AP EAMCET · PHYSICS · Gravitation
As shown in the figure, two spherical cavities are made in the uniform solid sphere of radius \(R\). The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the \(X\)-axis. The mass of the solid sphere before the cavities were created was \(M\). The gravitational force on a point mass \(m\) at a distance \(d\) away from the centre of the solid sphere is

- A \(\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{2 d}\right)^2}\right]\)
- B \(\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{d}\right)^2}\right]\)
- C \(\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{d}{R}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{d}{R}\right)^2}\right]\)
- D \(\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{d}{R}\right)^2}+\frac{1}{8} \frac{1}{\left(1-\frac{d}{R}\right)^2}\right]\)
Answer & Solution
Correct Answer
(A) \(\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{2 d}\right)^2}\right]\)
Step-by-step Solution
Detailed explanation
Given situation is as shown Radius of sphere given \(=R\) Mass of sphere \(=M\) Density of sphere, \(d=\frac{M}{\frac{4}{3} \pi R^3}\) Radius of each of cavity \(=R / 2\) Mass of each of portion removed to create a cavity \(=\) density \(\times\) volume…
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