A circular disc of mass M and radius R is rotating about its axis with angular speed $\omega _{1}$. if the another stationary disc having radius $\frac{R}{2}$ and same mass M is dropped co-axially on the rotating disc. Gradually both the discs attain constant angular speed $\omega _{2}$. The energy lost in the process is p% of the initial energy. Value of p is_________ Option: 1 10 Option: 2 20 Option: 3 30 Option: 4 40

Let the moment of inertia of bigger disc is $I=\frac{M R^{2}}{2}$

$\Rightarrow \text{ MOI of small disc} =I_{2}=\frac{M\left(\frac{R}{2}\right)^{2}}{2}=\frac{I}{4}$
by angular momentum conservation
$\mathrm{I} \omega_{1}+\frac{\mathrm{I}}{4}(0)=\mathrm{I} \omega_{2}+\frac{\mathrm{I}}{4} \omega_{2} \Rightarrow \omega_{2}=\frac{4 \omega_{1}}{5}$
initial kinetic energy =$\mathrm{K}_{1}=\frac{1}{2} \mathrm{I} \omega_{1}^{2}$
final kinetic energy

$\mathrm{K}_{2} =\frac{1}{2}\left(I+\frac{I}{4}\right)\left(\frac{4 \omega_{1}}{5}\right)^{2}=\frac{1}{2} I \omega_{1}^{2}\left(\frac{4}{5}\right) \\$

$\mathrm{P} \%=\frac{\mathrm{K}_{1}-\mathrm{K}_{2}}{\mathrm{~K}_{1}} \times 100 \%=\frac{1-4 / 5}{1} \times 100=20 \%$