A) \[[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]\]
B) \[[ML{{T}^{-1}}]\]
C)
D) \[[M{{L}^{-2}}{{T}^{-2}}]\]
Correct Answer: A
Solution :
[a] \[F=\frac{GMm}{{{R}^{2}}}\] \[\therefore \,\,\,\,\,[L]=\frac{F{{R}^{2}}}{Mm}\Rightarrow G=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]\]
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