The length of a metal wire is $l_{1},$ when the tension in it is $T_{1}$ and is $l_{2}$ when the tension is $T_{2}$ . The natural length of the wire is

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$\frac{l_{1} T_{2} + l_{2} T_{1}}{T_{2} + T_{1}}$

$\sqrt{l_{1} l_{2}}$

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

$\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$

answer is B.

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Detailed Solution

Let the actual length of wire is L and area of crossection isA. Let Y is the Young’s modulus. Use the formula of Young’s Modulus

$Y = \frac{T L}{A . Δ l}$

Where $Δ l$ is change in length

So, $\frac{T_{1}}{A} = \frac{Y (l_{1} - L)}{L} ..... (i); \frac{T_{2}}{A} = \frac{Y (l_{2} - L)}{L} ...... (i i)$

From equation (i) and (ii) we get

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{l_{1} - L}{l_{2} - L} \Rightarrow L = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$

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