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Simple Pendulum

A wire of len...

Question

A wire of length $L$ with Young's modulus $Y$ is hanging from a fixed support. The length of wire becomes $L_{1}$ when mass $M_{1}$ is suspended, and it becomes $L_{2}$ when mass $M_{2}$ is suspended at its free end. Then, the original length $L$ is equal to -

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

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\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{1} + M_{2}}

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\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}

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

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Solution

The correct option is C $\frac{L_{1} M_{2} - L_{2} M_{1}}{M_{2} - M_{1}}$
$Young's modulus = \frac{Stress}{Strain}$

$\Rightarrow Y = \frac{T_{1} L}{A Δ L_{1}} = \frac{T_{2} L}{A Δ L_{2}}$

$\Rightarrow \frac{T_{1}}{Δ L_{1}} = \frac{T_{2}}{Δ L_{2}}$

$∴ \frac{T_{1}}{L_{1} - L} = \frac{T_{2}}{L_{2} - L}$

$\Rightarrow \frac{M_{1} g}{L_{1} - L} = \frac{M_{2} g}{L_{2} - L}$

$\Rightarrow M_{1} (L_{2} - L) = M_{2} (L_{1} - L)$

$\Rightarrow L = \frac{(L_{1} M_{2} - L_{2} M_{1})}{(M_{2} - M_{1})}$

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