A) \[\sqrt{\frac{GM}{R}(1+2\sqrt{2})}\]
B) \[\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}\]
C) \[\sqrt{\frac{GM}{R}}\]
D) \[\sqrt{2\sqrt{2}\frac{GM}{R}}\]
Correct Answer: B
Solution :
[b] Net force on any one particle \[=\frac{G{{M}^{2}}}{{{(2R)}^{2}}}+\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ +\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ \] \[=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1}{4}+\frac{1}{\sqrt{{}}} \right]\] This force will be equal to centripetal force so \[\frac{M{{u}^{2}}}{R}=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1+2\sqrt{2}}{4} \right]\] \[u=\sqrt{\frac{GM}{R}[1+2\sqrt{2}]}\] \[=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}\]You need to login to perform this action.
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