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 5 ~cm . Another wire of the same material of length 4 L and radius 4 r is pulled by a force 4 F under same conditions. The increase in length of this wire is \underline\hspace2cm cm .

Properties of Matter PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q63 with Solution

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Correct answer: 5

For a wire under tensile force,

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

where:

Since both wires are made of the same material, $Y$ remains the same.

Given that for the first wire,

\Delta L_1 = 5\text{ cm}

And

\Delta L_1 = \frac{F L}{\pi r^2 Y}

For the second wire:

So its area is

A_2 = \pi (4r)^2 = 16\pi r^2

Hence extension is

\Delta L_2 = \frac{(4F)(4L)}{(16\pi r^2)Y}

Simplifying,

\Delta L_2 = \frac{16FL}{16\pi r^2 Y} = \frac{FL}{\pi r^2 Y}

But this is exactly equal to the extension of the first wire:

\Delta L_2 = \Delta L_1

Therefore,

\Delta L_2 = 5\text{ cm}

The increase in length of the second wire is

\boxed{5}

cm.

Stored correct answer: $5$

Our derived answer matches the stored answer.