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 figure) and the force becomes $F_2$. The value of $F_1 : F_2$ is :

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

Hint:

For cavity problems, use the {Principle of Superposition}: $F_{rem} = F_{total} - F_{removed\_part}$. Remember that mass scales with the cube of the radius ($M \propto R^3$) for solid spheres.

Answer 1

The Correct Option is B

Step 1: $F_1 = \frac{G M m}{(3R)^2} = \frac{GMm}{9R^2}$.
Step 2: Mass of removed part $m' = M(r_{cav}/R)^3 = M(1/2)^3 = M/8$.
Step 3: Distance of cavity center from test particle $= 3R - R/2 = 2.5R = 5R/2$.
Step 4: Force from removed part $F' = \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{8 \times \frac{25R^2}{4}} = \frac{GMm}{50R^2}$.
Step 5: $F_2 = F_1 - F' = \frac{GMm}{R^2} [1/9 - 1/50] = \frac{GMm}{R^2} [\frac{50-9}{450}] = \frac{41GMm}{450R^2}$.
Step 6: Ratio $F_1/F_2 = (1/9) / (41/450) = \frac{1}{9} \times \frac{450}{41} = 50/41$.