The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3R$ from the centre of the sphere is $F_1$. A spherical hole of radius $\dfrac{R}{2}$ is now made in the sphere as shown in the figure. The sphere with the hole now exerts a force $F_2$ on the same particle. The ratio $\dfrac{F_1}{F_2}$ is
Answer: (A) $\dfrac{50}{41}$
Concept. Sphere with a hole $=$ full sphere $-$ the removed sphere (superposition).
Step 1. Removed piece. Mass $\dfrac M8$, centre $\dfrac R2$ from O towards the particle, so $\dfrac{5R}{2}$ from it.
Step 2. Forces.
$$F_1 = \frac{GMm}{9R^2}, \qquad F_2 = \frac{GMm}{R^2}\left(\frac19 - \frac{1}{50}\right) = \frac{GMm}{R^2} \cdot \frac{41}{450}$$
Step 3. Ratio. $\dfrac{F_1}{F_2} = \dfrac{50}{41}$.
Answer. $\dfrac{50}{41}$ (Option A).
Solution by Sreeraj P, M.Sc Physics