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JEE Main 2014 · 11 Apr · Shift 1, Physics question 7

From a sphere of mass M and radius R, a smaller sphere of radius R2\frac R2 is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :

A large shaded sphere with a spherical cavity of radius R/2 touching its surface, between its centre and its edge; the removed small sphere sits on the same line to the right, its centre 3R from the large sphere's centre.
Choose one option
Answer and worked solution

Official answer: A

Treat the remaining body as the full sphere MM minus the cavity, a sphere of mass M8\frac M8 centred R2\frac R2 from O. The removed sphere has mass M8\frac M8, at 3R3R from O and 2.5R2.5R from the cavity's centre.
F=GM⋅M8(3R)2−G(M8)2(2.5R)2=GM2R2(172−1400)=41GM23600R2.F=\frac{GM\cdot\frac M8}{(3R)^2}-\frac{G\left(\frac M8\right)^2}{(2.5R)^2}=\frac{GM^2}{R^2}\left(\frac1{72}-\frac1{400}\right)=\frac{41GM^2}{3600R^2}.

Topics:Newton's law of gravitation
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