A small point of mass m is placed at a distance 2R from the centre ‘O’ of a big uniform solid sphere of mass M and radius R. The gravitational force on ‘m’ due to M is F₁. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to
remaining part of M is found to be F₂. The value of ratio F₁ : F₂ is

To determine the ratio $F_1 : F_2$, we analyze the gravitational force acting on the mass $m$ before and after removing the smaller spherical part.

Step 1: Gravitational Force Before Removing the Small Sphere

The force $F_1$ is due to the entire sphere of mass $M$ and radius $R$, and the mass $m$ is located at a distance $2R$ from the center.

The gravitational force due to a sphere acts as if the entire mass were concentrated at the center:

$$F_1 = \frac{G M m}{(2R)^2}$$

 $$F_1 = \frac{G M m}{4R^2}$$

Step 2: Gravitational Effect of the Removed Sphere

The removed sphere has a radius of $R/3$. The mass of a sphere is proportional to the volume, so the mass $M'$ of the removed part is:

$$M' = M \times \left(\frac{\text{volume of removed sphere}}{\text{volume of original sphere}}\right)$$

 $$M' = M \times \left(\frac{\frac{4}{3} \pi (R/3)^3}{\frac{4}{3} \pi R^3}\right)$$ 

$$M' = M \times \left(\frac{1}{27}\right) = \frac{M}{27}$$

Now, we calculate the gravitational force $F'$ due to this removed sphere on $m$. The center of the removed sphere was at a distance $R$ from the center of the big sphere. The force due to this removed mass at $2R$ is:

$F^{\text{'}}=\frac{G{M}^{\text{'}}m}{(R+{\displaystyle \frac{R}{3}}{)}^2}=\frac{G\left(M\mathrm{/}27\right)m}{{\displaystyle \frac{16}{9}}{R}^2}$

 $F^{\text{'}}=\frac{GMm}{48R^2}$

Step 3: Gravitational Force After Removing the Small Sphere

By the superposition principle, the net force $F_2$ due to the remaining part is:

$$F_2 = F_1 - F'$$

 $F_2=\frac{GMm}{4R^2}-\frac{GMm}{48R^2}$

 $F_2=\frac{44}{4\times 48}\times \frac{GMm}{R^2}$

Since $F_1=\frac{GMm}{4R^2}$, the ratio $F_1 : F_2$ is:

$F_1:F_2=\frac{4\times 48}{4\times 44}=\frac{12}{11}$