question_answer

A uniform ring of mass \[m\] and radius r is placed directly above a uniform sphere of mass \[M\] and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance \[r\sqrt{3}\]\[r\] as shown in the figure. The gravitational force exerted by the sphere on the ring will be

A) \[\frac{GMm}{8{{r}^{2}}}\]   

B) \[\frac{GMm}{4{{r}^{2}}}\]

C) \[\sqrt{3}\frac{GMm}{8{{r}^{2}}}\]   

D) \[\frac{GMm}{8{{r}^{2}}\sqrt{3}}\]

Correct Answer: C

Solution :

[c] \[dF=G\frac{Mdm}{4{{r}^{2}}}\]  \[F=\Sigma dF\cos \theta \] \[=\Sigma \frac{GMdm}{4{{r}^{2}}}\cos \theta \] \[=\frac{GM}{4{{r}^{2}}}\times \frac{\sqrt{3r}}{2r}\Sigma dm\] \[=\frac{\sqrt{3}GMm}{8{{r}^{2}}}\] Alternative solution: The gravitational field due to the ring at a distance \[E=\frac{Gm(\sqrt{3r})}{{{[{{r}^{2}}+{{(\sqrt{3}r)}^{2}}]}^{3/2}}}\]or E=\[\frac{\sqrt{3}Gm}{8{{r}^{2}}}\] The required force is EM, i.e., \[(\sqrt{3}Gm)M/8{{r}^{2}}\].