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 $R/2$ 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 placed at $A$. The ratio $F_2/F_1$ will be

• A. $1/2$
• B. $3$
• C. $7$
• D. $7/9$

Answer 1

Answer: (A), (D)

$1/2$
$7/9$

Let the density of the sphere is $\rho$ and the mass be M

And, let the mass of the particle at point A is m

Force due the solid sphere ($F_1$) is

$F_1=\frac{GMm}{{\left(2R\right)}^2}=\frac{GMm}{4R^2}$

Consider sphere of radius R/2, whose mass is M/8 and distance of its center from point A is 3R/2

Force exerted by the small sphere is $F_2$.

Consider, matter and anti-matter state,

Therefore, force exerted is

$F_2=\frac{GMm}{(2R{)}^2}-\frac{G\left(\frac{M}{8}\right)m}{{\left(\frac{3R}{2}\right)}^2}=\frac{7GMm}{36R^2}$

Hence, $\frac{F_2}{F_1}=\frac{7}{9}$