Two wires A and B are made of same material having ratio of lengths \fracL_AL_B=\frac13 and their diameters ratio \fracd_Ad_B=2 . If both the wires are stretched using same force, what would be the ratio of their respective elongations?

Properties of Matter PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q59 with Solution

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Correct answer: B

For a wire stretched by force $F$ , the elongation is

\Delta L = \frac{FL}{AY}

where:

Since both wires are made of the same material, $Y$ is same. Also, both are stretched by the same force, so $F$ is same.

Hence,

\Delta L \propto \frac{L}{A}

For a circular wire,

A = \frac{\pi d^2}{4}

So,

\Delta L \propto \frac{L}{d^2}

\frac{\Delta L_A}{\Delta L_B} = \frac{L_A/d_A^2}{L_B/d_B^2} = \frac{L_A}{L_B}\cdot \frac{d_B^2}{d_A^2}

Given,

\frac{L_A}{L_B} = \frac{1}{3}, \qquad \frac{d_A}{d_B}=2

Therefore,

\frac{d_B}{d_A} = \frac{1}{2}

So,

\frac{d_B^2}{d_A^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

Thus,

\frac{\Delta L_A}{\Delta L_B} = \frac{1}{3}\cdot \frac{1}{4} = \frac{1}{12}

Therefore, the ratio of elongations is

\Delta L_A : \Delta L_B = 1:12

So the correct option is B.