A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F , its length increases by l . If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

Properties of Matter PYQ — JEE Main 2024 · 29 Jan · Shift 2 · Q69 with Solution

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Correct answer: A

For a wire under तनाव (tension), the increase in length is

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

where:

Given initially,

l = \frac{F L}{\pi r^2 Y}

New force:

F' = \frac{F}{2}

New radius:

r' = \frac{r}{2}

So the new area becomes

A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}

Let the new increase in length be $l'$ .

l' = \frac{F' L}{A' Y} = \frac{\left(\frac{F}{2}\right)L}{\left(\frac{\pi r^2}{4}\right)Y}

Simplify:

l' = \frac{F L}{2} \cdot \frac{4}{\pi r^2 Y} = \frac{2 F L}{\pi r^2 Y}

But

l = \frac{F L}{\pi r^2 Y}

Hence,

l' = 2l

The increase in length becomes 2 times the original.

Therefore, the correct option is A.