From a sphere of mass M and radius R, a smaller sphere of radius R/2… | JEE Main 2014 (11 Apr, Shift 1)

Q19JEE Main 2014 · 11 APR · SHIFT 1Difficult

Newton's Law of Gravitation

From a sphere of mass M and radius R, a smaller sphere of radius $\frac{R}{2}$ 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 center of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is:

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Answer: (a)

Removed sphere: mass $\frac{M}{8}$ ; it sits $3R$ from O. The cavity centre is $\frac{R}{2}$ from O, so $3R - \frac{R}{2} = \frac{5R}{2}$ from the removed sphere.
Remaining part = full sphere $M$ minus cavity sphere $\frac{M}{8}$ .
$F = \frac{GM(M/8)}{(3R)^2} - \frac{G(M/8)(M/8)}{(5R/2)^2} = \frac{GM^2}{72R^2} - \frac{GM^2}{400R^2}$
$F = \frac{(50 - 9)GM^2}{3600R^2} = \frac{41GM^2}{3600R^2}$
So the answer is (a).

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