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 \% $