Question

A solid sphere of mass $M$ and radius $R$ has a spherical cavity of radius $\frac{R}{2}$ such that the centre of the cavity is at a distance of $\frac{R}{2}$ from the centre of the sphere. A point mass $m$ is placed inside the cavity at a distance of $\frac{R}{4}$ from the centre of the sphere. The gravitational pull between the sphere and the point mass $m$ is

• A. $\frac{11GMm}{R^2}$
• B. $\frac{14GMm}{R^2}$
• C. $\frac{GMm}{2R^2}$
• D. $\frac{GMm}{R^2}$

Answer 1

The correct option is C

$\frac{GMm}{2R^2}$

Step 1 : Given data

Mass of solid sphere $=M$

The radius of the sphere $=R$

Radius of a spherical cavity $=\frac{R}{2}$

Inside the sphere, a point mass $m$is placed

Step 2: Formula used

$F=G\frac{m_1{m}_2}{r^2}$

Where,

$F=$ force, $G=$ gravitational constant, $m_1=$ the mass of object $1$, $m_2=$mass of object $2$, $r=$ distance between centers of the masses.

Step 3 : To find the value gravitational pull

Field strength is uniform inside the cavity. Let us find at it's centre.

$F_t=F_r+F_c$

Where, $E$ is the gravitational field, $t=$Total time, $r=$Remaining , $c=$Cavity

$\Rightarrow$ $F_r=F_t-F_c$

$\Rightarrow$$\frac{GM}{R^3}\frac{R}{2}-0=\frac{GM}{2R^2}$

$F=m\times$gravitational field of remaining portion of the cavity

$$\begin{gathered} F=m{F}_R \\ =\frac{GMm}{2R^2} \end{gathered}$$

Therefore, the value of gravitational pull or acting force, $\mathit{F}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{G}\mathbf{M}\mathbf{m}}{\mathbf{2}{\mathbf{R}}^{\mathbf{2}}}$.

Hence, the correct answer is option (C).