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. $\frac{9}{50}$
• B. $\frac{41}{50}$
• C. $\frac{3}{25}$
• D. $\frac{22}{25}$

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, $F_1=\frac{\mathrm{GMm}}{(3\mathrm{R}{)}^2}=\frac{\mathrm{GMm}}{9{\mathrm{R}}^2}$
${\mathrm{F}}_{\mathrm{c}}=\frac{\mathrm{G}\left(\frac{\mathrm{M}}{8}\right)\mathrm{m}}{{\left(\frac{5}{2\mathrm{R}}\right)}^2}=\frac{\mathrm{GMm}}{50{\mathrm{R}}^2}$
$\therefore \frac{{\text{F}}_2}{{\text{F}}_1}=\frac{41}{50}$