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

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Solution

We know, $Y = \frac{F L}{A Δ L}$
Given $F, L and Y$ are constant
Hence, $Δ L \propto \frac{1}{A}$
As we know,
$Mass = Volume \times density$
$m = (A \times L) \times ρ$
[Same material, hence $ρ$ = constant]
or, $m \propto A$
$∴ Δ L \propto \frac{1}{m}$
$\frac{Δ L_{1}}{Δ L_{2}} = \frac{m_{2}}{m_{1}} = \frac{2}{3}$
So, $n$ will be 3.

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