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Young's Modulus of Elasticity

The length of...

Question

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 wire is

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

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

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\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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Solution

The correct option is D $\frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$
Let the natural length of wire be $l_{o}$ .
Using Hooke's law, $Y = \frac{T l_{o}}{A Δ l}$
where $Δ l = l - l_{o}$
We get $l - l_{o} = \frac{T l_{o}}{A Y}$
Case 1 : Tension is $T_{1}$ and length of wire $l = l_{1}$
$∴$ $l_{1} - l_{o} = \frac{T_{1} l_{o}}{A Y}$ .....(1)
Case 2 : Tension is $T_{2}$ and length of wire $l = l_{2}$
$∴$ $l_{2} - l_{o} = \frac{T_{2} l_{o}}{A Y}$ .....(2)
Dividing both equations $\frac{l_{1} - l_{o}}{l_{2} - l_{o}} = \frac{T_{1}}{T_{2}}$
$⟹$ $l_{o} = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{2} - T_{1}}$

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