A small point of mass $m$ is placed at a distance $2R$ from the centre $O$ of a big uniform solid sphere of mass $M$ and radius $R$. The gravitational force on $m$ due to $M$ is $F_1$. A spherical part of radius $R/3$ is removed from the big sphere as shown in the figure and the gravitational force on $m$ due to the remaining part of $M$ is found to be $F_2$. The value of ratio $F_1 : F_2$ is
Answer: (A) $12 : 11$
Full sphere: $F_1 = \dfrac{GMm}{(2R)^2} = \dfrac{GMm}{4R^2}$.
The removed sphere has radius $R/3$, so its mass is $M/27$. It touches the surface on the side of $m$, so its centre is at $2R/3$ from $O$ and at $2R - 2R/3 = 4R/3$ from $m$:
$$F_{rem} = \frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{48R^2}$$
$$F_2 = F_1 - F_{rem} = \frac{GMm}{R^2}\left(\frac{1}{4} - \frac{1}{48}\right) = \frac{11}{48}\frac{GMm}{R^2}$$
$$F_1 : F_2 = \frac{12}{48} : \frac{11}{48} = 12 : 11$$
Solution by Sreeraj P, M.Sc Physics