Explanation

(c) Kinetic energy $K = \frac{J^{2}}{2I}$
where $J$ is angular momentum and $I$ the moment of inertia.
$\therefore$ $K_{1} = \frac{J^{2}}{2I}\ ,\ \ \ \ \ K_{2} = \frac{\left( J + \frac{10}{100}\ J \right)^{2}}{2I}$
$\therefore$ $\frac{K_{1}}{K_{2}} = \frac{(100)^{2}}{(110)^{2}} = \frac{100}{121}$
% change $= \frac{K_{2} - K_{1}}{K_{1}} = \frac{K_{2}}{K_{1}} - 1$
$= \frac{121}{100} - 1 = 21\%$