JEE Physics Mechanical Properties Of Solids MCQs Set 05

Answer: (d) 4 x 10⁻³ m

Explanation:
1. \( \Delta L=\frac{FL}{AY}=\frac{250\times100}{6.25\times10^{-4}\times10^{10}}=\frac{25000}{6.25\times10^{6}}=4\times10^{-3}\ \text{m} \).

Teacher's Note:
1. Young's modulus Y = (F/A)/(ΔL/L), so ΔL = FL/(AY).
2. Check units: N, m, m² and N m⁻² give metres.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Steel has a large Young's modulus (it is more elastic than rubber or copper in the sense of resisting deformation) and a high elastic limit.
2. These properties let steel bear heavy loads without permanent deformation, so R explains A.

Teacher's Note:
1. A larger Young's modulus means a smaller strain for a given stress.
2. Structures must stay within the elastic limit of the material.

Answer: (a) 5

Explanation:
1. Slope \( \frac{\Delta L}{F}=\tan45^\circ=1\ \text{m/N} \).
2. \( Y=\frac{FL}{A\,\Delta L}=\frac{L}{A\times\text{slope}}=\frac{0.628}{\pi(2\times10^{-3})^2}=\frac{0.628}{1.2566\times10^{-5}}=5\times10^{4}\ \text{N m}^{-2} \).
3. So x = 5.

Teacher's Note:
1. The slope of extension against load is L/(AY).
2. Radius is half of the diameter: 2 mm.

Answer: (a) \( \sigma=\frac{3K-2\eta}{6K+2\eta} \)

Explanation:
1. Using \( Y=3K(1-2\sigma)=2\eta(1+\sigma) \).
2. \( 3K-6K\sigma=2\eta+2\eta\sigma\Rightarrow\sigma(6K+2\eta)=3K-2\eta \).
3. \( \sigma=\frac{3K-2\eta}{6K+2\eta} \).

Teacher's Note:
1. Write Young's modulus in two ways and eliminate it.
2. For a stable solid, Poisson's ratio lies between -1 and 0.5.

Answer: (d) 6.9 x 10⁻² mm

Explanation:
1. \( \Delta L=\frac{FL}{AY}\propto\frac{L}{d^2} \) (same F and Y).
2. \( \Delta L_B=0.2\times\frac{2}{(2.4)^2}=\frac{0.4}{5.76}=0.0694\ \text{mm}=6.9\times10^{-2}\ \text{mm} \).

Teacher's Note:
1. Area is proportional to the square of the diameter.
2. Same material and same force, so only L and d change.

Answer: (a) 60 kg

Explanation:
1. Thermal contraction \( \Delta L=\alpha L\Delta T=2\times10^{-5}\times1\times50=10^{-3}\ \text{m} \).
2. The load must stretch the rod by the same amount: \( \frac{Mg\,L}{AY}=10^{-3} \).
3. \( Mg=10^{-3}\times3\times10^{-6}\times2\times10^{11}=600\ \text{N}\Rightarrow M=60\ \text{kg} \).

Teacher's Note:
1. The rod regains its original length, so contraction equals extension.
2. Thermal stress = Y α ΔT, so Mg/A = Y α ΔT.

Answer: (d) 6.32 x 10³ m s⁻¹

Explanation:
1. \( v=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{3.2\times10^{11}}{8\times10^{3}}}=\sqrt{4\times10^{7}}=6.32\times10^{3}\ \text{m/s} \).

Teacher's Note:
1. Longitudinal waves in a rod travel at √(Y/ρ).
2. √(4 x 10⁷) = 2 x 3162 = 6324.

Answer: (b) 1 : 1

Explanation:
1. \( \Delta L=\frac{FL}{AY}\Rightarrow Y\propto\frac{L}{A} \) for equal F and equal \( \Delta L \).
2. \( \frac{Y_A}{Y_B}=\frac{5/2.5\times10^{-5}}{6/3\times10^{-5}}=\frac{2\times10^5}{2\times10^5}=1 \).

Teacher's Note:
1. With the same force and extension, Y is proportional to L/A.
2. Both L/A values come out equal.

Answer: (a) 1

Explanation:
1. Bulk modulus \( B=\frac{P}{\Delta V/V} \): for the same pressure, B is inversely proportional to the fractional compression.
2. \( \frac{B_w}{B_l}=\frac{0.03}{0.01}=3=\frac3x \), so x = 1.

Teacher's Note:
1. Use the fractional (percentage) change in volume, not the absolute change.
2. The volumes 1 L and 2 L only matter if absolute changes are used.

Answer: (c) 0.1 mm

Explanation:
1. Planet gravity \( g_p=2.5\ \text{m/s}^2 \), so the force is \( F=4\times2.5=10\ \text{N} \).
2. \( \Delta L=\frac{FL}{AY}=\frac{10\times6}{3\times10^{-6}\times2\times10^{11}}=\frac{60}{6\times10^5}=10^{-4}\ \text{m}=0.1\ \text{mm} \).

Teacher's Note:
1. Weight on the planet is mg/4.
2. Convert 3 mm² into 3 x 10⁻⁶ m².