A) \[1.21\sqrt{\frac{GM}{a}}\]
B) \[1.41\sqrt{\frac{GM}{a}}\]
C) \[1.16\sqrt{\frac{GM}{a}}\]
D) \[1.35\sqrt{\frac{GM}{a}}\]
Correct Answer: C
Solution :
Net force on particle towards centre of circle is\[{{F}_{C}}=\frac{G{{M}^{2}}}{2{{a}^{2}}}+\frac{G{{M}^{2}}}{{{a}^{2}}}\sqrt{2}\]\[=\frac{G{{M}^{2}}}{{{a}^{2}}}\left( \frac{1}{2}+\sqrt{2} \right)\] This force will act as centripetal force. Distance of particle from centre of circle is\[\frac{a}{\sqrt{2}}.\] \[r=\frac{a}{\sqrt{2}},{{F}_{C}}=\frac{m{{\text{v}}^{2}}}{r}\] \[\frac{m{{\text{v}}^{2}}}{\frac{a}{\sqrt{2}}}=\frac{G{{M}^{2}}}{{{a}^{2}}}\left( \frac{1}{2}+\sqrt{2} \right)\] \[{{\text{v}}^{2}}=\frac{GM}{a}\left( \frac{1}{2\sqrt{2}}+1 \right)\] \[{{\text{v}}^{2}}=\frac{GM}{a}(1.35)\] \[\text{v}=1.16\sqrt{\frac{GM}{a}}\]You need to login to perform this action.
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