JEE Physics Mechanical Properties of Solids MCQs Set 18

Physics Objective Questions and Answers: Mechanical Properties Of Solids

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Multiple Choice Questions

Question 1: A weight suspended from a long uniform wire increases its length by 1 cm. If the same weight is suspended from another wire of the same material and length but with half the diameter, the increase in length will be
(a) 0.5 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm
Show Answer & Explanation

Answer: (c) 4 cm

Explanation:
1. \( \Delta l=\dfrac{FL}{AY}\propto\dfrac1{d^2} \).
2. Halving the diameter makes the area one quarter, so the extension becomes 4 times.
3. \( \Delta l=4 \) cm.

Teacher's Note:
1. For the same load, length and material, extension varies inversely with the square of the diameter.

Question 2: The ratio of diameters of two wires of the same material is n : 1 and the length of each is 4 m. On applying the same load, the increase in length of the thin wire will be (n > 1)
(a) \( n^2\ \text{times that of the thick wire} \)
(b) \( n\ \text{times that of the thick wire} \)
(c) \( 2n\ \text{times that of the thick wire} \)
(d) \( (2n+1)\ \text{times that of the thick wire} \)
Show Answer & Explanation

Answer: (a) \( n^2\ \text{times that of the thick wire} \)

Explanation:
1. \( \Delta l\propto1/d^2 \).
2. \( \dfrac{\Delta l_{thin}}{\Delta l_{thick}}=\dfrac{n^2}{1} \).

Teacher's Note:
1. The thin wire has an area smaller by n squared, so its extension is n squared times larger.

Question 3: A thick rubber rope of density \( 1.5\times10^3 \) kg/m\(^3\) and Young's modulus \( 5\times10^6 \) N/m\(^2\), 8 m long, is hung from the ceiling of a room. The increase in its length due to its own weight is (g = 10 m/s\(^2\))
(a) \( 9.6\times10^{-2} \) m
(b) \( 19.2\times10^{-2} \) m
(c) \( 9.6\times10^{-3} \) m
(d) 9.6 m
Show Answer & Explanation

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

Explanation:
1. For a hanging rope, \( \Delta l=\dfrac{\rho gL^2}{2Y} \).
2. \( \Delta l=\dfrac{1.5\times10^3\times10\times64}{2\times5\times10^6}=9.6\times10^{-2} \) m.

Teacher's Note:
1. The tension varies along the rope, so the extension is half that of a rope loaded only at the bottom with the full weight.

Question 4: A copper wire of negligible mass, 1 m long and of cross-section \( 10^{-6} \) m\(^2\), lies on a smooth horizontal table with one end fixed. A ball of mass 1 kg is attached to the other end, and the wire and ball rotate with angular velocity 20 rad/s. If the elongation in the wire is \( 10^{-3} \) m, Young's modulus is
(a) \( 6\times10^{11}\ \text{N/m}^2 \)
(b) \( 4\times10^{11}\ \text{N/m}^2 \)
(c) \( 8\times10^{11}\ \text{N/m}^2 \)
(d) \( 10\times10^{11}\ \text{N/m}^2 \)
Show Answer & Explanation

Answer: (b) \( 4\times10^{11}\ \text{N/m}^2 \)

Explanation:
1. Tension supplies the centripetal force: \( T=m\omega^2L=1\times400\times1=400 \) N.
2. \( Y=\dfrac{TL}{A\Delta l}=\dfrac{400\times1}{10^{-6}\times10^{-3}} \).
3. \( Y=4\times10^{11} \) N/m\(^2\).

Teacher's Note:
1. The wire tension provides the centripetal force.
2. Then apply the definition of Young's modulus.

Question 5: A steel cable with a radius of 1.5 cm supports a chair lift. If the maximum stress is not to exceed \( 10^8 \) N/m\(^2\), the maximum load the cable can support is
(a) \( 7\times10^5\ \text{N} \)
(b) \( 7\times10^6\ \text{N} \)
(c) \( 7\times10^4\ \text{N} \)
(d) \( 9\times10^5\ \text{N} \)
Show Answer & Explanation

Answer: (c) \( 7\times10^4\ \text{N} \)

Explanation:
1. \( A=\pi r^2=\pi(1.5\times10^{-2})^2=7.07\times10^{-4} \) m\(^2\).
2. \( F=\text{stress}\times A=10^8\times7.07\times10^{-4}=7.07\times10^4 \) N.

Teacher's Note:
1. Maximum load equals the permitted stress times the cross-sectional area.
2. Convert the radius to metres.

Question 6: A 1 m long wire is stretched without tension at 30°C between two rigid supports. The strain produced in the wire when the temperature falls to 0°C is (\( \alpha=12\times10^{-6}\,\text{K}^{-1} \))
(a) \( 36\times10^{-5} \)
(b) \( 64\times10^{-5} \)
(c) 0.78
(d) 0.32
Show Answer & Explanation

Answer: (a) \( 36\times10^{-5} \)

Explanation:
1. The wire would contract by \( \alpha\Delta T \) but the supports prevent it.
2. Strain \( =12\times10^{-6}\times30=36\times10^{-5} \).

Teacher's Note:
1. The thermal strain that is prevented appears as mechanical strain in the wire.

Question 7: If x is the longitudinal strain produced in a wire of Young's modulus Y, the energy stored per unit volume of the wire is
(a) \( \tfrac12Yx^2 \)
(b) \( Yx^2 \)
(c) \( 2Yx^2 \)
(d) \( \tfrac12Y^2x \)
Show Answer & Explanation

Answer: (a) \( \tfrac12Yx^2 \)

Explanation:
1. Energy density \( =\tfrac12\times\text{stress}\times\text{strain} \).
2. With stress \( =Yx \): \( u=\tfrac12Yx^2 \).

Teacher's Note:
1. Elastic energy density is half the product of stress and strain.

Question 8: Two identical wires hang from the same rigid support, one of copper and the other of iron. Young's modulus of iron is three times that of copper. The weights to be added to the copper and iron wires so that their ends are on the same level must be in the ratio
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 1 : 3
Show Answer & Explanation

Answer: (d) 1 : 3

Explanation:
1. Same length and area, so \( \Delta l=\dfrac{WL}{AY} \) must be equal.
2. \( W\propto Y \), so \( W_{Cu}:W_{Fe}=Y_{Cu}:Y_{Fe}=1:3 \).

Teacher's Note:
1. Equal extensions in identical wires need loads proportional to Young's modulus.

Question 9: The temperature of a wire of length 1 m and cross-section 1 cm\(^2\) is increased from 0°C to 100°C. If the wire is not allowed to expand, the force required is (\( \alpha=10^{-5}\,/^\circ\)C, \( Y=10^{11} \) N/m\(^2\))
(a) \( 10^4\ \text{N} \)
(b) \( 10^3\ \text{N} \)
(c) \( 10^5\ \text{N} \)
(d) \( 10^2\ \text{N} \)
Show Answer & Explanation

Answer: (a) \( 10^4\ \text{N} \)

Explanation:
1. \( F=YA\alpha\Delta T \).
2. \( F=10^{11}\times10^{-4}\times10^{-5}\times100=10^4 \) N.

Teacher's Note:
1. Preventing thermal expansion creates a thermal stress equal to Y times alpha times the temperature change.

Question 10: The dimensions of four wires of the same material are given below. The increase in length for the same load will be maximum in the wire with
(a) length 50 cm, diameter 0.5 mm
(b) length 100 cm, diameter 1 mm
(c) length 200 cm, diameter 2 mm
(d) length 300 cm, diameter 3 mm
Show Answer & Explanation

Answer: (a) length 50 cm, diameter 0.5 mm

Explanation:
1. \( \Delta l\propto L/d^2 \).
2. Values: 100/1 = 100; 200/4 = 50; 300/9 = 33; 50/0.25 = 200.
3. The last wire gives the largest value.

Teacher's Note:
1. Compare the ratio of length to diameter squared; the thinnest wire dominates here.

Question 11: The length of a steel wire of cross-section 2 mm\(^2\) and length 2 m increases by 0.5 mm. The force required is (Y for steel = \( 2.2\times10^{11} \) N/m\(^2\))
(a) \( 1.1\times10^5\ \text{N} \)
(b) \( 1.1\times10^2\ \text{N} \)
(c) \( 1.1\times10^4\ \text{N} \)
(d) \( 1.1\times10^3\ \text{N} \)
Show Answer & Explanation

Answer: (b) \( 1.1\times10^2\ \text{N} \)

Explanation:
1. \( F=\dfrac{YA\Delta l}{L}=\dfrac{2.2\times10^{11}\times2\times10^{-6}\times0.5\times10^{-3}}{2} \).
2. \( F=1.1\times10^2 \) N.

Teacher's Note:
1. Use Y = F L / (A delta l) rearranged for the force, with all lengths in metres.

Question 12: Two wires of the same material and length are stretched by the same force. Their masses are in the ratio 3 : 2. Their elongations are in the ratio
(a) 3 : 2
(b) 9 : 4
(c) 2 : 3
(d) 4 : 9
Show Answer & Explanation

Answer: (c) 2 : 3

Explanation:
1. Same material and length, so mass is proportional to area: \( A_1:A_2=3:2 \).
2. \( \Delta l\propto1/A \), so \( \Delta l_1:\Delta l_2=2:3 \).

Teacher's Note:
1. Equal length and density make mass proportional to area, and extension is inversely proportional to area.

Question 13: A 100 N force stretches a hanging wire by 0.5 mm. The force required to stretch a wire of the same material and length but with four times the diameter by 0.5 mm is
(a) 100 N
(b) 1600 N
(c) 400 N
(d) 1200 N
Show Answer & Explanation

Answer: (b) 1600 N

Explanation:
1. For the same extension, \( F\propto A\propto d^2 \).
2. \( F=100\times16=1600 \) N.

Teacher's Note:
1. For equal extension in the same material and length the force is proportional to the area.

Question 14: Two wires of the same length and material, with radii in the ratio 1 : 2, are stretched by unequal forces to produce equal elongation. The ratio of the forces is
(a) 1 : 4
(b) 1 : 1
(c) 1 : 2
(d) 2 : 3
Show Answer & Explanation

Answer: (a) 1 : 4

Explanation:
1. \( F\propto A\propto r^2 \) for equal elongation.
2. \( F_1:F_2=1:4 \).

Teacher's Note:
1. Force for a given elongation is proportional to area, so radius squared.

Question 15: Two wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio 1 : 2. If they are stretched by equal forces, the increases in their lengths are in the ratio
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show Answer & Explanation

Answer: (c) 2 : 1

Explanation:
1. \( \Delta l\propto L/r^2 \).
2. \( \dfrac{\Delta l_1}{\Delta l_2}=\dfrac{1/1}{2/4}=2 \).

Teacher's Note:
1. Extension is directly proportional to length and inversely proportional to radius squared.

Question 16: A wire of length L and radius r is clamped at one end. When the other end is pulled by a force F, its length increases by l. Another wire of the same material of length 4L and radius 4r is pulled by a force 4F. The increase in length will be
(a) l/2
(b) l
(c) 2l
(d) 4l
Show Answer & Explanation

Answer: (b) l

Explanation:
1. \( \Delta l=\dfrac{FL}{\pi r^2Y} \).
2. New: \( \dfrac{(4F)(4L)}{\pi(16r^2)Y}=\dfrac{FL}{\pi r^2Y}=l \).

Teacher's Note:
1. The factor 16 in the numerator cancels the factor 16 from the squared radius.

Question 17: A weight of 5 kg is suspended from a copper wire of length 30 m and diameter 0.5 mm, and the length increases by 2.4 cm. If the diameter is doubled, the extension produced is
(a) 1.2 cm
(b) 0.6 cm
(c) 0.3 cm
(d) 0.15 cm
Show Answer & Explanation

Answer: (b) 0.6 cm

Explanation:
1. \( \Delta l\propto1/d^2 \).
2. Doubling the diameter divides the extension by 4: \( 2.4/4=0.6 \) cm.

Teacher's Note:
1. Same load, material and length: extension depends only on the diameter, as one over diameter squared.

Question 18: The length of a wire increases by 1 mm on applying a given load. In a wire of the same material but with length and radius twice that of the first, the extension under the same load is
(a) 0.5 mm
(b) 0.25 mm
(c) 2 mm
(d) 4 mm
Show Answer & Explanation

Answer: (a) 0.5 mm

Explanation:
1. \( \Delta l\propto L/r^2 \).
2. Factor \( =2/4=\tfrac12 \), so \( \Delta l=0.5 \) mm.

Teacher's Note:
1. Doubling the length doubles the extension, but doubling the radius divides it by four.

Question 19: An aluminium rod (Young's modulus \( 7.0\times10^9 \) N/m\(^2\)) has a breaking strain of 0.2%. The minimum cross-sectional area of the rod needed to support a load of \( 10^4 \) N is
(a) \( 7.1\times10^{-4}\ \text{m}^2 \)
(b) \( 1\times10^{-2}\ \text{m}^2 \)
(c) \( 1.4\times10^{-3}\ \text{m}^2 \)
(d) \( 1.0\times10^{-3}\ \text{m}^2 \)
Show Answer & Explanation

Answer: (a) \( 7.1\times10^{-4}\ \text{m}^2 \)

Explanation:
1. Breaking stress \( =Y\times\text{strain}=7\times10^9\times2\times10^{-3}=1.4\times10^7 \) N/m\(^2\).
2. \( A=\dfrac{F}{\text{stress}}=\dfrac{10^4}{1.4\times10^7}=7.1\times10^{-4} \) m\(^2\).

Teacher's Note:
1. Convert the breaking strain to a breaking stress, then divide the load by it.

Question 20: A material breaks under a stress of \( 10^6 \) N/m\(^2\). If its density is \( 3\times10^3 \) kg/m\(^3\), the length of a wire of this substance that will break under its own weight when hung vertically is (g = 10 m/s\(^2\))
(a) 66.6 m
(b) 60.0 m
(c) 30.0 m
(d) 33.3 m
Show Answer & Explanation

Answer: (d) 33.3 m

Explanation:
1. At the top the stress is \( \rho gL \).
2. \( L=\dfrac{10^6}{3\times10^3\times10}=33.3 \) m.

Teacher's Note:
1. The stress at the support is the full weight over the area, which is density times g times length.

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