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.

A thin circular disc of mass $$\mathrm{M}$$ and radius $$\mathrm{R}$$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $$\omega$$. If another disc of same dimensions but of mass $$\mathrm{M} / 2$$ is placed gently on the first disc co-axially, then the new angular velocity of the system is :