The length of a rod under longitudinal tension $T_{1}$ is $L_{1}$ and that under longitudinal tension $T_{2}$ is $L_{2} .$ What is the actual length of the rod, in the absence of tensions?

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

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

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

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

answer is D.

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Detailed Solution

Let, the normal length of wire be $L_{o}$
Using Hooke’s law,
$Y = \frac{T L_{o}}{A Δ L} ... (i)$
Where, Y is Young’s modulus of elasticity, T is temperature, A is area and $Δ L$ is change in length, i.e. $Δ L = L - L_{o}$
From Eq. (i) we get
$L - L_{o} = \frac{T L_{o}}{A Y}$
Case I When tension is $T_{1}$ and length of wire $L = L_{1},$
$L_{1} = L_{o} = \frac{T_{1} L_{o}}{A Y} .... (i i)$
Case II When tension is $T_{2}$ and length of wire $L = L_{2},$
$L_{2} = L_{o} = \frac{T_{2} L_{o}}{A Y} .... (i i i)$
Dividing Eq. (ii) by Eq. (iii) we get
$\frac{L_{1} - L_{o}}{L_{2} - L_{o}} = \frac{T_{1}}{T_{2}} \Rightarrow L_{o} = \frac{L_{1} T_{2} - L_{2} T_{1}}{T_{2} - T_{1}}$