Q 11-07-142JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1$. Now a spherical cavity of radius $\left(\dfrac{R}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1 : F_2$ is:
Answer: (B) 50 : 41
$F_1 = \dfrac{GMm}{(3R)^2} = \dfrac{GMm}{9R^2}$.
The removed sphere has mass $\dfrac{M}{8}$ (radius halved) and its centre B is at $\dfrac{R}{2}$ from O, so it is at $3R - \dfrac{R}{2} = \dfrac{5R}{2}$ from the particle. Its attraction would have been
$$\frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{50R^2}$$
$$F_2 = GMm\left(\frac{1}{9R^2} - \frac{1}{50R^2}\right) = \frac{41\,GMm}{450R^2}$$
$$\frac{F_1}{F_2} = \frac{50}{41}$$
Solution by Sreeraj P, M.Sc Physics