JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Elasticity & SHM

The value of tension in a long thin metal wire has been changed from

T_{1}

T_{2}

. The lengths of the metal wire at two different values of tension

T_{1}

and

T_{2}

are

ℓ_{1}

and

ℓ_{2}

respectively. The actual length of the metal wire is :

Answer: (A)

Solution

Let

L

be the unstretched length and

A

the area of cross-section. Young's modulus gives the extension under a tension

T

\frac{T L}{A Y}

, so the stretched lengths are

ℓ_{1} = L + \frac{T _{1} L}{A Y}

and

ℓ_{2} = L + \frac{T _{2} L}{A Y}

. Hence

\frac{ℓ _{1} - L}{T _{1}} = \frac{ℓ _{2} - L}{T _{2}} = \frac{L}{A Y}

Cross-multiplying,

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

, so

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

and the actual length of the wire is

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

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option