The length of metallic wire is $\ell_{1}$ when tension in it is $T_{1}$. It is $\ell_{2}$ when the tension is $T_{2}$. The original length of the wire will be :

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

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

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

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

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Elasticity & SHM

The length of metallic wire is

ℓ_{1}

when tension in it is

T_{1}

. It is

ℓ_{2}

when the tension is

T_{2}

. The original length of the wire will be :

Answer: (C)

Let the unstretched length be

ℓ_{0}

and the area of cross-section be

A

. Young's modulus written for each measurement,

Y = \frac{T _{1} ℓ _{0}}{A ( ℓ _{1} - ℓ _{0} )} = \frac{T _{2} ℓ _{0}}{A ( ℓ _{2} - ℓ _{0} )}

uses the same

Y

A

and

ℓ_{0}

, so

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

. Cross-multiplying gives

T_{1} ℓ_{2} - T_{1} ℓ_{0} = T_{2} ℓ_{1} - T_{2} ℓ_{0}

, hence

ℓ_{0} (T_{2} - T_{1}) = T_{2} ℓ_{1} - T_{1} ℓ_{2}

. The original length of the wire is

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

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option