Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Exercises - Page 306: 37

$1.5\times10^{5}Nm$

$\omega=\omega_{0}+at$, where a is angular acceleration. $2=0+a12$ $a=\frac{1}{6}rad/s^{2}$ Now, $I=\frac{1}{2}MR^{2}=\frac{1}{2}\times2000\times30^{2}=900000kgm^{2}$ Torque = $900000\times\frac{1}{6}=1.5\times10^{5}Nm$