JEE Physics Mechanical Properties Of Solids MCQs Set 09

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Multiple Choice Questions

Question 1: A steel ball initially at a pressure of \( 10^{5}\ \text{Pa} \) is heated from 20 °C to 120 °C keeping its volume constant. The final pressure inside the ball is closest to (\( \alpha=1.1\times10^{-5}\ /{}^\circ\text{C} \), bulk modulus \( B=1.6\times10^{11}\ \text{N/m}^2 \))
(a) \( 1.1\times10^{5}\ \text{Pa} \)
(b) \( 1.76\times10^{8}\ \text{Pa} \)
(c) \( 5.28\times10^{7}\ \text{Pa} \)
(d) \( 5.28\times10^{8}\ \text{Pa} \)
Show Answer & Explanation

Answer: (d) \( 5.28\times10^{8}\ \text{Pa} \)

Explanation:
1. Pressure needed to hold the volume fixed: \( \Delta P=B\gamma\Delta T=3\alpha\Delta T\,B \).
2. \( \Delta P=3(1.1\times10^{-5})(100)(1.6\times10^{11})=5.28\times10^{8}\ \text{Pa} \).
3. Final pressure \( =10^5+5.28\times10^8\approx5.28\times10^8\ \text{Pa} \).

Teacher's Note:
1. The initial \( 10^5 \) Pa is negligible compared with the thermal pressure.
2. Use \( \gamma=3\alpha \) for isotropic solids.

Question 2: When a certain weight is suspended from a long uniform wire, its length increases 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) 1 cm
(d) 4 cm
Show Answer & Explanation

Answer: (d) 4 cm

Explanation:
1. \( \Delta L\propto 1/d^2 \).
2. Half the diameter gives one-fourth the area, so four times the extension: 4 cm.

Teacher's Note:
1. Extension scales inversely with area.
2. Check units: only ratios matter here.

Question 3: Two wires of the same diameter and material have lengths \( \ell \) and \( 2\ell \). If the same force \( F \) is applied to each, the ratio of the work done in stretching them is
(a) 1 : 4
(b) 1 : 2
(c) 2 : 1
(d) 1 : 1
Show Answer & Explanation

Answer: (b) 1 : 2

Explanation:
1. Work \( W=\tfrac12F\,\Delta L \) and \( \Delta L=\dfrac{FL}{AY}\propto L \).
2. So \( W\propto L \) and the ratio is \( \ell:2\ell=1:2 \).

Teacher's Note:
1. With the same force the longer wire stretches more and stores more energy.
2. Do not use \( \tfrac12 F\Delta L \) with different forces.

Question 4: A rod of length 2 m is at 20 °C. It is prevented from expanding and its temperature is raised to 50 °C. Find the stress produced. (\( Y=2\times10^{11}\ \text{N/m}^2,\ \alpha=15\times10^{-6}\ /{}^\circ\text{C} \))
(a) \( 9\times10^{6}\ \text{N/m}^2 \)
(b) \( 4.5\times10^{7}\ \text{N/m}^2 \)
(c) \( 1.8\times10^{8}\ \text{N/m}^2 \)
(d) \( 9\times10^{7}\ \text{N/m}^2 \)
Show Answer & Explanation

Answer: (d) \( 9\times10^{7}\ \text{N/m}^2 \)

Explanation:
1. Thermal stress \( =Y\alpha\Delta T \).
2. \( =(2\times10^{11})(15\times10^{-6})(30)=9\times10^{7}\ \text{N/m}^2 \).

Teacher's Note:
1. The length of the rod does not appear in thermal stress.
2. The stress is compressive here.

Question 5: A brass wire 1.8 m long is held taut at 27 °C with little tension between two rigid supports. The wire is cooled to \( -39\,^\circ\text{C} \). Find the tension developed in the wire if its diameter is 2.0 mm. (\( \alpha=2.0\times10^{-5}\ /{}^\circ\text{C},\ Y=0.91\times10^{11}\ \text{Pa} \))
(a) \( 3.8\times10^{2}\ \text{N} \)
(b) \( 1.9\times10^{2}\ \text{N} \)
(c) \( 3.8\times10^{3}\ \text{N} \)
(d) \( 7.6\times10^{2}\ \text{N} \)
Show Answer & Explanation

Answer: (a) \( 3.8\times10^{2}\ \text{N} \)

Explanation:
1. Temperature change \( \Delta T=27-(-39)=66\,^\circ\text{C} \).
2. Area \( A=\pi(10^{-3})^2=3.14\times10^{-6}\ \text{m}^2 \).
3. Tension \( F=YA\alpha\Delta T=(0.91\times10^{11})(3.14\times10^{-6})(2\times10^{-5})(66)\approx3.8\times10^{2}\ \text{N} \).

Teacher's Note:
1. Use the radius, not the diameter, when finding the area.
2. The length is not needed for thermal tension.

Question 6: A bob of mass \( m \) hangs by a string from the ceiling of a smooth trolley moving with constant acceleration \( a \). The string has Young's modulus \( Y \), radius \( r \) and length \( l \). When the bob is at rest relative to the trolley, the stress in the string is
(a) \( \dfrac{m\sqrt{g^2+a^2}}{\pi r^2} \)
(b) \( \dfrac{m(g+a)}{\pi r^2} \)
(c) \( \dfrac{mg}{\pi r^2} \)
(d) \( \dfrac{ma}{\pi r^2} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{m\sqrt{g^2+a^2}}{\pi r^2} \)

Explanation:
1. In the trolley frame the bob is in equilibrium under gravity \( mg \), the pseudo force \( ma \) (horizontal) and the tension \( T \).
2. Hence \( T=m\sqrt{g^2+a^2} \).
3. Stress \( =T/A=\dfrac{m\sqrt{g^2+a^2}}{\pi r^2} \).

Teacher's Note:
1. The effective gravity is the vector sum of \( g \) and \( -a \).
2. The extension would then be stress times \( l/Y \).

Question 7: Two opposite forces \( F_1=120\ \text{N} \) and \( F_2=80\ \text{N} \) act on the two ends of an elastic plank of Young's modulus \( 2\times10^{11}\ \text{N/m}^2 \), length \( 1\ \text{m} \) and cross-section \( 0.5\ \text{m}^2 \) lying on a smooth horizontal surface. The change in length of the plank is \( x\times10^{-11}\ \text{m} \). The value of \( x \) is
(a) 200
(b) 40
(c) 50
(d) 100
Show Answer & Explanation

Answer: (d) 100

Explanation:
1. The tension varies linearly along the plank from \( F_2=80 \) N at one end to \( F_1=120 \) N at the other (the plank accelerates).
2. Mean tension \( =100\ \text{N} \).
3. \( \Delta l=\dfrac{(\bar F)\,l}{AY}=\dfrac{100\times1}{0.5\times2\times10^{11}}=10^{-9}\ \text{m}=100\times10^{-11}\ \text{m} \).

Teacher's Note:
1. For a body accelerating under unequal end forces use the average tension.
2. Integrating \( dT/dx \) along the length gives the same result.

Question 8: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. The elastic energy stored in the wire is
(a) 10 J
(b) 0.2 J
(c) 20 J
(d) 0.1 J
Show Answer & Explanation

Answer: (d) 0.1 J

Explanation:
1. Energy \( =\tfrac12F\,\Delta L=\tfrac12(200)(10^{-3})=0.1\ \text{J} \).

Teacher's Note:
1. Energy stored is half the work done by the final force acting over the whole stretch.
2. Convert mm to m.

Question 9: A wire fixed at the upper end stretches by length \( l \) when a force \( F \) is applied. The work done in stretching is
(a) \( Fl/2 \)
(b) \( Fl \)
(c) \( 2Fl \)
(d) \( F/2l \)
Show Answer & Explanation

Answer: (a) \( Fl/2 \)

Explanation:
1. The force rises linearly from 0 to \( F \) while the extension goes from 0 to \( l \).
2. Work \( =\tfrac12Fl \).

Teacher's Note:
1. Work equals the area of the triangle under the force-extension line.
2. Using \( Fl \) would assume a constant force.

Question 10: If \( S \) is the stress and \( Y \) is Young's modulus of the material of a wire, the energy stored in the wire per unit volume is
(a) \( 2S^2Y \)
(b) \( \dfrac{2Y}{S^2} \)
(c) \( \dfrac{S^2}{2Y} \)
(d) \( \dfrac{S}{2Y} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{S^2}{2Y} \)

Explanation:
1. \( u=\tfrac12\times\text{stress}\times\text{strain}=\tfrac12S\cdot\dfrac{S}{Y}=\dfrac{S^2}{2Y} \).

Teacher's Note:
1. Substitute strain \( =S/Y \).
2. Dimensions of \( S^2/Y \) are those of energy per volume.

Question 11: A wire elongates by \( l \) mm when a load \( W \) is hung from it. If the wire is passed over a pulley and two weights \( W \) each are hung at the two ends, the elongation of the wire will be (in mm)
(a) \( l/2 \)
(b) \( 2l \)
(c) \( l \)
(d) zero
Show Answer & Explanation

Answer: (c) \( l \)

Explanation:
1. Over a frictionless pulley the tension throughout the wire is \( W \).
2. Each part of the wire is under tension \( W \), exactly as when \( W \) is hung from one end with the other end fixed.
3. So the elongation is \( l \).

Teacher's Note:
1. The fixed support in the original case also supplies a reaction equal to \( W \).
2. Tension, not the total of the two loads, determines the stretch.

Question 12: Two wires are made of the same material and have the same volume. Wire 1 has cross-sectional area \( A \) and wire 2 has cross-sectional area \( 3A \). If the length of wire 1 increases by \( \Delta x \) on applying a force \( F \), how much force is needed to stretch wire 2 by the same amount?
(a) \( 4F \)
(b) \( 9F \)
(c) \( F \)
(d) \( 6F \)
Show Answer & Explanation

Answer: (b) \( 9F \)

Explanation:
1. Same volume: wire 2 has length \( L/3 \) if wire 1 has length \( L \).
2. Extension \( \Delta x=\dfrac{FL}{AY} \) for wire 1 and \( \Delta x=\dfrac{F_2(L/3)}{3AY}=\dfrac{F_2L}{9AY} \) for wire 2.
3. Equating: \( F_2=9F \).

Teacher's Note:
1. The force constant \( YA/L \) is multiplied by 3 for the area and by 3 for the shorter length.
2. Equal volume links the length to the area.

Question 13: A metal rod of Young's modulus \( Y \) and coefficient of thermal expansion \( \alpha \) is held at its two ends so that its length remains invariant. If its temperature is raised by \( t \) °C, the linear stress developed in it is
(a) \( Y\alpha t \)
(b) \( \dfrac{Y}{\alpha t} \)
(c) \( \dfrac{\alpha t}{Y} \)
(d) \( \dfrac{1}{Y\alpha t} \)
Show Answer & Explanation

Answer: (a) \( Y\alpha t \)

Explanation:
1. Prevented thermal strain \( =\alpha t \).
2. Stress \( =Y\times\alpha t=Y\alpha t \).

Teacher's Note:
1. Thermal stress = modulus times thermal strain.
2. The unit of \( Y\alpha t \) is N/m².

Question 14: A wooden wheel of radius \( R \) is made of two semicircular parts held together by a ring made of a metal strip of cross-sectional area \( S \) and length \( L \), where \( L \) is slightly less than \( 2\pi R \). To fit the ring on the wheel it is heated through \( \Delta T \) so that it just steps over the wheel. As it cools to the surrounding temperature it presses the semicircular parts together. If the coefficient of linear expansion of the metal is \( \alpha \) and its Young's modulus is \( Y \), the force that one part of the wheel applies on the other part is
(a) \( SY\alpha\Delta T \)
(b) \( 2SY\alpha\Delta T \)
(c) \( 2\pi SY\alpha\Delta T \)
(d) \( \pi SY\alpha\Delta T \)
Show Answer & Explanation

Answer: (b) \( 2SY\alpha\Delta T \)

Explanation:
1. On cooling, the ring is stretched by the wheel by the strain \( \alpha\Delta T \), so the ring tension is \( T=SY\alpha\Delta T \).
2. Consider half of the ring: two tensions \( T \) act at its ends and balance the force \( F \) exerted by the wheel half across the diameter.
3. So \( F=2T=2SY\alpha\Delta T \).

Teacher's Note:
1. For a ring or hoop under internal pressure, the force across a diameter equals twice the tension.
2. The strain needed is the thermal strain \( \alpha\Delta T \).

Question 15: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is (for steel, Young's modulus is \( 2\times10^{11}\ \text{N m}^{-2} \) and the coefficient of thermal expansion is \( 1.1\times10^{-5}\ \text{K}^{-1} \))
(a) \( 2.2\times10^{8}\ \text{Pa} \)
(b) \( 2.2\times10^{6}\ \text{Pa} \)
(c) \( 2.2\times10^{9}\ \text{Pa} \)
(d) \( 2.2\times10^{7}\ \text{Pa} \)
Show Answer & Explanation

Answer: (a) \( 2.2\times10^{8}\ \text{Pa} \)

Explanation:
1. Required stress \( =Y\alpha\Delta T=(2\times10^{11})(1.1\times10^{-5})(100)=2.2\times10^{8}\ \text{Pa} \).

Teacher's Note:
1. The length (10 cm) is a distractor and does not enter.
2. Thermal stress is independent of wire length.

Question 16: A pendulum made of a uniform wire of cross-sectional area \( A \) has time period \( T \). When an additional mass \( M \) is added to its bob, the time period changes to \( T_M \). If Young's modulus of the material of the wire is \( Y \), then \( 1/Y \) is equal to (\( g \) = gravitational acceleration)
(a) \( \left[1-\left(\dfrac{T_M}{T}\right)^2\right]\dfrac{A}{Mg} \)
(b) \( \left[1-\left(\dfrac{T}{T_M}\right)^2\right]\dfrac{A}{Mg} \)
(c) \( \left[\left(\dfrac{T_M}{T}\right)^2-1\right]\dfrac{A}{Mg} \)
(d) \( \left[\left(\dfrac{T_M}{T}\right)^2-1\right]\dfrac{Mg}{A} \)
Show Answer & Explanation

Answer: (c) \( \left[\left(\dfrac{T_M}{T}\right)^2-1\right]\dfrac{A}{Mg} \)

Explanation:
1. Adding \( M \) stretches the wire by \( \Delta L=\dfrac{MgL}{AY} \), so the pendulum length becomes \( L+\Delta L \).
2. Since \( T\propto\sqrt{L} \): \( \left(\dfrac{T_M}{T}\right)^2=\dfrac{L+\Delta L}{L}=1+\dfrac{Mg}{AY} \).
3. Hence \( \dfrac1Y=\left[\left(\dfrac{T_M}{T}\right)^2-1\right]\dfrac{A}{Mg} \).

Teacher's Note:
1. The period increases when the wire stretches, so the bracket must be positive.
2. Dimension check: \( A/Mg \) has the dimension of 1/modulus.

Question 17: A man grows into a giant such that all his linear dimensions increase by a factor of 9. Assuming his density remains the same, the stress in his legs will change by a factor of
(a) \( 1/81 \)
(b) \( 1/9 \)
(c) 81
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. Weight \( \propto\text{volume}\propto L^3 \) so it grows by \( 9^3 \).
2. Leg cross-section \( \propto L^2 \), so it grows by \( 9^2 \).
3. Stress \( \propto L^3/L^2=L \): it increases 9 times.

Teacher's Note:
1. This is the square-cube law: strength grows as area, weight as volume.
2. It explains why large animals need thicker legs.

Question 18: An external pressure \( P \) is applied on a cube at 0 °C so that it is equally compressed from all sides. \( K \) is the bulk modulus of the material of the cube and \( \alpha \) is its coefficient of linear expansion. The temperature to which the cube should be raised to bring it back to its original size is
(a) \( 3PK\alpha \)
(b) \( \dfrac{P}{\alpha K} \)
(c) \( \dfrac{P}{3\alpha K} \)
(d) \( \dfrac{3\alpha}{PK} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{P}{3\alpha K} \)

Explanation:
1. Volume compression \( \dfrac{\Delta V}{V}=\dfrac{P}{K} \).
2. Volume expansion on heating \( \dfrac{\Delta V}{V}=3\alpha\Delta T \).
3. Equate: \( \Delta T=\dfrac{P}{3\alpha K} \).

Teacher's Note:
1. Volume expansion coefficient is \( 3\alpha \).
2. Compression and thermal expansion act in opposite senses here.

Question 19: A granite rod of length 60 cm is clamped at its middle point and set into longitudinal vibrations. The density of granite is \( 2.7\times10^{3}\ \text{kg/m}^3 \) and its Young's modulus is \( 9.27\times10^{10}\ \text{Pa} \). The fundamental frequency of the longitudinal vibrations is
(a) 10 kHz
(b) 2.5 kHz
(c) 7.5 kHz
(d) 5 kHz
Show Answer & Explanation

Answer: (d) 5 kHz

Explanation:
1. Speed of longitudinal waves \( v=\sqrt{Y/\rho}=\sqrt{\dfrac{9.27\times10^{10}}{2.7\times10^{3}}}\approx5.86\times10^{3}\ \text{m/s} \).
2. The middle is a node and the two free ends are antinodes, so \( L=\dfrac{\lambda}{2} \) and \( \lambda=1.2\ \text{m} \).
3. Frequency \( f=\dfrac{v}{\lambda}\approx4.9\times10^{3}\ \text{Hz}\approx5\ \text{kHz} \).

Teacher's Note:
1. A rod clamped in the middle is the same as a pipe open at both ends with a node at the centre.
2. A free end of a rod is always an antinode of displacement.

Question 20: A solid sphere of radius \( r \) made of a soft material of bulk modulus \( K \) is surrounded by a liquid in a cylindrical container. A massless piston of area \( a \) floats on the surface of the liquid, covering the entire cross-section of the container. When a mass \( m \) is placed on the piston to compress the liquid, the fractional decrement in the radius of the sphere, \( \dfrac{dr}{r} \), is
(a) \( \dfrac{mg}{3Ka} \)
(b) \( \dfrac{Ka}{3mg} \)
(c) \( \dfrac{mg}{Ka} \)
(d) \( \dfrac{Ka}{mg} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{mg}{3Ka} \)

Explanation:
1. Extra pressure on the liquid and the sphere \( P=\dfrac{mg}{a} \).
2. Volume strain \( \dfrac{dV}{V}=-\dfrac{P}{K} \) and \( V\propto r^3\Rightarrow\dfrac{dV}{V}=3\dfrac{dr}{r} \).
3. Hence \( \dfrac{dr}{r}=\dfrac{P}{3K}=\dfrac{mg}{3Ka} \) (magnitude of the decrease).

Teacher's Note:
1. The fractional change in radius is one-third of the fractional change in volume.
2. The piston is massless, so the pressure comes only from the added mass.

Question 21: A steel wire of length \( L \) at 40 °C is suspended from the ceiling and a mass \( m \) is hung from its free end. The wire is cooled from 40 °C to 30 °C to regain its original length \( L \). The coefficient of linear thermal expansion of steel is \( 10^{-5}/{}^\circ\text{C} \), Young's modulus of steel is \( 10^{11}\ \text{N/m}^2 \) and the radius of the wire is 1 mm. Assume \( L \) is much larger than the diameter and take \( g=10\ \text{m/s}^2 \). The value of \( m \) in kg is nearly
(a) 6
(b) 10
(c) 3
(d) 1
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Cooling shortens the wire by strain \( \alpha\Delta T=10^{-5}\times10=10^{-4} \); the load stretches it by the same strain.
2. Hence \( \dfrac{mg}{AY}=\alpha\Delta T\Rightarrow m=\dfrac{AY\alpha\Delta T}{g} \).
3. \( m=\dfrac{\pi(10^{-3})^2(10^{11})(10^{-4})}{10}\approx3.14\ \text{kg}\approx3\ \text{kg} \).

Teacher's Note:
1. The wire regains its length when the thermal contraction equals the elastic stretch.
2. Here the length \( L \) cancels.

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