Question

A cavity of radius $\frac{R}{2}$is made inside a solid sphere of radius $R$. The centre of the cavity is located at a distance $\frac{R}{2}$ from the centre of the sphere. The gravitational field intensity at a point $\text{P}$, as shown in the figure is:

[Here $g=\frac{GM}{R^2}$, where $M$ is the mass of the solid sphere without cavity]

• A. $0$
• B. $\frac{g}{4}$, towards Right
  
• C. $\frac{g}{2}$, towards left
  
• D. $\frac{g}{4}$, towards Right
  

Answer 1

The correct option is A $0$

The net gravitational field at point $\text{P}$ can be calculated by the superposition principle.


Given, $g=\frac{GM}{R^2}$

Since, the density of sphere is uniform throughout, $m\propto v$

So, the mass of the cavity portion , $M_2=\frac{M}{8}$

Gravitational field intensity at any point inside a sphere is given by,

$g=\frac{GMr}{R^3}$

Here, $r$ is distance of the given point from the center of the sphere, $M$ and $R$ are mass and radius of the sphere, respectively.

Also, the gravitational field intensity is directed towards the center of the sphere, as the gravitationaal force is attractive in nature.

$\therefore$Gravitational field intensity at point $\text{P}$ due to the entire sphere is given by:

$g_1=\frac{GM\left(\frac{R}{4}\right)}{R^3}=\frac{1}{4}\frac{GM}{R^2}$

$g_1=\frac{g}{4}$, directed towards left.

Now, gravitational field intensity at the point $\text{P}$ due to cavity is given by,

$g_2=\frac{G\left(\frac{M}{8}\right)\left(\frac{R}{4}\right)}{{\left(\frac{R}{2}\right)}^3}$

$g_2=\frac{g}{4}$, directed towards right.

$\therefore$ Net gravitational field due to solid sphere with cavity will be,

$g_{net}=g_1-g_2$

$=\frac{g}{4}-\frac{g}{4}=0$

Hence, option $\left(A\right)$ is the correct answer.