JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 1)
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 :
$2 l$
$4 l$
$l$
$l / 2$
Explanation
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 $l$.
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 $l$.
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