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 $2R$ from the centre of the sphere. A spherical cavity of radius $\frac{R}{2}$ 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 $\frac{F_1}{F_2}$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{7}{8}$
• D. $\frac{9}{7}$

Answer 1

The correct option is D $\frac{9}{7}$

Let the density of the sphere be $\rho$ and the mass of sphere be $M$ and let the mass of the particle at point $\text{A}$ be $\text{m}$.

Mass of complete sphere $=M$

Force on the mass $m$ due the solid sphere is $$F_1=\frac{GMm}{(2R{)}^2}$$$\Rightarrow F_1=\frac{GMm}{4R^2}$

Mass of cavity $M_C=\rho \times$Volume of cavity

$\Rightarrow M_C=\frac{M}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3$

$\therefore M_C=\frac{M}{8}$

By principle of super position, force acting the particle of mass $m$ by the sphere with cavity is

$F_2=F_{complete}-F_{cavity}$

Where,
$F_{complete}=F_1=$ Force due to whole solid sphere.
$F_{cavity}=$Force due to the mass of spherical cavity.

Substituting the values,

$\Rightarrow F_2=\frac{GMm}{(2R{)}^2}-\frac{GMm}{8{\left(\frac{3R}{2}\right)}^2}$

$\Rightarrow F_2=\frac{GMm}{4R^2}-\frac{GMm}{18R^2}$

$\Rightarrow F_2=\frac{7}{36}\frac{GMm}{R^2}$

$\therefore F_2=\frac{7}{36}\frac{GMm}{R^2}$



Ratio of the forces will be,

$\frac{F_1}{F_2}=\frac{1}{4}\times \frac{36}{7}$

$\therefore \frac{F_1}{F_2}=\frac{9}{7}$

Why this question: To make the student familier with use of the principle of superposition when the mass of body is changed. This can be applied for any kind of body shape.