Two fly wheels A and B are mounted side by side with frictionless bearings on a common shaft. Their moments…

When wheel B, initially stationary is coupled to A with the help of a clutch, both will have a common angular velocity and let that be $\omega$.
Since there is no external torque, angular momentum will be conserved.

From conservation of angular momentum,$I\omega =I_1{\omega }_1+I_{\mathit{2}}{\omega }_{\mathit{2}}$

$\Rightarrow I\omega =5\times 10+20\times 0$

$\Rightarrow I\omega =50$

$\Rightarrow \left(5+20\right)\omega =50$

$\Rightarrow \omega =\frac{50}{\left(5+20\right)}$

$\Rightarrow \omega =2 \mathrm{rps}$