Gravitation

JEE Main level, single correct, medium

153 of 223

Question 153 of 223

From a sphere of mass MM and radius R\mathrm{R}, a smaller sphere of radius R2\frac{\mathrm{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 centre of the original sphere and the removed sphere is 3R, the gravitational force between the two sphere is:

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Answer: (A) 41GM23600R2\frac{41 \mathrm{GM}^2}{3600 \mathrm{R}^2}

Volume of removed sphere
Vremo =43π(R2)3=43πR3(18)V_{\text {remo }}=\frac{4}{3} \pi\left(\frac{R}{2}\right)^3=\frac{4}{3} \pi R^3\left(\frac{1}{8}\right)
Volume of the sphere (remaining)
Vremain =43πR3−43πR3(18)=43πR3(78)\begin{aligned} &V_{\text {remain }}=\frac{4}{3} \pi R^3-\frac{4}{3} \pi R^3\left(\frac{1}{8}\right) \\ &=\frac{4}{3} \pi R^3\left(\frac{7}{8}\right) \end{aligned}
Therefore mass of sphere carved and remaining sphere are at respectively 18M\frac{1}{8} \mathrm{M} and 78M\frac{7}{8} \mathrm{M}

Therefore, gravitational force between these two sphere,
F=GMmr2F=\frac{G \mathrm{M} m}{r^2}

=G7M8×18M(3R)2=764×9GM2R2≃413600GM2R2\begin{aligned} &=\frac{G \frac{7 M}{8} \times \frac{1}{8} M}{(3 R)^2}=\frac{7}{64 \times 9} \frac{G M^2}{R^2} \\ &\simeq \frac{41}{3600} \frac{\mathrm{GM}^2}{\mathrm{R}^2} \end{aligned}

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