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: (1) 16 : 9 (2) 11 : 10 (3) 12 : 11 (4) 12 : 9

Key Concepts

1. Inverse-square law: $F = G \frac{M m}{r^{2}}$
2. Shell theorem: For any point outside a uniform solid sphere, the entire mass can be considered at the centre.
3. Superposition (addition & subtraction): The net gravitational field equals the vector sum of fields produced by each mass element. If mass is removed, treat it as adding a negative mass at its own centre.

Logical chain to solve

1. Whole sphere (before removal) Distance from point mass to centre O: $2R$ $F_1 = G \frac{M m}{(2R)^2}=G \frac{M m}{4R^2}$

2. Describe the cavity Radius of cavity: $R/3$
   Volume ratio: $(R/3)^3 : R^3 = 1:27$
   Mass removed: $M/27$ (since density is uniform).

3. Locate the cavity centre (given/figure): on the line joining O and $m$, at a distance $\dfrac{2R}{3}$ from O.
   Distance from cavity centre to $m$: $\bigl(2R - \frac{2R}{3}\bigr)=\frac{4R}{3}$

4. Force due to the removed part (take its mass as negative):

   $F_{\text{cav}}= G\frac{(M/27)m}{\left(\dfrac{4R}{3}\right)^2}=G\frac{M m}{48R^2}$ (direction is the same as $F_1$ toward O).

5. Subtract to find the new force

   $F_2 = F_1 - F_{\text{cav}} = G\frac{M m}{4R^2}-G\frac{M m}{48R^2}=G\frac{M m}{R^2}\left(\frac{1}{4}-\frac{1}{48}\right)=G\frac{M m}{R^2}\frac{11}{48}$

6. Ratio

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

What’s happening?

Imagine you have a huge, perfectly round laddu (the big sphere) that pulls a tiny chocolate chip (the small mass $m$) toward its centre because of gravity.

1. First case: The laddu is whole, so the pull (force) is $F_1$.
2. Second case: You scoop out a small ball–shaped bite (a cavity) from the laddu. Now the remaining laddu pulls the chip with a slightly different force, $F_2$.

Because every bit of the laddu attracts the chip, removing some dough means subtracting that little bit of attraction. We use the idea that a sphere attracts as if all its weight was concentrated at its centre. The scoop you removed also acted like a mini-laddu whose pull you now have to ‘take away’.
So, $F_2$ is just

pull of whole laddu minus pull of the scooped-out bite.

Do the maths carefully and the numbers turn out so that the first force ($F_1$) and the new force ($F_2$) are in the ratio 12 : 11.

Step-by-step Calculation

1. Force before removing the cavity

   $F_1 = G\frac{M m}{(2R)^2}=G\frac{M m}{4R^{2}}$

2. Mass of the removed sphere

   $M_{\text{cav}} = \rho \cdot \left(\frac{4\pi}{3}\right)\left(\frac{R}{3}\right)^3 = \frac{1}{27}\,M$

3. Distance of point mass from cavity centre

   $r_{\text{cav}} = 2R - \frac{2R}{3} = \frac{4R}{3}$

4. Force that the cavity alone would exert

   $F_{\text{cav}} = G\frac{\left(\frac{M}{27}\right)m}{\left(\dfrac{4R}{3}\right)^2} = G\frac{M m}{48R^{2}}$

5. Net force after cavity is removed

   $F_2 = F_1 - F_{\text{cav}} = G\frac{M m}{4R^{2}} - G\frac{M m}{48R^{2}} = G\frac{M m}{R^{2}}\left(\frac{1}{4}-\frac{1}{48}\right)=G\frac{M m}{R^{2}}\frac{11}{48}$

6. Ratio

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

Hence the correct option is (3) 12 : 11.