A wire of length L is hanging from a fixed support. The length changes to $L_{1}$ and $L_{2}$ when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :

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$\frac{L_{1} + L_{2}}{2}$

$\sqrt{L_{1} L_{2}}$

$3 L_{1} - 2 L_{2}$

$2 L_{1} - L_{2}$

answer is C.

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Detailed Solution

By Hooke's Law
so $F α ΔL$
$\begin{array}{l} \frac{F_{1}}{F_{2}} = \frac{{ΔL}_{1}}{{ΔL}_{2}} \\ \frac{10}{20} = \frac{(L_{1} - L)}{(L_{2} - L)} \\ L = 2 L_{1} - L_{2} \end{array}$

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