A wire of length ' $L$ ' and radius ' $r$ ' is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by ' $l$ '. Another wire of same material of length ' $2 \mathrm{~L}$ ' and radius ' $2 r$ ' is pulled by a force ' $2 f$ '. Then the increase in its length will be :

Let $A$ be the cross-sectional area of the first wire, and let $Y$ be its Young's modulus.<br/><br/> 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 $\sigma = \frac{f}{A}$. <br/><br/>According to Hooke's law, the stress is proportional to the strain, so we have $\sigma = Y \epsilon$. Solving for $f$, we get $f = \frac{YA}{l}$. <br/><br/> 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 $Y$. <br/><br/>When a force of $2f$ is applied to this wire, the stress in the wire is $\sigma = \frac{2f}{4A} = \frac{f}{2A}$.<br/><br/> Using Hooke's law again, we have $\sigma = Y \epsilon$. Solving for $\epsilon$, we get $\epsilon = \frac{\sigma}{Y} = \frac{f}{2AY}$. <br/><br/> The increase in length of the second wire is given by $\Delta l = \epsilon \cdot 2L = \frac{f \cdot 2L}{2AY}$. <br/><br/>Substituting the expression for $f$ that we derived earlier, we get $\Delta l = \frac{YL \cdot 2A \cdot l}{2AY \cdot L} = \boxed{l}$. <br/><br/> Therefore, the increase in length of the second wire is the same as the increase in length of the first wire, which is $l$.