JEE Physics 2025 Question with Solution

Given: Mass $m = 50 \, \text{kg}$, acceleration due to gravity $g = 3\pi \, \text{m/s}^2$, length $L = 3 \, \text{m}$, radius $r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$, extension $\Delta L = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}$.

Find: The value of $P$ in $Y = P \times 10^{11} \, \text{N/m}^2$.

Use the formula for Young's modulus:

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

The applied force is:

$$F = mg = 50 \times 3\pi = 150\pi \, \text{N}$$

The cross-sectional area of the wire is:

$$A = \pi r^2 = \pi \times \left(3 \times 10^{-3}\right)^2 = 9\pi \times 10^{-6} \, \text{m}^2$$

Substitute in the formula:

$$Y = \frac{150\pi \times 3}{9\pi \times 10^{-6} \times 0.1 \times 10^{-3}}$$ $$Y = \frac{450\pi}{9\pi \times 10^{-10}}$$ $$Y = 50 \times 10^{10} \, \text{N/m}^2 = 5 \times 10^{11} \, \text{N/m}^2$$

Therefore, the value of $P$ is $5$. Hence, the correct option is D.

Given: The wire is stretched by the weight of the hanging mass.

Find: Young's modulus in the form $P \times 10^{11} \, \text{N/m}^2$.

1. Compute the load on the wire:

$$F = m g = 50 \times 3\pi = 150\pi \, \text{N}$$

2. Convert radius into SI unit and find area:

$$r = 3 \times 10^{-3} \, \text{m}$$ $$A = \pi r^2 = \pi \left(3 \times 10^{-3}\right)^2 = 9\pi \times 10^{-6} \, \text{m}^2$$

3. Convert extension into SI unit:

$$\Delta L = 0.1 \times 10^{-3} \, \text{m} = 10^{-4} \, \text{m}$$

4. Apply Young's modulus formula:

$$Y = \frac{F L}{A \Delta L}$$ $$Y = \frac{150\pi \times 3}{\left(9\pi \times 10^{-6}\right) \left(10^{-4}\right)}$$ $$Y = \frac{450\pi}{9\pi \times 10^{-10}}$$ $$Y = 50 \times 10^{10} = 5 \times 10^{11} \, \text{N/m}^2$$

So, in the form $P \times 10^{11} \, \text{N/m}^2$, we get $P = 5$. The correct option is D.