The value of tension in a long thin metal wire has been changed from T _ 1 to 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

Correct answer: A. T _ 1 ℓ _ 2 - T _ 2 ℓ _ 1 T _ 1 - T _ 2 Y = (FL)/(A ) ⇒ Y = T _ 1 ℓ _ 0 A( ℓ _ 1 - ℓ _ 0 ) = T _ 2 ℓ _ 0 A( ℓ _ 2 - ℓ _ 0 ) 1 = T _ 1 ( ℓ _ 2 - ℓ _ 0 ) T _ 2 ( ℓ _ 1 - ℓ _ 0 ) T _ 2 ℓ _ 1 - T _ 2 ℓ _ 0 = T _ 1 ℓ _ 2 - T _ 1 ℓ _ 0 ( T _ 1 - T _ 2 ) ℓ _ 0 = T _ 1 ℓ _ 2 - T _ 2 ℓ _ 1 ℓ _ 0 = ( T _ 1 ℓ _ 2 - T _ 2 ℓ _ 1 T _ 1 - T _ 2 )

JEE Main 202120 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main 2021 · 20 July · Morning Shift · Physics · Mechanical Properties of Solids

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

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:

Options

A $\dfrac{{T}_{1}{ℓ}_{2} - {T}_{2}{ℓ}_{1}}{{T}_{1} - {T}_{2}}$
B $\dfrac{{T}_{1}{ℓ}_{1} - {T}_{2}{ℓ}_{2}}{{T}_{1} - {T}_{2}}$
C $\dfrac{{ℓ}_{1} + {ℓ}_{2}}{2}$
D $\sqrt{{T}_{1}{T}_{2}{ℓ}_{1}{ℓ}_{2}}$

Correct answer

A. $\dfrac{{T}_{1}{ℓ}_{2} - {T}_{2}{ℓ}_{1}}{{T}_{1} - {T}_{2}}$

Step-by-step solution

Y = \dfrac{FL}{A\DeltaL}

⇒ Y = \dfrac{{T}_{1}{ℓ}_{0}}{A\left({ℓ}_{1} - {ℓ}_{0}\right)} = \dfrac{{T}_{2}{ℓ}_{0}}{A\left({ℓ}_{2} - {ℓ}_{0}\right)}

1 = \dfrac{{T}_{1}\left({ℓ}_{2} - {ℓ}_{0}\right)}{{T}_{2}\left({ℓ}_{1} - {ℓ}_{0}\right)}

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

\left({T}_{1} - {T}_{2}\right){ℓ}_{0} = {T}_{1}{ℓ}_{2} - {T}_{2}{ℓ}_{1}

<br />

{ℓ}_{0} = \left(\dfrac{{T}_{1}{ℓ}_{2} - {T}_{2}{ℓ}_{1}}{{T}_{1} - {T}_{2}}\right)

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