The dimensional formula for universal gravitational constant is

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$M L^{2} T^{- 1}$

$M L^{- 1} T^{- 2}$

$M L^{- 1} T^{2}$

$M^{- 1} L^{3} T^{- 2}$

answer is D.

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Detailed Solution

$F = \frac{{Gm}_{1} m_{2}}{r^{2}} \Rightarrow G = \frac{{Fr}^{2}}{m_{1} m_{2}} Thus dimensional formula of G is \frac{[{MLT}^{- 2}] [L]^{2}}{[M]^{2}} \Rightarrow G = M^{- 1} L^{3} T^{- 2}$ Theory

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