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 ~L ' and radius ' 2 r ' is pulled by a force ' 2 f '. Then the increase in its length will be :

Properties of Matter PYQ — JEE Main 2023 · 15 Apr · Shift 1 · Q45 with Solution

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Correct answer: C

For a wire under tension,

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}

So,

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

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

Length $= L$ , radius $= r$ , force $= f$ , extension $= l$ .

Area of cross-section:

A_1 = \pi r^2

Thus,

l = \frac{fL}{\pi r^2 Y}

Length $= 2L$ , radius $= 2r$ , force $= 2f$ .

Area of cross-section:

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

Extension $\Delta l_2$ is

\Delta l_2 = \frac{(2f)(2L)}{(4\pi r^2)Y}

Simplify:

\Delta l_2 = \frac{4fL}{4\pi r^2 Y} = \frac{fL}{\pi r^2 Y}

But from the first case,

\frac{fL}{\pi r^2 Y} = l

Hence,

\Delta l_2 = l

Therefore, the correct answer is C.