QJEEM23Ph7734: JEE Main 2023 Physics - Class 11 JEEM23Ph7734 Solution: A wire of length ' ' and radius ' ' is clamped rigidly at one…

Let $A$ be the cross-sectional area of the first wire, and let $Y$ be its Young's modulus.

The strain in the wire is given by $\epsilon = \frac{l}{L}$ , where $l$ is the increase in length. The stress in the wire is given by $\sigma = \frac{f}{A}$ .

According to Hooke's law, the stress is proportional to the strain, so we have $\sigma = Y \epsilon$ . Solving for $f$ , we get $f = \frac{YA}{l}$ .

The second wire has twice the length and four times the cross-sectional area of the first wire, so its cross-sectional area is $4A$ and its Young's modulus is still $Y$ .

When a force of $2f$ is applied to this wire, the stress in the wire is $\sigma = \frac{2f}{4A} = \frac{f}{2A}$ .

Using Hooke's law again, we have $\sigma = Y \epsilon$ . Solving for $\epsilon$ , we get $\epsilon = \frac{\sigma}{Y} = \frac{f}{2AY}$ .

The increase in length of the second wire is given by $\Delta l = \epsilon \cdot 2L = \frac{f \cdot 2L}{2AY}$ .

Substituting the expression for $f$ that we derived earlier, we get $\Delta l = \frac{YL \cdot 2A \cdot l}{2AY \cdot L} = \boxed{l}$ .

Therefore, the increase in length of the second wire is the same as the increase in length of the first wire, which is $l$ .