Verified Solution • 2025–26 Session moderate

QJEEM15Phb08f: JEE Main 2015 Physics - Class 11 JEEM15Phb08f Solution: A pendulum made of a uniform wire of cross sectional area has…

A pendulum made of a uniform wire of cross sectional area AA has time period T.T. When an additional mass MM is added to its bob, the time period changes to TM.{T_{M.}} If the Young's modulus of the material of the wire is YY then 1Y{1 \over Y} is equal to :
( g=g= gravitationalgravitational accelerationacceleration )

Multiple Choice Options

  • A [1−(TMT)2]AMg\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}
  • B [1−(TTM)2]AMg\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}
  • C [(TMT)2−1]AMg\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}
  • D [(TMT)2−1]MgA\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}

Step-by-Step Verified Solution

As we know, time period, T=2πℓgT = 2\pi \sqrt {{\ell \over g}}

When a additional mass MM is added then

TM=2πℓ+Δℓg{T_M} = 2\pi \sqrt {{{\ell + \Delta \ell } \over g}}

TMT=ℓ+Δℓℓ{{{T_M}} \over T} = \sqrt {{{\ell + \Delta \ell } \over \ell }}

or,   (TMT)2=ℓ+Δℓℓ\,\,{\left( {{{{T_M}} \over T}} \right)^2} = {{\ell + \Delta \ell } \over \ell }

or,   (TMT)2=1+MgAy\,\,{\left( {{{{T_M}} \over T}} \right)^2} = 1 + {{Mg} \over {Ay}}

[ \left[ \, \right. as Δℓ=MgℓAy ]\left. {\Delta \ell = {{Mg\ell } \over {Ay}}\,} \right]

∴\therefore 1y=[(TMT)2−1]AMg{1 \over y} = \left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}

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