Question

A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). 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 the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:

 

• A. 16 : 9
• B. 11 : 10
• C. 12 : 11
• D. 12 : 9

Hint:

Use the inverse square law for gravitational force and break the mass into parts when necessary.

Answer 1

The Correct Option is C

The gravitational force on \(m\) due to the whole sphere is: \[ F_1 = \frac{GMm}{(2R)^2} \quad \text{(1)} \] The gravitational force due to the remaining mass after removing the spherical part of radius \(R/3\) is: \[ F_2 = \frac{GMm}{(2R)^2} \times \left(\frac{M}{27} \times \frac{4R}{3} \right)^2 = \frac{11 GMm}{48R^2} \quad \text{(2)} \] Thus, the ratio is: \[ \frac{F_1}{F_2} = 12 : 11 \] Thus, the answer is \( \boxed{12 : 11} \).