JEE Physics 2023 Question with Solution

Given: Length of rod $\ell = 1 \, \text{m}$, area $A = 3 \times 10^{-6} \, \text{m}^2$, coefficient of linear expansion $\alpha = 2 \times 10^{-5} \, K^{-1}$, Young's modulus $Y = 2 \times 10^{11} \, N/m^2$, temperature fall $\Delta T = 210 - 160 = 50 \, \text{K}$, and $g = 10 \, \text{m/s}^2$.

Find: The mass $M$ required so that the rod regains length $1 \, \text{m}$ after cooling.

If $\Delta \ell$ is decease in length of rod due to decease in temperature,

$$\Delta \ell = \ell \alpha \Delta T$$ $$\Delta \ell = 1 \times 2 \times 10^{-5} \times 50 = 10^{-3} \, \text{m}$$

Young Modulus

$$Y = \frac{\frac{F}{A}}{\frac{\Delta \ell}{\ell}}$$

Using $F = Mg$ and $A = 3 \times 10^{-6} \, \text{m}^2$,

$$2 \times 10^{11} = \frac{\frac{Mg}{3 \times 10^{-6}}}{\frac{10^{-3}}{1}}$$ $$Mg = 2 \times 10^{11} \times 3 \times 10^{-9}$$ $$Mg = 6 \times 10^{2}$$ $$M \times 10 = 600$$ $$M = 60 \, \text{kg}$$

Therefore, the value of the mass is $60 \, \text{kg}$.

Given: The rod first contracts on cooling, and then stretches due to the attached mass. The final length again becomes the original $1 \, \text{m}$.

Find: Mass $M$ for which elastic extension equals thermal contraction.

The change in length due to temperature change is given by

$$\Delta \ell = \ell \alpha \Delta T$$

where

$$\alpha = 2 \times 10^{-5} \, \text{K}^{-1}, \quad \Delta T = 210^\circ \text{C} - 160^\circ \text{C} = 50 \, \text{K}$$

Hence,

$$\Delta \ell = 1 \times 2 \times 10^{-5} \times 50 = 10^{-3} \, \text{m}$$

Now use Young's modulus relation as written in the solution:

$$Y = \frac{F}{A} \times \frac{\Delta \ell}{\ell}$$

Substituting $F = Mg$,

$$2 \times 10^{11} = \frac{Mg}{3 \times 10^{-6}} \times \frac{10^{-3}}{1}$$

This gives

$$Mg = 2 \times 10^{11} \times 3 \times 10^{-6} \times 10^{-3}$$ $$Mg = 6 \times 10^{2}$$ $$M = \frac{600}{10} = 60 \, \text{kg}$$

Thus, the value of $M$ is $60$, so the required mass is $60 \, \text{kg}$.