A) \[{{R}^{n}}\]
B) \[{{R}^{-n}}\]
C) \[{{R}^{(n+1)2}}\]
D) \[{{R}^{(n-1)/2}}\]
Correct Answer: C
Solution :
\[T=\frac{2\pi R}{v}\]??.(1); \[V={{\left[ \frac{2GM}{{{R}^{n-1}}} \right]}^{1/2}}\]?.(2) \[E=\frac{1}{2}m{{v}^{2}}=\frac{GMm}{{{R}^{n-1}}}\] From (1) & (2), \[T=-\frac{2\pi R}{\sqrt{\frac{2GM}{{{R}^{n-1}}}}}=\frac{2\pi R}{\sqrt{\frac{2GM}{{{R}^{n-1}}}}}=\frac{2\pi }{\sqrt{2GM}}\times {{R}^{(n+1)/2}}.\] Therefore, \[T\propto {{R}^{(n+1)/2}}\]
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