JEE Physics Mechanical Properties Of Solids MCQs Set 17

Answer: (c) 2.00

Explanation:
1. Same force acts on both wires, same material, so \( \Delta L = \dfrac{FL}{\pi r^2 Y} \propto \dfrac{L}{r^2} \).
2. Thin wire: \( \dfrac{L}{R^2} \). Thick wire: \( \dfrac{2L}{4R^2} = \dfrac{L}{2R^2} \).
3. Ratio \( = \dfrac{L/R^2}{L/(2R^2)} = 2 \). Answer: 2.00.

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

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

Explanation:
1. Thermal stress \( = Y\alpha\,\Delta T \).
2. \( = 2\times10^{11}\times1.1\times10^{-5}\times100 = 2.2\times10^{8} \) Pa.

Teacher's Note:
1. A rod that is not allowed to expand develops stress \( Y\alpha\Delta T \), independent of its length.
2. The length 10 cm is only a distractor.

Answer: (d) 1.8 : 1

Explanation:
1. Same load and same extension: \( Y \propto \dfrac{L}{A} \).
2. \( \dfrac{Y_s}{Y_c} = \dfrac{4.7/3.0}{3.5/4.0} = \dfrac{1.567}{0.875} \approx 1.8 \).
3. Ratio is 1.8 : 1.

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (d) 0.02

Explanation:
1. Strain \( = \dfrac{\Delta L}{L} \) = 2% = 0.02.

Teacher's Note:
1. Strain is a pure ratio, so a 2% increase means 0.02.
2. The load value is extra information here.

Answer: (a) \( 0.5\,YX^{2} \)

Explanation:
1. Energy per unit volume \( = \tfrac12\times\text{stress}\times\text{strain} \).
2. Stress \( = YX \), so energy density \( = \tfrac12 YX^{2} \).

Teacher's Note:
1. Energy density \( = \tfrac12\sigma\varepsilon = \tfrac12 Y\varepsilon^2 = \sigma^2/2Y \).
2. Memorise all three forms.

Answer: (b) 1 mm

Explanation:
1. Tension in the wire \( T = \dfrac{2m_1m_2g}{m_1+m_2} = \dfrac{2\times1\times2\times10}{3} = \dfrac{40}{3} \) N.
2. Minimum area \( A = \dfrac{T}{\text{breaking stress}} = \dfrac{40/3}{(40/3\pi)\times10^{6}} = \pi\times10^{-6}\ \text{m}^2 \).
3. \( \pi r^2 = \pi\times10^{-6} \Rightarrow r = 10^{-3} \) m = 1 mm.

Teacher's Note:
1. In an Atwood arrangement the wire carries the tension \( T = 2m_1m_2g/(m_1+m_2) \), not the sum of the weights.
2. Radius follows from \( A = T/\text{stress} \).

Answer: (c) \( 2.2\times10^{8}\ \text{N/m}^2 \)

Explanation:
1. Stress \( = \dfrac{F}{\pi R^2} = \dfrac{6.2\times10^{4}}{\pi(9.5\times10^{-3})^2} \).
2. \( = \dfrac{6.2\times10^{4}}{2.835\times10^{-4}} \approx 2.2\times10^{8}\ \text{N/m}^2 \).

Teacher's Note:
1. Convert the radius to metres before squaring.
2. The length of the rod is not needed for stress.

Answer: (d) \( L = 100 \) cm, \( r = 0.2 \) mm

Explanation:
1. \( \Delta L \propto \dfrac{L}{r^2} \).
2. (a) \( 400/0.64 = 625 \); (b) \( 300/0.36 = 833 \); (c) \( 200/0.16 = 1250 \); (d) \( 100/0.04 = 2500 \).
3. The largest value is for (d).

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (c) 1.36%

Explanation:
1. Pressure at depth \( = \rho g h = 1000\times10\times3000 = 3\times10^{7} \) Pa.
2. \( \dfrac{\Delta V}{V} = \dfrac{p}{K} = \dfrac{3\times10^{7}}{2.2\times10^{9}} = 0.0136 \) = 1.36%.

Teacher's Note:
1. Bulk strain \( = p/K \).
2. Take water density as 1000 kg/m\( ^3 \).

Answer: (a) 4

Explanation:
1. Wire acts as a spring of constant \( k = \dfrac{YA}{l} \), and \( \omega^2 = \dfrac{k}{m} \).
2. \( Y = \dfrac{m\omega^2 l}{A} = \dfrac{0.1\times19600\times1}{4.9\times10^{-7}} = 4\times10^{9} \).
3. So \( n = 4 \).

Teacher's Note:
1. A stretched wire behaves like a spring with \( k = YA/l \).
2. Combine this with \( \omega = \sqrt{k/m} \).

Answer: (d) \( 9F \)

Explanation:
1. Same volume: \( l_2 = l_1/3 \) when area is \( 3A \).
2. \( \Delta x = \dfrac{F l_1}{AY} \) for wire 1 and \( \Delta x = \dfrac{F_2 (l_1/3)}{3AY} = \dfrac{F_2 l_1}{9AY} \) for wire 2.
3. Equating, \( F_2 = 9F \).

Teacher's Note:
1. Equal volume means area up by 3 gives length down by 3.
2. The force scales as \( A^2 \).

Answer: (c) 4500 N

Explanation:
1. Breaking force \( \propto \) area \( \propto d^2 \).
2. \( F = 500\times\left(\dfrac{3}{1}\right)^2 = 4500 \) N.

Teacher's Note:
1. Breaking stress depends on the material, so breaking force goes as the cross-section.
2. Diameter ratio 3 gives area ratio 9.

Answer: (b) \( \dfrac{Mg}{3AK} \)

Explanation:
1. Extra pressure on the sphere \( p = \dfrac{Mg}{A} \).
2. Volume strain \( \dfrac{\Delta V}{V} = \dfrac{p}{K} = \dfrac{Mg}{AK} \).
3. \( V \propto R^3 \Rightarrow \dfrac{\Delta R}{R} = \dfrac13\dfrac{\Delta V}{V} = \dfrac{Mg}{3AK} \).

Teacher's Note:
1. For a sphere \( \Delta V/V = 3\,\Delta R/R \).
2. Pressure in a liquid is transmitted equally in all directions.

Answer: (d) 168 kPa

Explanation:
1. \( \dfrac{\Delta V}{V} = 0.008\% = 8\times10^{-5} \).
2. \( \Delta p = K\dfrac{\Delta V}{V} = 2.1\times10^{9}\times8\times10^{-5} = 1.68\times10^{5} \) Pa = 168 kPa.

Teacher's Note:
1. Do not use the 200 L volume; only the fractional change matters.
2. Watch the percentage-to-fraction conversion.

Answer: (a) \( L = 100 \) cm, \( r = 1 \) mm

Explanation:
1. \( \Delta L \propto L/r^2 \).
2. (a) 100; (b) \( 200/9 = 22.2 \); (c) \( 300/9 = 33.3 \); (d) \( 400/16 = 25 \).
3. Largest is (a).

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (d) \( Y = 2\eta(1+\sigma) \)

Explanation:
1. The standard relation between the elastic constants is \( Y = 2\eta(1+\sigma) \).
2. The form \( Y = 3K(1-2\sigma) \) is the one involving bulk modulus.

Teacher's Note:
1. Remember \( Y = 2\eta(1+\sigma) = 3K(1-2\sigma) \).
2. Do not mix the two.

Answer: (b) Steel

Explanation:
1. Elasticity is measured by the modulus of elasticity.
2. Young's modulus: steel about \( 2\times10^{11} \), copper about \( 1.2\times10^{11} \) Pa; rubber and sponge are far lower.
3. Steel is the most elastic.

Teacher's Note:
1. Elasticity means resisting deformation, not stretching easily.
2. Steel has a higher Young's modulus than copper.

Answer: (a) \( l \)

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

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (c) 2.3

Explanation:
1. Same \( L \), same load, same extension: \( Y d^2 = \) constant.
2. \( d_{Cu} = 3\sqrt{\dfrac{7}{12}} = 3\times0.764 = 2.29 \) mm.

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (b) 1 : 1

Explanation:
1. Strain \( = \dfrac{\text{stress}}{Y} = \dfrac{F}{AY} \).
2. Same force, area and material give the same strain in both wires, ratio 1 : 1.

Teacher's Note:
1. Strain depends on stress and Y, not on length.
2. Extension, not strain, depends on length.

Answer: (b) \( l \)

Explanation:
1. \( \Delta L \propto \dfrac{FL}{a^2} \).
2. Factor \( = \dfrac{2\times2}{4} = 1 \), so the increase is \( l \).

Teacher's Note:
1. Young's modulus links stress and strain: \( Y = \dfrac{FL}{A\,\Delta L} \).
2. Always compare wires through \( L/r^2 \) when material and force are the same.

Answer: (b) 60 kg

Explanation:
1. \( F = YA\varepsilon = 2\times10^{11}\times3\times10^{-6}\times10^{-3} = 600 \) N.
2. \( m = F/g = 60 \) kg.

Teacher's Note:
1. Maximum load \( = Y A\times\) maximum strain.
2. Divide by g to get the mass.

Answer: (d) Rubber, glass, copper, steel

Explanation:
1. Approximate Young's moduli: rubber \( \sim10^{7}\text{--}10^{8} \), glass \( \sim6\times10^{10} \), copper \( \sim1.2\times10^{11} \), steel \( \sim2\times10^{11} \) Pa.
2. Increasing order: rubber, glass, copper, steel.

Teacher's Note:
1. Steel is more elastic than copper.
2. Rubber is the least elastic in this sense because it stretches easily.

Answer: (b) \( 7.5\times10^{8}\ \text{N m}^{-2} \)

Explanation:
1. \( \Delta p = K\dfrac{\Delta V}{V} = 7.5\times10^{10}\times0.01 = 7.5\times10^{8}\ \text{N m}^{-2} \).

Teacher's Note:
1. Bulk modulus \( K = -\Delta p/(\Delta V/V) \).
2. 1% means 0.01.

Answer: (c) infinity

Explanation:
1. An incompressible liquid has \( \Delta V = 0 \) for any pressure change.
2. \( K = -V\Delta p/\Delta V \to \infty \).

Teacher's Note:
1. Large K means hard to compress.
2. An ideal incompressible liquid has infinite bulk modulus.

Answer: (b) \( \dfrac{\rho g L^2}{2Y} \)

Explanation:
1. Take a small element \( dx \) at distance \( x \) from the free end; the weight below it is \( \rho A x g \).
2. \( d(\Delta L) = \dfrac{\rho A x g\,dx}{AY} \), so \( \Delta L = \dfrac{\rho g}{Y}\int_0^L x\,dx = \dfrac{\rho g L^2}{2Y} \).

Teacher's Note:
1. The weight of the rod acts effectively at the mid-point, which gives the factor of 2.
2. The result does not depend on radius.

Answer: (b) remains constant

Explanation:
1. Within the elastic limit stress \( \propto \) strain.
2. The ratio is Young's modulus, a constant.

Teacher's Note:
1. Hooke's law holds only below the proportional limit.
2. The ratio is the modulus.