Verified Solution • 2025–26 Session moderate

QJEEM21Phe015: JEE Main 2021 Physics - Class 11 JEEM21Phe015 Solution: 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 T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is :

Multiple Choice Options

  • A l1+l22{{{l_1} + {l_2}} \over 2}
  • B T1T2l1l2\sqrt {{T_1}{T_2}{l_1}{l_2}}
  • C T1l2−T2l1T1−T2{{{T_1}{l_2} - {T_2}{l_1}} \over {{T_1} - {T_2}}}
  • D T1l1−T2l2T1−T2{{{T_1}{l_1} - {T_2}{l_2}} \over {{T_1} - {T_2}}}

Step-by-Step Verified Solution

Suppose, I0 be the actual length of metal wire and Y be its Young's modulus.

From Hooke's law,

Y=TI0AΔIY = {{T{I_0}} \over {A\Delta I}}

where, ΔI=I−I0\Delta I = I - {I_0}

⇒Y=TI0A(I−I0)\Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}} or I−I=TI0AYI - I = {{T{I_0}} \over {AY}}

∴\therefore I1−I0I2−I0=T1I0AY×AYT2I0=T1T2{{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}}

⇒I1T2−I0T2=I2T1−I0T1\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}

⇒I0=I1T2−I2T1T2−T1=T1I2−T2I1(T1−T2)\Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}

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