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If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of :

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$[M^{2}] [L^{- 1}] [T^{0}]$

$[M] [L^{0}] [T^{0}]$

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

$[M] [L^{- 1}] [T^{- 1}]$

answer is B.

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Detailed Solution

$\begin{array}{l} [E] = {ML}^{2} T^{- 2} \\ [G] = \frac{[{Fr}^{2}]}{[m_{1} m_{2}]} = [\frac{{MLT}^{- 2} L^{2}}{M^{2}}] = M^{- 1} L^{3} T^{- 2} \\ ∴ [\frac{E}{G}] = \frac{{ML}^{2} T^{- 2}}{M^{- 1} L^{3} T^{- 2}} = [M^{2} L^{- 1} T^{0}] \end{array}$

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