A) \[T=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\]
B) \[T=2\pi \sqrt{\frac{GM}{{{R}^{3}}}}\]
C) \[T=2\pi \sqrt{\frac{GM}{{{R}^{2}}}}\]
D) \[T=2\pi \sqrt{\frac{{{R}^{2}}}{GM}}\]
Correct Answer: A
Solution :
Taking \[T=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\] Substituting the dimensions, LHS \[T=[T]\] RHS \[2\pi \sqrt{\frac{{{R}^{3}}}{GM}}=\sqrt{\frac{[{{L}^{3}}]}{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}][M]}}\] \[=\sqrt{[{{T}^{2}}]}=[T]\] We get, LHS=RHS Hence, the equation \[T=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}\] is correct.
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