The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

The mass density of a spherical galaxy varies as $${K \over r}$$ over a large distance ‘r’ from its centre. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as :

Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are v_A and v_B, respectively, then $${{{v_A}} \over {{v_B}}} = {n \over 4}$$. The value of n is :

A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass $${m \over 2}$$ collides with A with a velocity which is half $$\left( {{{\overrightarrow v } \over 2}} \right)$$ the instantaneous velocity$${\overrightarrow v }$$ of A. The collision is completely inelastic. Then, the combined body :