QJEEM20Phf336: JEE Main 2020 Physics - Class 11 JEEM20Phf336 Solution: A circular disc of mass M and radius R is rotating about its…

Question

Verified Solution • 2025–26 Session moderate

A circular disc of mass M and radius R is rotating about its axis with angular speed

{\omega _1}

. If another stationary disc having radius

{R \over 2}

and same mass M is droped co-axially on to the rotating disc. Gradually both discs attain constant angular speed

{\omega _2}

the energy lost in the process is p% of the initial energy. Value of p is __________.

Accepted Answer

{I_f}{\omega _f} = {I_i}{\omega _i}

{I_i} = {{M{R^2}} \over 2}

{I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}

= {5 \over 4}.{{M{R^2}} \over 2}

\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega

\Rightarrow

\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega

\omega = {4 \over 5}\omega

loss of K.E. =

{{Loss} \over {{K_i}}} \times 100

=

{{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}

\times

100

=

{{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}

\times

100 =

\left( {1 - {{80} \over {100}}} \right) \times 100

= 20%

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