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QJEEM19Ph6bbb: JEE Main 2019 Physics - Class 11 JEEM19Ph6bbb Solution: Two coaxial discs, having moments of inertia I 1 and I 1 /2,…

Two coaxial discs, having moments of inertia I1 and I1/2, are rotating with respective angular velocities ω\omega 1 and ω\omega 1/2 , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef - Ei) is:

Multiple Choice Options

  • A I1ω1224{{{I_1}\omega _1^2} \over {24}}
  • B I1ω1212{{{I_1}\omega _1^2} \over {12}}
  • C 38I1ω12{3 \over 8}{I_1}\omega _1^2
  • D I1ω126{{{I_1}\omega _1^2} \over {6}}

Step-by-Step Verified Solution

Ei=12II×ω12+12I2×ω124{E_i} = {1 \over 2}{I_I} \times \omega _1^2 + {1 \over 2}{I \over 2} \times {{\omega _1^2} \over 4}

=I1ω122(98)=916I1ω12= {{{I_1}\omega _1^2} \over 2}\left( {{9 \over 8}} \right) = {9 \over {16}}{I_1}\omega _1^2

I1ω1+I1ω14=3I12ω;54I1ω1=3I12ω{I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _1} = {{3{I_1}} \over 2}\omega

ω=56ω1;Ef=12×3I12×2536ω12\omega = {5 \over 6}{\omega _1};{E_f} = {1 \over 2} \times {{3{I_1}} \over 2} \times {{25} \over {36}}\omega _1^2

=2548I1ω12= {{25} \over {48}}{I_1}\omega _1^2

⇒Ef−Ei=I1ω122549−−248I2ω12\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2{{25} \over {49}} - {{ - 2} \over {48}}{I_2}\omega _1^2

=2548I1ω12= {{25} \over {48}}{I_1}\omega _1^2

⇒Ef−Ei=I1ω12(2548−916)=−248I1ω12\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {{{25} \over {48}} - {9 \over {16}}} \right) = {{ - 2} \over {48}}{I_1}\omega _1^2

=−I1ω1224= {{ - {I_1}\omega _1^2} \over {24}}

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