JEE Physics Mechanical Properties Of Solids MCQs Set 19

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Multiple Choice Questions

Question 1: A force of 200 N is applied at one end of a wire of length 2 m and cross-section \( 10^{-2}\ \text{cm}^2 \). The other end is rigidly fixed. The coefficient of linear expansion is \( 8\times10^{-6}\,/^\circ\text{C} \) and \( Y=2.2\times10^{11} \) N/m\(^2\). If the temperature is raised by \( 5^\circ \)C, the increase in the tension of the wire is
(a) 4.2 N
(b) 4.4 N
(c) 2.4 N
(d) 8.8 N
Show Answer & Explanation

Answer: (d) 8.8 N

Explanation:
1. Cross-section \( A=10^{-2}\ \text{cm}^2=10^{-6}\ \text{m}^2 \).
2. Increase in tension \( =YA\alpha\Delta T=2.2\times10^{11}\times10^{-6}\times8\times10^{-6}\times5=8.8 \) N.

Teacher's Note:
1. A fixed wire cannot expand, so heating builds up tension.
2. The applied 200 N does not enter the extra tension.

Question 2: For silver, Young's modulus is \( 7.25\times10^{10} \) N/m\(^2\) and bulk modulus is \( 11\times10^{10} \) N/m\(^2\). Its Poisson's ratio is
(a) \( -1 \)
(b) 0.5
(c) 0.39
(d) 0.25
Show Answer & Explanation

Answer: (c) 0.39

Explanation:
1. \( 1-2\sigma=\dfrac{Y}{3K}=\dfrac{7.25}{33}=0.22 \).
2. \( \sigma=0.39 \).

Teacher's Note:
1. Poisson's ratio of materials lies between \( -1 \) and 0.5.

Question 3: Young's modulus of rubber is \( 10^4 \) N/m\(^2\) and the area of cross-section is \( 2\ \text{cm}^2 \). If a force of \( 2\times10^5 \) dyne is applied along its length, its initial length \( l \) becomes
(a) \( 3l \)
(b) \( 4l \)
(c) \( 2l \)
(d) none of the above
Show Answer & Explanation

Answer: (c) \( 2l \)

Explanation:
1. \( F=2\times10^5 \) dyne \( =2 \) N and \( A=2\times10^{-4} \) m\(^2\).
2. Strain \( =\dfrac{2}{2\times10^{-4}\times10^4}=1 \), so the new length is \( 2l \).

Teacher's Note:
1. Convert dyne to newton and cm\(^2\) to m\(^2\).

Question 4: In steel, Young's modulus is \( 2\times10^{11} \) N/m\(^2\) and the strain at the breaking point is 0.15. The stress at the breaking point is
(a) \( 1.33\times10^{11} \) N/m\(^2\)
(b) \( 1.33\times10^{12} \) N/m\(^2\)
(c) \( 7.5\times10^{-13} \) N/m\(^2\)
(d) \( 3\times10^{10} \) N/m\(^2\)
Show Answer & Explanation

Answer: (d) \( 3\times10^{10} \) N/m\(^2\)

Explanation:
1. \( \sigma=Y\varepsilon=2\times10^{11}\times0.15 \).

Teacher's Note:
1. Hooke's law is assumed up to the breaking point here.

Question 5: The force required to stretch a steel wire of \( 1\ \text{cm}^2 \) cross-section to 1.1 times its length is (\( Y=2\times10^{11} \) N/m\(^2\))
(a) \( 2\times10^{6} \) N
(b) \( 2\times10^{3} \) N
(c) \( 2\times10^{-6} \) N
(d) \( 2\times10^{-7} \) N
Show Answer & Explanation

Answer: (a) \( 2\times10^{6} \) N

Explanation:
1. Strain \( =0.1 \); \( A=10^{-4} \) m\(^2\).
2. \( F=2\times10^{11}\times10^{-4}\times0.1=2\times10^6 \) N.

Teacher's Note:
1. Final length 1.1 L means extension 0.1 L.

Question 6: A copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied, which stretches their combined length by 1 cm. The two wires will have
(a) different stresses and strains
(b) the same stress and strain
(c) the same strain but different stresses
(d) the same stress but different strains
Show Answer & Explanation

Answer: (d) the same stress but different strains

Explanation:
1. Stress \( =F/A \) is the same in both wires.
2. Strain \( =\text{stress}/Y \) differs as \( Y \) differs.

Teacher's Note:
1. Wires in series carry the same force.

Question 7: A steel ring of radius \( r \) and cross-section \( A \) is fitted on to a wooden disc of radius \( R\ (R>r) \). If Young's modulus is \( E \), the force with which the steel ring is expanded is
(a) \( AE\dfrac Rr \)
(b) \( AE\Big(\dfrac{R-r}{r}\Big) \)
(c) \( \dfrac EA\Big(\dfrac{R-r}{r}\Big) \)
(d) \( \dfrac{Er}{AR} \)
Show Answer & Explanation

Answer: (b) \( AE\Big(\dfrac{R-r}{r}\Big) \)

Explanation:
1. Circumference changes from \( 2\pi r \) to \( 2\pi R \).
2. Strain \( =\dfrac{R-r}{r} \), so tension \( =AE\dfrac{R-r}{r} \).

Teacher's Note:
1. Strain of a ring equals the fractional change in radius.

Question 8: A wire extends by 1 mm when a force is applied. If the force is doubled on another wire of the same material and length but half the radius of cross-section, the elongation in mm will be
(a) 8
(b) 4
(c) 2
(d) 1
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. \( \Delta L=\dfrac{FL}{\pi r^2Y} \).
2. Force doubles and area becomes one-fourth, so the extension becomes 8 times.

Teacher's Note:
1. Halving radius quarters the area.

Question 9: Two wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio \( 1:\sqrt2 \). If they are stretched by equal forces, the increase in their lengths will be in the ratio
(a) \( 2:\sqrt2 \)
(b) \( \sqrt2:2 \)
(c) 1 : 1
(d) 1 : 2
Show Answer & Explanation

Answer: (c) 1 : 1

Explanation:
1. \( \dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L_1}{L_2}\cdot\dfrac{r_2^2}{r_1^2}=\dfrac12\times2=1 \).

Teacher's Note:
1. Elongation is proportional to length and inversely proportional to the square of the radius.
2. Compute both factors separately before taking the ratio.

Question 10: A weight of 10 kg suspended from a copper wire of length 3 m and diameter 0.4 mm produces an extension of 2.4 mm. If the diameter of the wire is doubled, the extension will be
(a) 9.6 mm
(b) 4.8 mm
(c) 1.2 mm
(d) 0.6 mm
Show Answer & Explanation

Answer: (d) 0.6 mm

Explanation:
1. Extension \( =\dfrac{4FL}{\pi d^2Y}\propto\dfrac1{d^2} \).
2. Doubling the diameter divides the extension by 4: \( 2.4/4=0.6 \) mm.

Teacher's Note:
1. Check units carefully: the extension is given in mm.

Question 11: A force of \( 10^3 \) N stretches the length of a hanging wire by 1 mm. The force required to stretch a wire of the same material and length but having four times the diameter by 1 mm is
(a) \( 4\times10^3 \) N
(b) \( 16\times10^3 \) N
(c) \( \tfrac14\times10^3 \) N
(d) \( \tfrac1{16}\times10^3 \) N
Show Answer & Explanation

Answer: (b) \( 16\times10^3 \) N

Explanation:
1. \( F=\dfrac{YA\Delta L}{L} \).
2. Four times the diameter means 16 times the area.

Teacher's Note:
1. Same extension and length keep \( \Delta L/L \) fixed.

Question 12: The ratio of the two specific heats \( C_p/C_v \) for argon is 1.6 and for hydrogen is 1.4. The adiabatic elasticity of argon at pressure \( P \) is \( E \). The adiabatic elasticity of hydrogen will also be equal to \( E \) at the pressure
(a) \( P \)
(b) \( \dfrac87P \)
(c) \( \dfrac78P \)
(d) \( 1.4P \)
Show Answer & Explanation

Answer: (b) \( \dfrac87P \)

Explanation:
1. Adiabatic bulk modulus of a gas is \( \gamma P \).
2. Equate \( 1.6P=1.4P' \).

Teacher's Note:
1. Elasticity of gases is proportional to pressure.

Question 13: The relation between Young's modulus \( Y \), modulus of rigidity \( \eta \) and bulk modulus \( K \) is
(a) \( \dfrac1\eta=\dfrac1{3Y}+\dfrac1{9K} \)
(b) \( \dfrac1K=\dfrac1{3Y}+\dfrac1{9\eta} \)
(c) \( \dfrac1Y=\dfrac1{3K}+\dfrac1{9\eta} \)
(d) \( \dfrac1Y=\dfrac1{3\eta}+\dfrac1{9K} \)
Show Answer & Explanation

Answer: (d) \( \dfrac1Y=\dfrac1{3\eta}+\dfrac1{9K} \)

Explanation:
1. Combining \( Y=2\eta(1+\sigma) \) and \( Y=3K(1-2\sigma) \) eliminates \( \sigma \).
2. This gives \( \dfrac9Y=\dfrac3\eta+\dfrac1K \), i.e. \( \dfrac1Y=\dfrac1{3\eta}+\dfrac1{9K} \).

Teacher's Note:
1. Check by the case \( \sigma=0.5 \) for an incompressible solid: \( K\to\infty \), \( Y=3\eta \).

Question 14: A wire of cross-sectional area \( 3\ \text{mm}^2 \) is first stretched between two fixed points at \( 20^\circ \)C. Determine the tension when the temperature falls to \( 10^\circ \)C. Take \( \alpha=10^{-5}\,^\circ\text{C}^{-1} \), \( Y=2\times10^{11} \) N/m\(^2\)
(a) 20 N
(b) 30 N
(c) 60 N
(d) 120 N
Show Answer & Explanation

Answer: (c) 60 N

Explanation:
1. \( T=2\times10^{11}\times3\times10^{-6}\times10^{-5}\times10=60 \) N.

Teacher's Note:
1. The fall in temperature tries to contract the wire.

Question 15: A wire is stretched by 0.01 m by a certain force \( F \). Another wire of the same material whose diameter and length are double those of the first is stretched by the same force. Then its elongation will be
(a) 0.005 m
(b) 0.01 m
(c) 0.02 m
(d) 0.002 m
Show Answer & Explanation

Answer: (a) 0.005 m

Explanation:
1. \( \Delta L=\dfrac{4FL}{\pi d^2Y} \).
2. Factor \( \dfrac{2}{2^2}=\dfrac12 \).

Teacher's Note:
1. Diameter enters as a square.

Question 16: A rod is fixed between two points at \( 20^\circ \)C. The coefficient of linear expansion of the material of the rod is \( 1.1\times10^{-5}\,/^\circ\)C and Young's modulus is \( 1.2\times10^{11} \) N/m\(^2\). The stress developed in the rod if the temperature rises to \( 30^\circ \)C is
(a) \( 1.32\times10^7 \) N/m\(^2\)
(b) \( 1.10\times10^{15} \) N/m\(^2\)
(c) \( 1.32\times10^8 \) N/m\(^2\)
(d) \( 1.10\times10^6 \) N/m\(^2\)
Show Answer & Explanation

Answer: (a) \( 1.32\times10^7 \) N/m\(^2\)

Explanation:
1. \( 1.2\times10^{11}\times1.1\times10^{-5}\times10=1.32\times10^7 \) N/m\(^2\).

Teacher's Note:
1. Thermal stress is independent of the rod's length and area.

Question 17: A rubber pipe of density \( 1.5\times10^3 \) kg/m\(^3\) and Young's modulus \( 5\times10^6 \) N/m\(^2\) is suspended from the roof. If the length of the pipe is 8 m, the change in length due to its own weight is (take g = 10 m/s\(^2\))
(a) 9.6 m
(b) \( 9.6\times10^{3} \) m
(c) \( 19.2\times10^{-2} \) m
(d) \( 9.6\times10^{-2} \) m
Show Answer & Explanation

Answer: (d) \( 9.6\times10^{-2} \) m

Explanation:
1. Weight of a hanging uniform rod acts at the mid-point, giving \( \Delta L=\dfrac{\rho gL^2}{2Y} \).
2. \( =\dfrac{1.5\times10^3\times10\times64}{2\times5\times10^6}=0.096 \) m.

Teacher's Note:
1. The factor 2 arises from the varying tension along the length.

Question 18: A force F is applied on a wire of radius \( r \) and length \( L \), and the change in length is \( l \). If the same force is applied on a wire of the same material and radius \( 2r \) and length \( 2L \), the change in length of the other wire is
(a) \( l \)
(b) \( 2l \)
(c) \( l/2 \)
(d) \( 4l \)
Show Answer & Explanation

Answer: (c) \( l/2 \)

Explanation:
1. Length doubles, area quadruples.

Teacher's Note:
1. Compare length and area changes separately.
2. Doubling the radius quadruples the area while doubling the length doubles the extension factor.

Question 19: A steel wire of 1 m length and \( 1\ \text{mm}^2 \) area of cross-section is hung from a rigid support. When a weight of 1 kg is hung from it, the change in length will be (\( Y=2\times10^{11} \) N/m\(^2\), g = 9.8 m/s\(^2\))
(a) 0.5 mm
(b) 0.25 mm
(c) 0.05 mm
(d) 5 mm
Show Answer & Explanation

Answer: (c) 0.05 mm

Explanation:
1. \( \Delta L=\dfrac{1\times9.8\times1}{10^{-6}\times2\times10^{11}}=4.9\times10^{-5} \) m = 0.05 mm.

Teacher's Note:
1. Weight in newton is mg.

Question 20: Two similar wires under the same load yield elongations of 0.1 mm and 0.05 mm respectively. If the area of cross-section of the first wire is \( 4\ \text{mm}^2 \), then the area of cross-section of the second wire is
(a) \( 6\ \text{mm}^2 \)
(b) \( 8\ \text{mm}^2 \)
(c) \( 10\ \text{mm}^2 \)
(d) \( 12\ \text{mm}^2 \)
Show Answer & Explanation

Answer: (b) \( 8\ \text{mm}^2 \)

Explanation:
1. Same load, length and material: \( \Delta L\propto1/A \).
2. \( A_2=A_1\dfrac{0.1}{0.05}=8 \) mm\(^2\).

Teacher's Note:
1. Half the elongation needs double the area.

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