c
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 \(K _{2}\)

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

\(P \%=\frac{ K _{1}- K _{2}}{ K _{1}} \times 100 \%=\frac{1-4 / 5}{1} \times 100=20 \%\)