JEE Physics 2023 Question with Solution

Given: radius of each wire is $0.2 \, \text{cm}$. The upper wire is steel of length $1.6 \, \text{m}$ and the lower wire is brass of length $1 \, \text{m}$. The attached masses are $2 \, \text{kg}$ and $1.14 \, \text{kg}$.

Find: elongation of the steel wire.

From the loading shown, the tension in the steel wire is the total load supported above the lower section:

$$T_2 = T_1 + 20 = 20 + 11.4 = 31.4 \, \text{N}$$

The elongation of the steel wire is given by:

$$\Delta L = \frac{T_2 L}{A Y}$$

where

• $T_2 = 31.4 \, \text{N}$
• $L = 1.6 \, \text{m}$
• $A = \pi (0.2 \times 10^{-2})^2 \, \text{m}^2$
• $Y = 2 \times 10^{11} \, \text{Pa}$

Substituting these values:

$$\Delta L = \frac{31.4 \cdot 1.6}{\pi (0.2 \times 10^{-2})^2 \cdot 2 \times 10^{11}}$$

Simplifying:

$$\Delta L = \frac{50.24}{\pi \cdot 4 \times 10^{-8} \cdot 2 \times 10^{11}}$$ $$\Delta L = 2 \times 10^{-5} \, \text{m}$$

Therefore,

$$\Delta L = 20 \times 10^{-6} \, \text{m} = 20 \, \mu\text{m}$$

So, the elongation of the steel wire is $20 \times 10^{-6} \, \text{m}$.

Given: elongation is to be found for the steel part only.

Use the relation for elongation under tensile stress:

$$\Delta L = \frac{FL}{AY}$$

The steel segment supports both loads, so the force in it is $31.4 \, \text{N}$ as obtained in the working.

Its cross-sectional area is:

$$A = \pi r^2 = \pi (0.2 \times 10^{-2})^2 \, \text{m}^2$$

Now substitute in the formula:

$$\Delta L = \frac{31.4 \times 1.6}{\pi (0.2 \times 10^{-2})^2 \times 2 \times 10^{11}}$$

This evaluates to:

$$\Delta L = 2 \times 10^{-5} \, \text{m}$$

Hence, the required numerical value is 20 when written as $20 \times 10^{-6} \, \text{m}$.