A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $\theta(t)=5 t^2-8 t$, where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque, when $t=2 \mathrm{~s}$ ?

A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle $30^{\circ}$ with the horizontal. The speed of the particle at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant. Then the new speed of the sphere at the bottom of the plane is $v_2$. The ratio $v_1^2: v_2^2$ is

The torque due to the force $(2 \hat{i}+\hat{j}+2 \hat{k})$ about the origin, acting on a particle whose position vector is $(\hat{i}+\hat{j}+\hat{k})$, would be

A uniform circular disc of radius ' $\mathrm{R}^{\prime}$ and mass ' $\mathrm{M}^{\prime}$ is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius $R / 2$ is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.