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., \[(\sqrt{3}Gm)M/8{{r}^{2}}\].You need to login to perform this action.
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