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If E and G respectively denote energy and gravitational constant, then \(E\over G\) has the dimensions of: 

(1) [M2 ] [L–2 ] [T–1 ] (2) [M2 ] [L–1 ] [T0 ] (3) [M] [L–1 ] [T–1 ] (4) [M] [L0 ] [T0 ]

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Correct answer: (2) [M2] [L-1] [T0]

Explanation:

E = energy = [ML2T-2]

G = Gravitational constant = [M-1L3T-2]

So \(\frac{E}{G} = \frac{[E]}{[G]}\)

\(ML^2T^{−2}\over M^{−1}L^3T^{−2}\) = [M2] [L-1] [T0]

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