Q: The angular momentum of rotating body is increased by 20 %. What will be the increase in its rotational kinetic energy ?

Sol: Kinetic Energy $\large K = \frac{L^2}{2 I}$

$\large K \propto L^2 $

$\large \frac{\Delta K}{K} = (\frac{120}{100})^2$

$\large \frac{\Delta K}{K} = 0.44 $

$\large \frac{\Delta K}{K} \times 100 = 44 \% $