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 :

Consider two solid spheres of radii R₁ = 1m, R₂ = 2m and masses M₁ and M₂, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $${{{M_1}} \over {{M_2}}}$$ is :

A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms^–2 at the north pole and the radius of the earth = 6400 km) :