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

(C) $\frac{\ell_{1} T_{2}-\ell_{2} T_{1}}{T_{2}-T_{1}}$

(B) $\frac{\ell_{1}+\ell_{2}}{2}$

(D) $\frac{\ell_{1} T_{2}+\ell_{2} T_{1}}{T_{2}+T_{1}}$

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Elasticity & SHM

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 :

Answer: (C)

Let the natural length be

ℓ_{0}

and the cross-sectional area be

A

. Hooke's law makes the extension under a tension

T

equal to

\frac{T ℓ _{0}}{A Y}

, so

ℓ_{1} = ℓ_{0} + \frac{T _{1} ℓ _{0}}{A Y}, ℓ_{2} = ℓ_{0} + \frac{T _{2} ℓ _{0}}{A Y}

Dividing the two extensions eliminates

A Y

\frac{ℓ _{1} - ℓ _{0}}{ℓ _{2} - ℓ _{0}} = \frac{T _{1}}{T _{2}} \Rightarrow T_{2} ℓ_{1} - T_{2} ℓ_{0} = T_{1} ℓ_{2} - T_{1} ℓ_{0}

Collecting the terms in

ℓ_{0}

, the natural length of the wire is

\frac{ℓ _{1} T _{2} - ℓ _{2} T _{1}}{T _{2} - T _{1}}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option