Explanation

(a) According t work-energy theorem,
Work done=change in rotational kinetic energy
$W = \left( \mathrm{\Delta}{KE}_{r} \right)_{1} - \left( \mathrm{\Delta}{KE}_{r} \right)_{2}$ …(i)
But rotational Kinetic energy
K=${\frac{1}{2}I\omega}^{2}$
From Eq.(i),We get
$W = \frac{1}{2}I\omega_{1}^{2} - \frac{1}{2}I\omega_{2}^{2}$
$= \frac{1}{2}I{(\omega}_{I}^{2} - \omega_{2}^{2})$
As, $\omega = 2\pi\eta$
Hence, We get
$W = \frac{1}{2}I\ \left\lbrack {\ \left( 2\pi n_{1} \right)}^{2} - \left( 2\pi n_{2} \right)^{2}\ \right\rbrack$
$= \frac{1}{2}I \times 4\pi^{2}\left( n_{1}^{2} - n_{2}^{2} \right)\ \ \ \ldots\ldots\ldots\ldots..(ii)$
Given $I = \frac{9.8}{\pi^{2}}kg - m^{2}$
$n_{1} = 600rpm = 10rps$
$n_{2} = 300rpm = 5rps$
From Eq.(ii),We get
$W = \frac{1}{2} \times \frac{9.8}{\pi^{2}} \times 4\pi^{2}\left( 10^{2} - 5^{2} \right) = 1467\ J$