Moment of inertia of a disc of mass $M$ and radius 'R' about any of its diameter is $MR^2/4$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $(x/2) MR^2$. The value of $x$ is

A thin uniform rod of length $2$ m cross-sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$. If the value of $\omega$ in terms of its rotational kinetic energy $E$ is $(\alpha E/Ad)^{1/2}$, then the value of $\alpha$ is

A body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity 'V'. If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of the plane is

A disc of radius 0.4 m and mass 1 kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is 10 rad/s$^2$. The tangential force applied to the rim of the disc is