Question

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio $3:2$. If their elongations are in the ratio $2:n$, then $n$ is

Answer 1

We know, $Y=\frac{FL}{A\Delta L}$
Given $F,L\text{and}Y$ are constant
Hence, $\Delta L\propto \frac{1}{A}$
As we know,
$\text{Mass}=\text{Volume}\times \text{density}$
$m=(A\times L)\times \rho$
[Same material, hence $\rho$ = constant]
or, $m\propto A$
$\therefore \Delta L\propto \frac{1}{m}$
$\frac{\Delta L_1}{\Delta L_2}=\frac{m_2}{m_1}=\frac{2}{3}$
So, $n$ will be 3.