Question

A solid sphere of uniform density and radius $R$ applies a gravitational force of attraction equal to $F_{1}$ on a particle placed at a distance $3R$ from the centre of the sphere. A spherical cavity of the radius $R/2$ is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force $F_{2}$ on the same particle. The ratio $\frac{F_{2}}{F_{1}}$ is

• A. $\frac{9}{5 0}$
• B. $\frac{4 1}{5 0}$
• C. $\frac{3}{2 5}$
• D. $\frac{2 2}{2 5}$

Answer 1

Answer: (B)

From the superposition principle,
$F_{1}=F_{r}+F_{c}$
Here, F_r = force due to remaining part = F₂
and F_c = force due to cavity
Now, $\quad F_1=\frac{\mathrm{GMm}}{(3 \mathrm{R})^2}=\frac{\mathrm{GMm}}{9 \mathrm{R}^2}$
$\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{G}\left(\frac{\mathrm{M}}{8}\right) \mathrm{m}}{\left(\frac{5}{2 \mathrm{R}}\right)^2}=\frac{\mathrm{GMm}}{50 \mathrm{R}^2}$
$\therefore \frac{\text{F}_{2}}{\text{F}_{1}} = \frac{4 1}{5 0}$