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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 $\frac{ R }{2}$ and same mass $M$ is dropped 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
$25$
$27$
$20$
$15$
Solution
Let moment of inertia of bigger disc is $I =\frac{ MR ^{2}}{2}$
$\Rightarrow$ $MOI$ of small disc $I_{2}=\frac{M\left(\frac{R}{2}\right)^{2}}{2}=\frac{I}{4}$
by angular momentum conservation
$I \omega_{1}+\frac{ I }{4}( D )= I \omega_{2}+\frac{ I }{4} \omega_{2} \Rightarrow \omega_{2}=\frac{4 \omega_{1}}{5}$
initial kinetic energy $K _{1}=\frac{1}{2} I \omega_{1}^{2}$
final kinetic energy $K _{2}$
$=\frac{1}{2}\left( I +\frac{ I }{4}\right)\left(\frac{4 \omega_{1}}{5}\right)^{2}=\frac{1}{2} I \omega_{1}^{2}\left(\frac{4}{5}\right)$
$P \%=\frac{ K _{1}- K _{2}}{ K _{1}} \times 100 \%=\frac{1-4 / 5}{1} \times 100=20 \%$
