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Longitudinal Strain

The length of...

Question

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

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

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

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

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

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Solution

The correct option is A $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
If l is the original length of wire, then change in length of first wire, $Δ l_{1} = (l_{1} - l)$
Change in length of second wire, $Δ l_{2} = (l_{2} - l)$
Now, $Y = \frac{T_{1}}{A} \times \frac{l}{Δ l_{1}} = \frac{T_{2}}{A} \times \frac{l}{Δ l_{2}}$
or, $\frac{T_{1}}{Δ l_{1}} = \frac{T_{2}}{Δ l_{2}}$ , or, $\frac{T_{1}}{l_{2} - l} = \frac{T_{2}}{l_{2} - l}$
Thus, $T_{1} l_{2} - T_{1} l = T_{2} l_{1} - l T_{2}$
$\Rightarrow l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

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