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When a metall...

Question

When a metallic wire is stretched with a tension $T_{1}$ its length is $l_{1}$ and with a tension $T_{2}$ its length is $l_{2}$ . The original length of the wire is:

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

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

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

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\frac{l_{1} l_{2}}{2}

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Solution

The correct option is A $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
From the relation between stress and strain, we get,

$\frac{F}{A} = Y \frac{Δ l}{l}$

Thus $\frac{T_{1}}{A} = Y \frac{l_{1} - l}{l}$ and $\frac{T_{2}}{A} = Y \frac{l_{2} - l}{l}$

Dividing the two equations, we get,

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

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

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