The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3 R$ from the centre of the sphere is $F_1$. A spherical hole of radius $(R / 2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. Ratio of $F_1$ and $F_2$ is

• Correct answer
  A

  $\frac{50}{41}$
• B

  $\frac{41}{50}$
• C

  $\frac{41}{42}$
• D

  $\frac{25}{41}$

Solution

Gravitational force due to solid sphere is $$ F_1=\frac{G M m}{(3 R)^2}=\frac{G M m}{9 R^2} $$
where, $M$ and $m$ are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity $$ \begin{aligned} F_2 & =\frac{G M m}{9 R^2}-\frac{G\left(\frac{M}{8}\right) m}{(5 R / 2)^2} \\ & =\frac{G M m}{R^2}\left[\frac{1}{9}-\frac{4}{8 \times 25}\right] \\ & =\frac{G M m}{R^2}\left[\frac{41}{50 \times 9}\right] \\ \therefore \quad \frac{F_1}{F_2} & =\frac{50}{41} \end{aligned} $$

Hence, the correct answer is option: 'A'.