Question

A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1$. Now, a spherical cavity of radius $R/2$ is made in the sphere as shown in the figure and the force becomes $F_2$. The value of $F_1:F_2$ is -

• A. $41:50$
• B. $36:25$
• C. $50:41$
• D. $25:36$

Answer 1

The correct option is C $50:41$


Gravitational field intensity before making cavity at $A$,

$g_1=\frac{GM}{(3R{)}^2}=\frac{GM}{9R^2}....\left(1\right)$

Gravitational field intensity after making cavity at $A$,

$g_2=\frac{GM}{(3R{)}^2}-\frac{G\left(M/8\right)}{(2R+R/2{)}^2}=\frac{41GM}{450R^2}....\left(2\right)$

So,

$F_1:F_2=m{g}_1:m{g}_2=g_1:g_2$

$\Rightarrow F_1:F_2=\frac{GM}{9R^2}:\frac{41GM}{450R^2}$

$[$From $\left(1\right)$ and $\left(2\right)]$

$\Rightarrow F_1:F_2=50:41$