}$$

Hence, the work done in stretching the wire by an amount x is $\frac{\text{YA}\text x^2}{2\text L}$.

The elastic potential energy stored in the stretched wire is

$\text U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

Since stress = Y × strain = Yα, we have

$\text U = \frac{1}{2}\text Y \alpha^2 \times \text{volume of the wire}$

The energy density is the elastic potential energy per unit volume of the wire,

$\text u = \frac{\text{elastic P.E.}}{\text{volume of wire}} = \frac{1}{2}\text Y \alpha^2$

Therefore,

$\frac{\text{elastic P.E.}}{\text{energy density}} = \frac{\frac{1}{2}\text Y \alpha^2 \times \text{volume}}{\frac{1}{2}\text Y \alpha^2} = \text{volume of the wire}$

Hence, the ratio of the elastic potential energy to the energy density is equal to the volume of the wire.

The work done in stretching a spring by applying a force F is

$\text W = \frac{\text F^2}{2\text k} \quad \text{where} \quad \text k = \frac{\text{YA}}{\text L}$

For identical springs pulled by equal forces, $\text W \propto \frac{1}{\text Y}$.

Since the Young's modulus of steel is greater than that of copper, the steel spring will be stretched to a smaller extent and less work will be done on it.

Hence, more work will have to be done on the copper spring.

Let the original length and diameter of the cylinder be L and D respectively. When it is subjected to a force, let its length increase to L + l and its diameter decrease from D to D − d.

Since the body is perfectly incompressible, its volume remains unchanged. Equating the initial and the final volumes,

$\frac{\pi \text D^2}{4} \times \text L = \frac{\pi}{4}(\text D - \text d)^2 (\text L + l)$

or

$\text D^2 \text L = \text D^2 \text L\left(1 - \frac{\text d}{\text D}\right)^2\left(1 + \frac{l}{\text L}\right)$

or

$1 = \left(1 - \frac{\text d}{\text D}\right)^2\left(1 + \frac{l}{\text L}\right)$

Expanding and neglecting the square of the small quantity $\frac{\text d}{\text D}$,

$$\begin{aligned}1 = \left(1 - \frac{2\text d}{\text D}\right)\left(1 + \frac{l}{\text L}\right) \\[1em] = 1 + \frac{l}{\text L} - \frac{2\text d}{\text D} - \frac{2\text dl}{\text D\text L}\end{aligned}$$

Now, since $\frac{2\text dl}{\text D\text L}$ is the product of two very small quantities, it can be neglected. Therefore

$$\begin{aligned}0 = \frac{l}{\text L} - \frac{2\text d}{\text D} \\[1em] \Rightarrow \quad \frac{2\text d}{\text D} = \frac{l}{\text L} \quad \Rightarrow \quad \frac{\text d/\text D}{l/\text L} = \frac{1}{2}\end{aligned}$$

But $\frac{\text d/\text D}{l/\text L}$ is the ratio of the lateral strain to the longitudinal strain, which is the Poisson's ratio σ.

$\Rightarrow \quad \sigma = \frac{1}{2} = 0.5$

Hence, for a perfectly incompressible body the Poisson's ratio is 0.5.

(i) The stiffness of the material under tensile stress.

Young's modulus measures the stiffness of a material when it is subjected to tensile stress. For the bridge's cables, a higher Young's modulus indicates less elongation under tension, which ensures the structural integrity of the bridge.

(ii) It indicates how much the pillars will compress under load.

The bulk modulus measures a material's resistance to uniform compression. For the concrete pillars it is important because it tells engineers how much the pillars will compress under the weight of the bridge and other forces.

(iii) It affects the deck's response to shear forces.

The modulus of rigidity, or shear modulus, measures how a material responds to shear stress. For the bridge's deck this modulus ensures that it can resist shear forces without significant deformation, maintaining the integrity of the deck.

(iv) the lateral contraction or expansion when stretched or compressed.

Poisson's ratio is the ratio of the lateral strain to the longitudinal strain. It is important for predicting how much a material will contract or expand laterally when it is stretched or compressed, which is critical in ensuring that the materials perform as expected under load.

(v) It determines the energy lost in the material during cyclic loading.

Elastic hysteresis refers to the energy loss that occurs when a material is subjected to repeated cycles of loading and unloading. A material with low hysteresis will better maintain its shape and strength over time, which is important for the long-term durability of the bridge.

(i) It measures the stiffness of the fuselage under tensile stress.

Young's modulus is crucial for determining how much the fuselage will deform under the tensile stress arising from pressurisation and aerodynamic forces. A high Young's modulus ensures that the fuselage remains stiff and maintains its shape.

(ii) It determines the wings' resistance to bending and twisting.

The modulus of rigidity, or shear modulus, is crucial for the wings because it measures their resistance to shear forces, such as those encountered during flight manoeuvres. This ensures that the wings do not deform excessively, maintaining the stability of the aircraft.

(iii) It measures the materials resistance to volume change under pressure.

The bulk modulus indicates how much a material will compress under pressure. For an aircraft, a high bulk modulus ensures that the materials can withstand pressure changes without significant volume changes, which is critical for maintaining the structural integrity of the aircraft.

(iv) predicts lateral and axial deformations under stress.

Poisson's ratio is used to predict the relationship between the lateral and the axial deformations in materials. This is important in aircraft design to ensure that the materials behave predictably under stress, maintaining the structural integrity and performance of the aircraft.

(v) It indicates the energy stored in materials during deformation.

Elastic potential energy is the energy stored in a material when it is deformed. Managing this energy is crucial to prevent sudden failures in the material, which could compromise the safety of the aircraft during flight.

(i) Limit of elasticity : Elastic bodies recover their original state when the deforming forces are removed, but they show this property only upto a certain value of the deforming force. The maximum deforming force upto which a body retains its property of elasticity is called the limit of elasticity of the material of the body.

(ii) Suppose two wires of the same length and the same radius, one of steel and the other of rubber, are suspended from a rigid support and equal weights are attached to their lower ends. The force required to produce a given change in length is more for steel than for rubber, that is,

$\frac{\text Y_S}{\text Y_R} = \frac{\Delta \text L_R}{\Delta \text L_S}$

and experimentally ΔL_R > ΔL_S, so Y_S > Y_R. Greater is the force required to produce a given change in the size or shape of a body, the more elastic the body is said to be.

Hence, steel is more elastic than rubber.

(iii) Ivory is more elastic than rubber, while wet-clay is almost plastic.

Hence, the ivory ball will rise to the greatest height and the wet-clay ball will rise to a smaller height.

Stress : When a body is acted upon by an external (deforming) force, the intermolecular separation corresponding to the equilibrium distance is affected and internal forces arise which tend to restore the body to its original shape and size. These are called restoring forces. The restoring force acting per unit area within a body is termed as stress.

$\text{stress} = \frac{\text F}{\text A}$

Its S.I. unit is the pascal (Pa) and its dimensional formula is [M L^-1 T^-2]. Stress is classified as follows :

(a) Normal stress, produced when the applied force is perpendicular to the cross-section of the material. It is further divided into tensile stress, which occurs when equal and opposite forces are applied at the two ends of a material tending to elongate it, and compressive stress, which occurs when equal and opposite forces are applied towards each other, tending to shorten it.

(b) Tangential or shearing stress, produced when two equal and opposite forces act along the tangents to the surfaces of two opposite faces of an object, so that one face gets displaced relative to the other.

(c) Volumetric or bulk stress, produced when a body is subjected to uniform pressure from all directions, so that its volume changes without any change in shape.

Strain : Strain is a measure of how much a body has changed compared to its original size. It is the fraction of change, and not the actual change. Strain has no unit, being a pure number. It is classified as follows :

(a) Longitudinal strain : the ratio of the change in length to the initial length of a body under a tensile or compressive stress,

$\text{longitudinal strain} = \frac{\Delta \text L}{\text L}$

(b) Volumetric strain : the ratio of the change in volume to the original volume when a body is subjected to uniform pressure from all sides,

$\text{volumetric strain} = \frac{\Delta \text V}{\text V}$

(c) Shearing strain : the angle θ through which a face of a body originally perpendicular to the fixed face is turned when it is under a shearing stress.

Young's modulus of elasticity : When a body whose length is very large as compared to its breadth and thickness is acted upon by two equal and opposite forces along the direction of its length, the length of the body changes. The change in length per unit length is called the longitudinal strain and the force applied per unit area of cross-section is called the longitudinal stress.

Within the elastic limit upto which Hooke's law is applicable, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Young's modulus of the material of the body. It is denoted by Y.

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} = \frac{\text{Mg}/\pi \text r^2}{\Delta \text L/\text L} = \frac{\text{Mg}\text L}{\pi \text r^2\Delta \text L}$

Its S.I. unit is newton/metre² (N m^-2) or pascal (Pa) and its dimensional formula is [M L^-1 T^-2]. Y can be determined only for solids and is a characteristic of the material of a solid.

Limit of elasticity : Elastic bodies recover their original state when the deforming forces are removed, but they show this property only upto a certain value of the deforming force. If we go on increasing the deforming force, a stage will be reached when on removing the force the body will not return to its original state.

For example, if a small load is applied at the lower end of a wire suspended vertically from a rigid support, the length of the wire is increased; when the load is removed, the wire acquires its original length. If we increase the load step by step, a stage is reached when on removing the load the wire does not recover its original length but its length is permanently increased. Thus its property of elasticity is destroyed.

The maximum deforming force upto which a body retains its property of elasticity is called the 'limit of elasticity' of the material of the body.

Hooke's law : When a material is stretched, compressed, twisted or otherwise deformed slightly, it resists that deformation. Hooke's law describes the relationship between the stress and the strain so produced.

Statement : Within the elastic limit of a material, stress is directly proportional to strain.

$\text{stress} \propto \text{strain} \quad \Rightarrow \quad \text{stress} = \text E \times \text{strain}$

Here E is a constant of proportionality called the modulus of elasticity, that is, the ratio of stress to the corresponding strain within the elastic limits. Its S.I. unit is the pascal (N m^-2) and its dimensional formula is [M L^-1 T^-2].

The modulus of elasticity E is a constant for a given material, though it differs from one material to another. It is independent of the dimensions of the body — such as its length, area of cross-section or volume. For example, two iron wires, one 1 cm long and the other 1 km long, will have the same modulus of elasticity.

Longitudinal stress : When a body whose length is very large compared to its breadth and thickness is acted upon by two equal and opposite forces along the direction of its length, the force applied per unit area of cross-section of the body is called the longitudinal stress.

$\text{longitudinal stress} = \frac{\text{force (weight suspended)}}{\text{area of cross-section}} = \frac{\text{Mg}}{\pi \text r^2}$

Longitudinal strain : The change in length per unit length of the body is called the longitudinal strain.

$\text{longitudinal strain} = \frac{\text{increase in length}}{\text{original length}} = \frac{\Delta \text L}{\text L}$

Being a ratio of two lengths, it is a pure number and has no unit.

Young's modulus : Within the elastic limit upto which Hooke's law is applicable, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Young's modulus of the material of the body. It is denoted by Y.

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} = \frac{\text{Mg}\text L}{\pi \text r^2\Delta \text L}$

Its S.I. unit is newton/metre² (N m^-2) or pascal (Pa).

Stress : When a body is acted upon by an external deforming force, internal restoring forces arise within it which tend to restore it to its original shape and size. The restoring force acting per unit area within a body is termed as stress, that is, $\text{stress} = \frac{\text F}{\text A}$. Its S.I. unit is the pascal (Pa).

Types of stress :

(a) Normal stress, in which the applied force is perpendicular to the cross-section. It is of two kinds — tensile stress, which tends to elongate the body, and compressive stress, which tends to shorten it.

(b) Tangential or shearing stress, which is the ratio of the force acting along the tangent to a surface to the area of that surface. It causes sliding between adjacent layers of the material without changing the overall volume.

(c) Volumetric or bulk stress, which is the change in pressure acting uniformly on a body per unit area over which it acts. It compresses the material equally from all sides.

Strain : Strain tells how much a body has changed compared to its original size. Strain is about the fraction of change, and not the actual change. It has no unit.

Types of strain :

(a) Longitudinal strain, $\frac{\Delta \text L}{\text L}$ — the fractional change in length under a tensile or compressive stress.

(b) Volumetric strain, $\frac{\Delta \text V}{\text V}$ — the fractional change in volume under uniform pressure from all sides.

(c) Shearing strain, θ — the angular deformation produced when a body is subjected to a tangential stress.

Comparison of elasticity with inertia :

Elasticity	Inertia
It is the property by which a body opposes any change in its shape or size caused by an external force.	It is the property by which a body opposes any change in its state of rest or of uniform motion.
It is measured by the modulus of elasticity of the material.	It is measured by the mass of the body.
It depends on the nature of the material of the body.	It depends only on the mass of the body and not on its material.
The body regains its original configuration when the deforming force is removed.	The body continues in its state of rest or of uniform motion unless an external force acts on it.

When a wire is subjected to a gradually increasing load, its extension changes in a characteristic manner, which is represented by the graph shown above.

Region OA : In this initial stage the graph is a straight line, showing that the stress is directly proportional to the strain. This is the region where Hooke's law is obeyed, and the point A is called the limit of proportionality.

Point B — elastic limit : Beyond A the curve deviates from the straight line, indicating that the stress is no longer exactly proportional to the strain. However, if the load is removed anywhere between O and B, the wire regains its original length and the curve is retraced back to the origin. The point B is therefore the elastic limit or yield point, and is marked on the graph.

Region BC : Here the strain increases more rapidly than the stress. If the load is removed within this region the wire does not return to its original length, leaving a permanent extension. This marks the start of plastic deformation.

Point D : As the load increases beyond C, local constrictions called necks develop in the wire and it finally breaks at D, the fracture point.

The straight part OA of the graph is related to the Young's modulus of the material of the wire, since it is only in this linear region that stress is proportional to strain, and the slope of this part gives the Young's modulus.

When a wire is stretched, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy.

Suppose the length of a wire is L and its area of cross-section is A. Suppose that on applying a force F along the length of the wire, its length increases by x. Then

$\text{longitudinal stress} = \frac{\text F}{\text A} \quad \text{and} \quad \text{longitudinal strain} = \frac{\text x}{\text L}$

The Young's modulus of the material of the wire is

$\text Y = \frac{\text{stress}}{\text{strain}} = \frac{\text F/\text A}{\text x/\text L} = \frac{\text F\text L}{\text A\text x}$

Thus the force necessary to increase the length of the wire by x is

$\text F = \frac{\text{YA}}{\text L}\text x$

Now, if the wire is further increased by an infinitesimally small length dx, then the work done is

$\text{dW} = \text F \times \text{dx} = \frac{\text{YA}}{\text L}\text x\text{dx}$

Therefore, in increasing the original length L to L + l, that is, from x = 0 to x = l, the total work done is obtained by integration,

$$\begin{aligned}\text W = \int_0^{l}\frac{\text{YA}}{\text L}\text x\text{dx} = \frac{\text{YA}}{\text L}\left[\frac{\text x^2}{2}\right]_0^{l} \\[1em] = \frac{1}{2}\text{YA}\frac{l^2}{\text L}\end{aligned}$$

This may be rewritten as

$$\begin{aligned}\text W = \frac{1}{2}\left(\text Y\frac{l}{\text L}\right)\left(\frac{l}{\text L}\right)(\text{AL}) \\[1em] = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume of the wire}\end{aligned}$$

since Y × strain = stress and AL is the volume of the wire.

This work remains stored in the wire in the form of elastic potential energy U. Thus,

$\text U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

Therefore the elastic potential energy per unit volume of the wire is

$\text u = \frac{1}{2} \times \text{stress} \times \text{strain}$

Hence, the elastic potential energy per unit volume stored in a stretched wire is $\frac{1}{2}$ × stress × strain.

Further, since stress = Young's modulus × strain,

$\text u = \frac{1}{2} \times \text{Young's modulus} \times \text{strain}^2$

Given,

• Length of the wire = L
• Area of cross-section = A
• Modulus of elasticity = Y
• Increase in length = x

The force necessary to increase the length of the wire by x is

$\text F = \frac{\text{YA}}{\text L}\text x$

The work done in a further infinitesimal increase dx is dW = F dx. Hence the total work done in increasing the length from 0 to x is

$$\begin{aligned}\text W = \int_0^{\text x}\frac{\text{YA}}{\text L}\text x\text{dx} = \frac{\text{YA}}{\text L}\left[\frac{\text x^2}{2}\right]_0^{\text x} \\[1em] = \frac{1}{2}\frac{\text{YA}\text x^2}{\text L}\end{aligned}$$

Hence, the work done on the wire is $\text W = \frac{1}{2}\frac{\text{YA}\text x^2}{\text L}$.

In stretching the wire this work is done against the inter-atomic forces between the atoms of the wire. It is not lost, but remains stored in the stretched wire in the form of elastic potential energy, which is recovered when the deforming force is removed.

Given,

• Mass suspended, M = 1.0 kg
• Area of cross-section, A = 10 mm² = 10 × 10^-6 m² = 10^-5 m²
• g = 9.8 m/s²

The longitudinal stress produced in the wire is the force per unit area of cross-section,

$\text{stress} = \frac{\text{Mg}}{\text A}$

Substituting the values,

$$\begin{aligned}\text{stress} = \frac{1.0 \times 9.8}{10^{-5}} \\[1em] = 9.8 \times 10^{5}\ \text{N/m}^2\end{aligned}$$

Hence, the stress produced in the wire is 9.8 × 10⁵ N/m².

Given,

• Longitudinal strain = 2 × 10^-4
• Longitudinal stress = 2.4 × 10⁷ N/m²

The Young's modulus of the material of the wire is

$\text Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}$

Substituting the values,

$$\begin{aligned}\text Y = \frac{2.4 \times 10^{7}}{2 \times 10^{-4}} \\[1em] = 1.2 \times 10^{11}\ \text{N/m}^2\end{aligned}$$

Hence, the Young's modulus of the material of the wire is 1.2 × 10¹¹ N/m².

Given,

• Length of the wire, L = 4.0 m
• Area of cross-section, A = 1.2 cm² = 1.2 × 10^-4 m²
• Force applied, F = 4.8 × 10³ N
• Young's modulus, Y = 1.2 × 10¹¹ N/m²

Stress : The stress produced in the wire is

$$\begin{aligned}\text{stress} = \frac{\text F}{\text A} = \frac{4.8 \times 10^{3}}{1.2 \times 10^{-4}} \\[1em] = 4 \times 10^{7}\ \text{N/m}^2\end{aligned}$$

Increase in length : From the definition of the Young's modulus,

$\text Y = \frac{\text{stress}}{\Delta \text L/\text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{stress} \times \text L}{\text Y}$

Substituting the values,

$$\begin{aligned}\Delta \text L = \frac{(4 \times 10^{7}) \times 4.0}{1.2 \times 10^{11}} \\[1em] = 13.3 \times 10^{-4}\ \text m\end{aligned}$$

Hence, the stress produced is 4 × 10⁷ N/m² and the increase in length of the wire is 13.3 × 10^-4 m.

Given,

• Young's modulus of steel, Y = 2.0 × 10¹¹ N/m²
• Length of the wire, L = 2.0 m
• Diameter of the wire = 1.0 mm, so radius r = 0.5 × 10^-3 m
• Increase in length, ΔL = 1.0 mm = 1.0 × 10^-3 m
• g = 9.8 m/s²

The area of cross-section of the wire is

$\text A = \pi \text r^2 = 3.14 \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}\ \text m^2$

The Young's modulus of the material of the wire is

$\text Y = \frac{\text{Mg}\text L}{\text A\Delta \text L} \quad \Rightarrow \quad \text{Mg} = \frac{\text{YA}\Delta \text L}{\text L}$

Substituting the values,

$$\begin{aligned}\text{Mg} = \frac{(2.0 \times 10^{11}) \times (7.85 \times 10^{-7}) \times (1.0 \times 10^{-3})}{2.0} \\[1em] = 78.5\ \text N\end{aligned}$$

Therefore the mass to be suspended is

$\text M = \frac{78.5}{9.8} = 8.0\ \text{kg}$

Hence, a weight of 8.0 kg should be suspended from the steel wire.

Given,

• Length of the brass bar, L = 0.5 m
• Area of cross-section, A = 1.0 cm² = 1.0 × 10^-4 m²
• Compressing load, Mg = 5 × 10 = 50 N
• Young's modulus of brass, Y = 1.0 × 10¹¹ N/m²

The elastic potential energy of the compressed bar is

$\text U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

The stress in the bar is

$\text{stress} = \frac{\text{Mg}}{\text A} = \frac{50}{1.0 \times 10^{-4}} = 5 \times 10^{5}\ \text{N/m}^2$

The strain in the bar is

$\text{strain} = \frac{\text{stress}}{\text Y} = \frac{5 \times 10^{5}}{1.0 \times 10^{11}} = 5 \times 10^{-6}$

The volume of the bar is

$\text V = \text{AL} = (1.0 \times 10^{-4}) \times 0.5 = 5 \times 10^{-5}\ \text m^3$

Substituting these values,

$$\begin{aligned}\text U = \frac{1}{2} \times (5 \times 10^{5}) \times (5 \times 10^{-6}) \times (5 \times 10^{-5}) \\[1em] = 6.25 \times 10^{-5}\ \text J\end{aligned}$$

Hence, the increase in energy of the brass bar is 6.25 × 10^-5 J.

Given,

• Area of cross-section, A = 0.5 mm² = 0.5 × 10^-6 m²
• Length of the wire, L = 4.0 m
• Increase in length, ΔL = 1.0 mm = 1.0 × 10^-3 m
• Young's modulus, Y = 2.0 × 10¹¹ N/m²

The work done in stretching a wire is

$\text W = \frac{1}{2} \times \text Y \times (\text{strain})^2 \times \text{volume of the wire}$

The strain produced is

$\text{strain} = \frac{\Delta \text L}{\text L} = \frac{1.0 \times 10^{-3}}{4.0} = 2.5 \times 10^{-4}$

The volume of the wire is

$\text V = \text{AL} = (0.5 \times 10^{-6}) \times 4.0 = 2 \times 10^{-6}\ \text m^3$

Substituting these values,

$$\begin{aligned}\text W = \frac{1}{2} \times (2.0 \times 10^{11}) \times (2.5 \times 10^{-4})^2 \times (2 \times 10^{-6}) \\[1em] = \frac{1}{2} \times (2.0 \times 10^{11}) \times (6.25 \times 10^{-8}) \times (2 \times 10^{-6})\end{aligned}$$

$= 1.25 \times 10^{-2}\ \text J$

Hence, the work done to increase the length of the wire is 1.25 × 10^-2 J.

Given,

• Length of the wire, L = 1.0 m
• Area of cross-section, A = 2.5 mm² = 2.5 × 10^-6 m²
• Increase in length, ΔL = 1.0 mm = 1.0 × 10^-3 m
• Young's modulus, Y = 2.0 × 10¹¹ N m^-2

The elastic potential energy stored in a stretched wire is

$\text U = \frac{1}{2} \times \text Y \times (\text{strain})^2 \times \text{volume of the wire}$

The strain produced is

$\text{strain} = \frac{\Delta \text L}{\text L} = \frac{1.0 \times 10^{-3}}{1.0} = 1.0 \times 10^{-3}$

The volume of the wire is

$\text V = \text{AL} = (2.5 \times 10^{-6}) \times 1.0 = 2.5 \times 10^{-6}\ \text m^3$

Substituting these values,

$$\begin{aligned}\text U = \frac{1}{2} \times (2.0 \times 10^{11}) \times (1.0 \times 10^{-3})^2 \times (2.5 \times 10^{-6}) \\[1em] = \frac{1}{2} \times (2.0 \times 10^{11}) \times (2.5 \times 10^{-12}) = 0.25\ \text J\end{aligned}$$

Hence, the elastic potential energy stored in the wire is 0.25 J.

Given,

• Length of the wire, L = 3.0 m
• Area of cross-section, A = 1.0 mm² = 1.0 × 10^-6 m²
• Increase in length, ΔL = 0.2 mm = 0.2 × 10^-3 m
• Young's modulus, Y = 2.0 × 10¹¹ N/m²

The work done in stretching the wire is

$\text W = \frac{1}{2} \times \text Y \times (\text{strain})^2 \times \text{volume of the wire}$

The strain produced is

$\text{strain} = \frac{0.2 \times 10^{-3}}{3.0} = 6.67 \times 10^{-5}$

The volume of the wire is

$\text V = \text{AL} = (1.0 \times 10^{-6}) \times 3.0 = 3.0 \times 10^{-6}\ \text m^3$

Substituting these values,

$$\begin{aligned}\text W = \frac{1}{2} \times (2.0 \times 10^{11}) \times (6.67 \times 10^{-5})^2 \times (3.0 \times 10^{-6}) \\[1em] = \frac{1}{2} \times (2.0 \times 10^{11}) \times (4.44 \times 10^{-9}) \times (3.0 \times 10^{-6})\end{aligned}$$

$= 1.33 \times 10^{-3}\ \text J$

Hence, the work to be done in stretching the wire is 1.33 × 10^-3 J.

Given,

• Area of cross-section, A = 0.5 cm² = 0.5 × 10^-4 m²
• Young's modulus for steel, Y = 2.0 × 10¹¹ N/m²

From the definition of the Young's modulus,

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \text F = \text{YA}\left(\frac{\Delta \text L}{\text L}\right)$

(i) Length increased to 1.1 times : Here the final length is 1.1 L, so ΔL = 0.1 L and the strain is

$\frac{\Delta \text L}{\text L} = 0.1$

$$\begin{aligned}\text F = (2.0 \times 10^{11}) \times (0.5 \times 10^{-4}) \times 0.1 \\[1em] = 1.0 \times 10^{6}\ \text N\end{aligned}$$

(ii) Length increased to twice the initial value : Here the final length is 2 L, so ΔL = L and the strain is

$\frac{\Delta \text L}{\text L} = 1$

$$\begin{aligned}\text F = (2.0 \times 10^{11}) \times (0.5 \times 10^{-4}) \times 1 \\[1em] = 1.0 \times 10^{7}\ \text N\end{aligned}$$

Hence, the necessary force is (i) 1.0 × 10⁶ N and (ii) 1.0 × 10⁷ N.

Given,

• Young's modulus of copper, Y = 1.25 × 10¹¹ N/m²
• Diameter of the wire = 1 mm, so radius r = 0.5 × 10^-3 m
• Increase in length = 0.1%
• g = 9.8 m/s²

The strain to be produced is

$\frac{\Delta \text L}{\text L} = \frac{0.1}{100} = 1 \times 10^{-3}$

The area of cross-section of the wire is

$\text A = \pi \text r^2 = 3.14 \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}\ \text m^2$

From the definition of the Young's modulus,

$\text{Mg} = \text{YA}\left(\frac{\Delta \text L}{\text L}\right)$

Substituting the values,

$$\begin{aligned}\text{Mg} = (1.25 \times 10^{11}) \times (7.85 \times 10^{-7}) \times (1 \times 10^{-3}) \\[1em] = 98.1\ \text N\end{aligned}$$

Therefore the weight to be suspended is

$\text M = \frac{98.1}{9.8} = 10\ \text{kg}$

Hence, a weight of 10 kg-wt should be suspended from the wire.

Given,

• Weight suspended, Mg = 20 × 9.8 = 196 N
• Area of cross-section, A = 1 mm² = 1 × 10^-6 m²
• Stretched length = 6 m, unstretched length L = 5.995 m

The increase in the length of the wire is

$\Delta \text L = 6 - 5.995 = 0.005\ \text m$

The Young's modulus of the material of the wire is

$\text Y = \frac{\text{Mg}\text L}{\text A\Delta \text L}$

Substituting the values,

$$\begin{aligned}\text Y = \frac{196 \times 5.995}{(1 \times 10^{-6}) \times 0.005} \\[1em] = \frac{1175}{5 \times 10^{-9}} = 2.35 \times 10^{11}\ \text{N/m}^2\end{aligned}$$

Hence, the Young's modulus of the material of the wire is 2.35 × 10¹¹ N/m².

Given,

• Under a weight of 100 g the length becomes 60 cm
• Under a weight of 120 g the length becomes 70 cm

Let the initial (unstretched) length of the cord be L₀. The extension produced is directly proportional to the weight suspended, since

$\Delta \text L = \frac{\text{Mg}\text L_0}{\text{AY}} \quad \Rightarrow \quad \Delta \text L \propto \text{Mg}$

Therefore,

$\frac{100}{120} = \frac{60 - \text L_0}{70 - \text L_0}$

Cross-multiplying,

$$\begin{aligned}100(70 - \text L_0) = 120(60 - \text L_0) \\[1em] 7000 - 100\text L_0 = 7200 - 120\text L_0\end{aligned}$$

$20\text L_0 = 200 \quad \Rightarrow \quad \text L_0 = 10\ \text{cm}$

Hence, the initial length of the cord is 10 cm.

Weight for a length of 74 cm : Under a weight of 100 g the extension is 60 − 10 = 50 cm. For a length of 74 cm the extension required is 74 − 10 = 64 cm. Since the extension is proportional to the weight,

$$\begin{aligned}\text W = 100 \times \frac{64}{50} \\[1em] = 128\ \text g\end{aligned}$$

Hence, the length of the cord will become 74 cm under a weight of 128 g.

Given,

• Length of the wire, L = 1 m
• Area of cross-section, A = 0.5 mm² = 0.5 × 10^-6 m²
• Increase in length, ΔL = 0.1 mm = 1 × 10^-4 m
• Work done, W = 5 × 10^-4 J

The work done in stretching a wire is

$\text W = \frac{1}{2} \times \text Y \times \left(\frac{\Delta \text L}{\text L}\right)^2 \times (\text{AL}) = \frac{\text{YA}(\Delta \text L)^2}{2\text L}$

Therefore the Young's modulus of the material is

$\text Y = \frac{2\text W\text L}{\text A(\Delta \text L)^2}$

Substituting the values,

$$\begin{aligned}\text Y = \frac{2 \times (5 \times 10^{-4}) \times 1}{(0.5 \times 10^{-6}) \times (1 \times 10^{-4})^2} \\[1em] = \frac{1 \times 10^{-3}}{5 \times 10^{-15}} = 2.0 \times 10^{11}\ \text{N/m}^2\end{aligned}$$

Hence, the Young's modulus of the material of the wire is 2.0 × 10¹¹ N/m².

Given,

• Original volume of the fluid, V = 3.0 × 10^-3 m³
• Increase in pressure, p = 6.0 × 10⁶ N m^-2
• Decrease in volume, ΔV = 5.0 × 10^-7 m³

The bulk modulus of the fluid is the ratio of the normal stress to the volume strain,

$\text K = \frac{\text p}{\Delta \text V/\text V} = \frac{\text{pV}}{\Delta \text V}$

Substituting the values,

$$\begin{aligned}\text K = \frac{(6.0 \times 10^{6}) \times (3.0 \times 10^{-3})}{5.0 \times 10^{-7}} \\[1em] = \frac{1.8 \times 10^{4}}{5.0 \times 10^{-7}} = 3.6 \times 10^{10}\ \text{N/m}^2\end{aligned}$$

Hence, the bulk modulus of the fluid is 3.6 × 10¹⁰ N/m².

Given,

• Volume of water, V = 1.0 m³
• Depth inside the lake, h = 200 m
• Bulk modulus of water, K = 22000 atmospheres
• Density of water, ρ = 1.0 × 10³ kg/m³
• 1 atmosphere = 10⁵ N/m², g = 10 m/s²

The increase in pressure on going to a depth h inside the lake is

$$\begin{aligned}\text p = \text h\rho \text g = 200 \times (1.0 \times 10^{3}) \times 10 \\[1em] = 2.0 \times 10^{6}\ \text{N/m}^2\end{aligned}$$

The bulk modulus of water is

$\text K = 22000 \times 10^{5} = 2.2 \times 10^{9}\ \text{N/m}^2$

The bulk modulus of elasticity is

$\text K = \frac{\text{pV}}{\Delta \text V} \quad \Rightarrow \quad \Delta \text V = \frac{\text{pV}}{\text K}$

Substituting the values,

$$\begin{aligned}\Delta \text V = \frac{(2.0 \times 10^{6}) \times 1.0}{2.2 \times 10^{9}} \\[1em] = 9.1 \times 10^{-4}\ \text m^3\end{aligned}$$

Hence, the volume of the water decreases by 9.1 × 10^-4 m³.

Given,

• Weight of the metal piece = 2 kg-wt = 2 × 9.8 = 19.6 N
• Density of the oil, ρ = 0.7 × 10³ kg/m³
• Increase in length of the wire, ΔL = 1 mm = 1 × 10^-3 m
• Diameter of the wire = 0.6 mm, so radius r = 0.3 × 10^-3 m
• Young's modulus, Y = 2.0 × 10¹¹ N/m²
• Volume of the metal piece, V = 800 cm³ = 8 × 10^-4 m³

Since the metal piece is fully immersed in the oil, an upthrust acts on it equal to the weight of the oil displaced,

$$\begin{aligned}\text F_B = \text V\rho \text g = (8 \times 10^{-4}) \times (0.7 \times 10^{3}) \times 9.8 \\[1em] = 5.488\ \text N\end{aligned}$$

Therefore the effective stretching force acting on the wire is

$\text F = 19.6 - 5.488 = 14.112\ \text N$

The area of cross-section of the wire is

$\text A = \pi \text r^2 = 3.14 \times (0.3 \times 10^{-3})^2 = 2.826 \times 10^{-7}\ \text m^2$

The Young's modulus of the material of the wire is

$\text Y = \frac{\text F\text L}{\text A\Delta \text L} \quad \Rightarrow \quad \text L = \frac{\text{YA}\Delta \text L}{\text F}$

Substituting the values,

$$\begin{aligned}\text L = \frac{(2.0 \times 10^{11}) \times (2.826 \times 10^{-7}) \times (1 \times 10^{-3})}{14.112} \\[1em] = \frac{56.52}{14.112} = 4.0\ \text m\end{aligned}$$

Hence, the initial length of the wire is 4.0 m.

Given,

• Breaking stress = 10⁹ N/m²
• Density of the substance, ρ = 3 × 10³ kg/m³
• g = 9.8 m/s²

Let L be the length of the wire and A its area of cross-section. The weight of the wire is

$\text W = \text A \times \text L \times \rho \times \text g$

The stress due to this weight at the point of suspension is

$\frac{\text W}{\text A} = \text L\rho \text g$

The wire will break under its own weight when this stress becomes equal to the breaking stress,

$\text L\rho \text g = 10^{9} \quad \Rightarrow \quad \text L = \frac{10^{9}}{\rho \text g}$

Substituting the values,

$$\begin{aligned}\text L = \frac{10^{9}}{(3 \times 10^{3}) \times 9.8} \\[1em] = \frac{10^{9}}{2.94 \times 10^{4}} = 3.4 \times 10^{4}\ \text m\end{aligned}$$

Hence, a wire of length 3.4 × 10⁴ m will break under its own weight when suspended.

Given,

• Stress applied = 1 kg-wt/mm²
• Young's modulus, Y = 1.0 × 10¹¹ N/m²
• 1 kg-wt = 9.8 N

Expressing the stress in S.I. units,

$\text{stress} = \frac{9.8\ \text N}{1 \times 10^{-6}\ \text m^2} = 9.8 \times 10^{6}\ \text{N/m}^2$

The longitudinal strain produced is

$$\begin{aligned}\frac{\Delta \text L}{\text L} = \frac{\text{stress}}{\text Y} = \frac{9.8 \times 10^{6}}{1.0 \times 10^{11}} \\[1em] = 9.8 \times 10^{-5}\end{aligned}$$

Therefore the percentage increase in the length of the wire is

$$\begin{aligned}\frac{\Delta \text L}{\text L} \times 100 = (9.8 \times 10^{-5}) \times 100 \\[1em] = 0.0098%\end{aligned}$$

Hence, the percentage increase in the length of the wire is 0.0098%.

9 F

Reason — Given,

• Area of cross-section of wire-1, A₁ = A, and of wire-2, A₂ = 3A
• Both wires are of the same material (same Y) and of the same volume
• Elongation of wire-1 = Δx under a force F

Since the two wires have the same volume,

$\text A \times \text L_1 = 3\text A \times \text L_2 \quad \Rightarrow \quad \text L_2 = \frac{\text L_1}{3}$

From the definition of the Young's modulus,

$\text Y = \frac{\text F/\text A}{\Delta \text x/\text L} \quad \Rightarrow \quad \Delta \text x = \frac{\text F\text L}{\text A\text Y}$

For wire-1,

$\Delta \text x = \frac{\text F\text L_1}{\text A\text Y}$

For wire-2, let the required force be F′. For the same elongation Δx,

$\Delta \text x = \frac{\text F'\text L_2}{3\text A\text Y} = \frac{\text F'\left(\frac{\text L_1}{3}\right)}{3\text A\text Y} = \frac{\text F'\text L_1}{9\text A\text Y}$

Equating the two expressions for Δx,

$\frac{\text F\text L_1}{\text A\text Y} = \frac{\text F'\text L_1}{9\text A\text Y} \quad \Rightarrow \quad \text F' = 9\text F$

9

Reason — The stress in the leg is the weight of the body per unit area of cross-section of the leg,

$\text{stress} = \frac{\text{force}}{\text{area}} = \frac{\text{mg}}{\text A} = \frac{\text{volume} \times \text{density} \times \text g}{\text{area}}$

If the original linear dimension is L, then the volume is proportional to L³ and the area of cross-section to L². When the linear dimensions increase by a factor of 9, the new linear dimension is 9L, so

$\text{stress} = \frac{(9\text L)^3 \times \rho \times \text g}{(9\text L)^2} = 9\text L\rho \text g$

Since the density ρ remains the same, the stress is directly proportional to the linear dimension. Hence the stress in the leg becomes 9 times its original value.

$\frac{\text P}{3\text B}$

Reason — The bulk modulus of the object is

$\text B = \frac{\text P}{\Delta \text V/\text V} \quad \Rightarrow \quad \frac{\Delta \text V}{\text V} = \frac{\text P}{\text B}$

For a sphere of radius r, the volume is

$\text V = \frac{4}{3}\pi \text r^3 \quad \Rightarrow \quad \text V \propto \text r^3$

Taking the fractional change on both sides, the volume strain is three times the linear strain,

$\frac{\Delta \text V}{\text V} = 3\frac{\Delta \text r}{\text r}$

Therefore,

$3\frac{\Delta \text r}{\text r} = \frac{\text P}{\text B} \quad \Rightarrow \quad \frac{\Delta \text r}{\text r} = \frac{\text P}{3\text B}$

$\frac{\text {mg}}{3\text {Ka}}$

Reason — When a mass m is placed on the massless piston of area a, the increase in the pressure transmitted to the liquid, and hence to the sphere, is

$\text{dP} = \frac{\text{mg}}{\text a}$

The bulk modulus of the material of the sphere is

$\text K = -\frac{\text{dP}}{\text{dV}/\text V} \quad \Rightarrow \quad -\frac{\text{dV}}{\text V} = \frac{\text{dP}}{\text K} = \frac{\text{mg}}{\text{Ka}}$

For the solid sphere of radius r, $\text V = \frac{4}{3}\pi \text r^3$, so that

$\frac{\text{dV}}{\text V} = \frac{4\pi \text r^2\text{dr}}{\frac{4}{3}\pi \text r^3} = 3\frac{\text{dr}}{\text r}$

Therefore,

$-3\frac{\text{dr}}{\text r} = \frac{\text{mg}}{\text{Ka}} \quad \Rightarrow \quad \frac{\text{dr}}{\text r} = -\frac{\text{mg}}{3\text{Ka}}$

The negative sign indicates the decrease in the radius, so the fractional decrement in the radius is $\frac{\text{mg}}{3\text{Ka}}$.

$\frac{\text P}{3\alpha \text K}$

Reason — When the external pressure P is applied, the cube is compressed and the volume strain produced is given by

$\text P = -\text K\frac{\Delta \text V}{\text V}$

To bring the cube back to its original size, the same volume must be regained by heating. On heating through Δθ, the volume increases according to

$\frac{\Delta \text V}{\text V} = \gamma\Delta \theta = 3\alpha\Delta \theta$

since the coefficient of volume expansion γ = 3α.

Equating the volume strain produced by the pressure to that produced by the heating,

$\text P = \text K(3\alpha\Delta \theta) \quad \Rightarrow \quad \Delta \theta = \frac{\text P}{3\alpha \text K}$

1.16 mm

Reason — Given,

• Elastic limit of brass = 379 MPa = 379 × 10⁶ N/m²
• Load to be supported, F = 400 N

Let D_min be the minimum diameter of the brass rod. The stress in the rod is

$\text{stress} = \frac{\text F}{\text A} = \frac{\text F}{\pi \text r^2} = \frac{\text F}{\pi(\text D_{min}/2)^2} = \frac{4\text F}{\pi \text D_{min}^2}$

For the rod not to exceed its elastic limit, the stress must not exceed 379 × 10⁶ N/m². Taking the limiting case,

$$\begin{aligned}\text D_{min}^2 = \frac{4 \times 400}{\pi \times (379 \times 10^{6})} \\[1em] = \frac{1600}{1.19 \times 10^{9}} = 1.344 \times 10^{-6}\ \text m^2\end{aligned}$$

$\text D_{min} = 1.16 \times 10^{-3}\ \text m = 1.16\ \text{mm}$

1.7 mm

Reason — Given,

• $\frac{\text Y_A}{\text Y_B} = \frac{7}{4}$
• L_A = 2 m, radius of A = R
• L_B = 1.5 m, radius of B = 2 mm = 2 × 10^-3 m
• The two wires stretch by the same length l for the same load F

The Young's modulus of each wire is

$\text Y_A = \frac{\text F\text L_A}{\pi \text R^2l} \quad \text{and} \quad \text Y_B = \frac{\text F\text L_B}{\pi \text R_B^2l}$

Dividing,

$\frac{\text Y_A}{\text Y_B} = \frac{\text L_A}{\text L_B} \times \frac{\text R_B^2}{\text R^2}$

Substituting the values,

$\frac{7}{4} = \frac{2}{1.5} \times \frac{(2 \times 10^{-3})^2}{\text R^2}$

$\text R^2 = \frac{2}{1.5} \times \frac{4}{7} \times (4 \times 10^{-6}) = 3.04 \times 10^{-6}\ \text m^2$

$\text R = 1.74 \times 10^{-3}\ \text m \approx 1.7\ \text{mm}$

10⁶ N/m²

Reason — Given,

• Length of the rubber cord, L = 42 cm = 0.42 m
• Extension of the cord, l = 20 cm = 0.2 m
• Diameter of cross-section, D = 6 mm, so radius r = 3 × 10^-3 m
• Mass of the stone, m = 0.02 kg
• Velocity of the stone, v = 20 m/s

The work done in stretching the rubber cord is stored in it in the form of elastic potential energy, and on release this energy is given to the stone as its kinetic energy. Therefore,

$\frac{1}{2}\text Y\left(\frac{l}{\text L}\right)^2 \times \text A \times \text L = \frac{1}{2}\text{mv}^2$

$\Rightarrow \quad \text Y = \frac{\text{mv}^2\text L}{\text Al^2}$

The area of cross-section of the cord is

$\text A = \pi \text r^2 = 3.14 \times (3 \times 10^{-3})^2 = 2.826 \times 10^{-5}\ \text m^2$

Substituting the values,

$$\begin{aligned}\text Y = \frac{0.02 \times (20)^2 \times 0.42}{(2.826 \times 10^{-5}) \times (0.2)^2} \\[1em] = \frac{3.36}{1.13 \times 10^{-6}} = 3 \times 10^{6}\ \text{N/m}^2\end{aligned}$$

So the closest value of Y among the given options is 10⁶ N/m².

3.0 mm

Reason — Given,

• Increase in length in air, l₁ = 4.0 mm
• Relative density of the liquid, ρ_l = 2
• Relative density of the material of the load, ρ_b = 8

When the load of mass M is suspended in air, the extension of the wire is

$l_1 = \frac{\text{Mg}\text L}{\text{AY}} \qquad \ldots(\text i)$

When the load is fully immersed in the liquid, an upthrust F_B acts on it, so the stretching force becomes (Mg − F_B) and

$l_2 = \frac{(\text{Mg} - \text F_B)\text L}{\text{AY}} \qquad \ldots(\text{ii})$

The buoyant force is the weight of the liquid displaced,

$\text F_B = \text V\rho_l\text g = \frac{\text M}{\rho_b}\rho_l\text g$

Given that $\frac{\rho_l}{\rho_b} = \frac{2}{8} = \frac{1}{4}$, we get $\text F_B = \frac{1}{4}\text{Mg}$. Substituting this in equation (ii),

$l_2 = \frac{\left(\text{Mg} - \frac{1}{4}\text{Mg}\right)\text L}{\text{AY}} = \frac{3}{4}\frac{\text{Mg}\text L}{\text{AY}}$

Dividing this by equation (i),

$\frac{l_2}{l_1} = \frac{3}{4} \quad \Rightarrow \quad l_2 = \frac{3}{4} \times 4 = 3\ \text{mm}$

$\frac{\text F}{\text A \alpha \Delta \text T}$

Reason — When the temperature of the rod is increased by ΔT, the increase in its length would be

$\Delta \text L = \text L\alpha\Delta \text T \quad \Rightarrow \quad \frac{\Delta \text L}{\text L} = \alpha\Delta \text T \qquad \ldots(\text i)$

When the rod is subjected to the compressive force F, its Young's modulus is given by

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \frac{\Delta \text L}{\text L} = \frac{\text F}{\text{AY}} \qquad \ldots(\text{ii})$

Since it is given that the length of the rod does not change, the contraction produced by the force must exactly balance the expansion produced by the heating. Hence from equations (i) and (ii),

$\alpha\Delta \text T = \frac{\text F}{\text{AY}} \quad \Rightarrow \quad \text Y = \frac{\text F}{\text A\alpha\Delta \text T}$

3F/(πr²YT)

Reason — The Young's modulus of the material of the rod is

$\text Y = \frac{\text F/\pi \text r^2}{\Delta l/\text L} = \frac{\text F}{\pi \text r^2} \times \frac{\text L}{\Delta l}$

Therefore the contraction produced by the compressional force is

$\Delta l = \frac{\text F \times \text L}{\pi \text r^2\text Y}$

The change in length when the rod is heated through a temperature T is

$\Delta l = \text L\alpha\text T$

Since the length of the rod remains unchanged, the two changes must be equal,

$\text L\alpha\text T = \frac{\text{FL}}{\pi \text r^2\text Y} \quad \Rightarrow \quad \alpha = \frac{\text F}{\pi \text r^2\text{YT}}$

The coefficient of volume expansion is γ = 3α, so

$\gamma = \frac{3\text F}{\pi \text r^2\text{YT}}$

3.1 × 10⁶ N m^-2

Reason — Given,

• Radius of the steel wire, r = 2.0 mm = 2.0 × 10^-3 m
• Load carried, m = 4 kg
• g = 3.1 π m s^-2

The wire is stretched along its length by the weight of the load hanging from it. The tensile stress is the tensile force acting per unit area of cross-section of the wire,

$\text{tensile stress} = \frac{\text{tensile force}}{\text{area of cross-section}} = \frac{\text{mg}}{\pi \text r^2}$

Substituting the values,

$$\begin{aligned}\text{tensile stress} = \frac{4 \times 3.1\pi}{\pi \times (2.0 \times 10^{-3})^2} \\[1em] = \frac{12.4}{4 \times 10^{-6}} = 3.1 \times 10^{6}\ \text{N m}^{-2}\end{aligned}$$

$\frac{1}{2}$ Mgl

Reason —

When the block of mass M is suspended, the wire is extended through l. The stretching force increases uniformly from zero to Mg as the extension grows from zero to l, so the average force acting during the extension is $\frac{\text{Mg}}{2}$.

The elastic potential energy stored in the wire is therefore

$$\begin{aligned}\text{P.E.} = \text{average force} \times \text{extension} = \frac{\text{Mg}}{2} \times l \\[1em] = \frac{1}{2}\text{Mg}l\end{aligned}$$

that is, the elastic potential energy stored is half the work done by gravity in lowering the block through l.

4 mm

Reason — Given,

• Length of the steel wire, L = 1 m
• Elastic limit of steel, $\frac{\text F}{\text A}$ = 8 × 10⁸ N m^-2
• Young's modulus of steel, Y = 2 × 10¹¹ N m^-2

For a wire of length L and area of cross-section A subjected to a force F, the change in its length l is given by

$\text Y = \frac{\text F\text L}{\text Al} \quad \Rightarrow \quad l = \frac{\left(\frac{\text F}{\text A}\right)\text L}{\text Y}$

The maximum elongation occurs when the stress equals the elastic limit. Substituting the values,

$$\begin{aligned}l = \frac{(8 \times 10^{8}) \times 1}{2 \times 10^{11}} \\[1em] = 4 \times 10^{-3}\ \text m = 4\ \text{mm}\end{aligned}$$

Given,

• Diameter of the steel wire, d = 0.5 mm = 5 × 10^-4 m
• Young's modulus, Y = 2 × 10¹¹ N m^-2
• Length of the wire, L = 1.0 m
• Increase in load, M = 1.2 kg
• Least count of the main scale = 1.0 mm, g = 10 m s^-2, π = 3.2

The Young's modulus of the material of the wire is

$\text Y = \frac{\text{Mg}/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text{Mg}\text L}{\text{AY}} = \frac{\text{Mg}\text L}{\frac{\pi \text d^2}{4} \times \text Y}$

Substituting the values,

$$\begin{aligned}\Delta \text L = \frac{1.2 \times 10 \times 1.0}{\frac{3.2}{4} \times (5 \times 10^{-4})^2 \times (2 \times 10^{11})} \\[1em] = \frac{12}{0.8 \times (25 \times 10^{-8}) \times (2 \times 10^{11})}\end{aligned}$$

$= \frac{12}{40 \times 10^{3}} = 0.3 \times 10^{-3}\ \text m = 0.3\ \text{mm}$

The vernier constant (least count of the vernier scale) is

$\text{L.C.} = \frac{\text{least count of the main scale}}{\text{number of vernier divisions}} = \frac{1}{10} = 0.1\ \text{mm}$

Therefore the number of the vernier scale division that coincides with a main scale division is

$\frac{\Delta \text L}{\text{L.C.}} = \frac{0.3}{0.1} = 3$

Hence, the 3rd division of the vernier scale will coincide with a main scale division.

Given,

• Mass of the rod, M = 20 kg
• Area of cross-section, A = 0.4 m²
• Length of the rod, L = 20 m
• Young's modulus, Y = 2 × 10¹¹ N m^-2, g = 10 m s^-2

The weight of the rod acts at its centre of gravity, which lies at the middle of the rod. Hence the average tension in the rod is

$\text T_{avg} = \frac{\text{Mg}}{2}$

The Young's modulus of the material of the rod is

$\text Y = \frac{\text F/\text A}{\Delta \text L/\text L} \quad \Rightarrow \quad \Delta \text L = \frac{\text T_{avg}\text L}{\text{AY}} = \frac{\text{Mg}\text L}{2\text{AY}}$

Substituting the values,

$$\begin{aligned}\Delta \text L = \frac{20 \times 10 \times 20}{2 \times 0.4 \times (2 \times 10^{11})} \\[1em] = \frac{4000}{1.6 \times 10^{11}} = 2.5 \times 10^{-8}\ \text m\end{aligned}$$

$\Delta \text L = 25 \times 10^{-9}\ \text m$

Hence, the value of x is 25.

Given,

• Length of the wire, L = 1 m (exact)
• Extension, l = 0.4 mm with Δl = ± 0.02 mm
• Diameter, D = 0.4 mm with ΔD = ± 0.01 mm
• Load applied = 1 kg, so F = 1 × 10 = 10 N
• g = 10 m s^-2

The Young's modulus of the material of the wire is

$\text Y = \frac{\text F\text L}{\text Al} \quad \text{where} \quad \text A = \frac{\pi \text D^2}{4}$

Since Y is inversely proportional to D² and to l, the maximum relative error in Y is

$\frac{\Delta \text Y}{\text Y} = \frac{2\Delta \text D}{\text D} + \frac{\Delta l}{l}$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text Y}{\text Y} = \frac{2 \times 0.01}{0.4} + \frac{0.02}{0.4} \\[1em] = 0.05 + 0.05 = 0.1 = \frac{1}{10}\end{aligned}$$

The value of the Young's modulus is

$$\begin{aligned}\text Y = \frac{\text F\text L}{\pi \text r^2l} = \frac{10 \times 1}{3.14 \times (0.2 \times 10^{-3})^2 \times (0.4 \times 10^{-3})} \\[1em] = 1.988 \times 10^{11} \approx 2 \times 10^{11}\ \text{N m}^{-2}\end{aligned}$$

Therefore the error in the measurement of the Young's modulus is

$\Delta \text Y = \frac{\text Y}{10} = \frac{2 \times 10^{11}}{10} = 2 \times 10^{10}\ \text{N m}^{-2}$

Hence, the value of x is 2.

Given,

• Average depth of the ocean, h = 4000 m
• Bulk modulus of water, K = 2 × 10⁹ N m^-2
• Density of water, ρ = 1000 kg m^-3, g = 10 m s^-2

The pressure at the bottom of the ocean is obtained from the hydrostatic pressure formula,

$$\begin{aligned}\text P = \text h\rho \text g = 4000 \times 1000 \times 10 \\[1em] = 40 \times 10^{6}\ \text{N m}^{-2}\end{aligned}$$

The bulk modulus of water is

$\text K = \frac{\text P}{\Delta \text V/\text V} \quad \Rightarrow \quad \frac{\Delta \text V}{\text V} = \frac{\text P}{\text K}$

Substituting the values,

$$\begin{aligned}\frac{\Delta \text V}{\text V} = \frac{40 \times 10^{6}}{2 \times 10^{9}} \\[1em] = 2 \times 10^{-2}\end{aligned}$$

Comparing this with α × 10^-2,

Hence, the value of α is 2.

Given,

• Length of the wire, l = 2 m
• Young's modulus, Y = 2.0 × 10¹¹ N m^-2
• Poisson's ratio, σ = 0.2
• Transverse (lateral) strain, $\frac{\Delta \text r}{\text r}$ = 10^-3

The Poisson's ratio is the ratio of the lateral strain to the longitudinal strain,

$\sigma = \frac{\left(\frac{\Delta \text r}{\text r}\right)}{\left(\frac{\Delta l}{l}\right)} \quad \Rightarrow \quad \frac{\Delta l}{l} = \frac{1}{\sigma} \times \left(\frac{\Delta \text r}{\text r}\right)$

Substituting the values,

$\frac{\Delta l}{l} = \frac{1}{0.2} \times (10^{-3}) = 5 \times 10^{-3}$

The elastic potential energy density of a stretched wire is

$\text U_E = \frac{1}{2}\text Y\varepsilon_l^2 = \frac{1}{2}\text Y\left(\frac{\Delta l}{l}\right)^2$

Substituting the values,

$$\begin{aligned}\text U_E = \frac{1}{2} \times (2 \times 10^{11}) \times (5 \times 10^{-3})^2 \\[1em] = \frac{1}{2} \times (2 \times 10^{11}) \times (25 \times 10^{-6}) = 25 \times 10^{5}\end{aligned}$$

Hence, the elastic potential energy density of the wire is 25 × 10⁵ J m^-3.