JEE Mains · Physics · STD 11 - 7. gravitation
A small point of mass \(m\) is placed at a distance \(2 R\) from the centre ' \(O^{\prime}\) of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is \(\mathrm{F}_1\). A spherical part of radius \(\mathrm{R} / 3\) 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 \(\mathrm{F}_2\). The value of ratio \(\mathrm{F}_1: \mathrm{F}_2\) is _______.

- A \(12: 11\)
- B \(11: 10\)
- C \(12: 9\)
- D \(16: 9\)
Answer & Solution
Correct Answer
(A) \(12: 11\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2} .....(1)\\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}\right) \\ &…
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