JEE Physics 2025 Question with Solution

Given: Two wires A and B are of the same material, so Young's modulus is the same for both. Also, $\frac{L_A}{L_B} = \frac{1}{3}$, $\frac{d_A}{d_B} = 2$, and the stretching force is the same.

Find: The ratio of elongations $\Delta L_A : \Delta L_B$.

For a wire under tension,

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

where $F$ is the applied force, $L$ is the length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.

Since $F$ and $Y$ are the same for both wires,

$$\frac{\Delta L_A}{\Delta L_B} = \frac{F L_A / A_A Y}{F L_B / A_B Y} = \frac{L_A}{L_B} \times \frac{A_B}{A_A}$$

The cross-sectional area of a wire is

$$A = \frac{\pi d^2}{4}$$

So,

$$\frac{A_A}{A_B} = \left( \frac{d_A}{d_B} \right)^2 = 2^2 = 4$$

Hence,

$$\frac{A_B}{A_A} = \frac{1}{4}$$

Now substitute the given ratios:

$$\frac{\Delta L_A}{\Delta L_B} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$$

Therefore, the ratio of their elongations is $1 : 12$. The correct option is B.

Given: Same material and same stretching force for both wires.

Find: $\Delta L_A : \Delta L_B$.

Using strain,

$$\frac{\Delta L}{L} = \frac{\text{Stress}}{Y} = \frac{F}{AY}$$

So elongation is directly proportional to $\frac{L}{A}$ when $F$ and $Y$ are constant.

Thus,

$$\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \frac{A_B}{A_A}$$

Now,

$$\frac{L_A}{L_B} = \frac{1}{3}, \qquad \frac{A_A}{A_B} = \left(\frac{d_A}{d_B}\right)^2 = 4$$

Therefore,

$$\frac{\Delta L_A}{\Delta L_B} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$$

Hence, the required ratio is $1 : 12$.