Question

If E and G respectively denote energy and gravitational constant, then $\frac{E}{G}$ has the dimensions of: 

Answer 1

Correct answer: (2) [M²] [L^-1] [T⁰]

Explanation:

E = energy = [ML²T^-2]

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

So $\frac{E}{G}=\frac{[E]}{[G]}$

= $\frac{M{L}^2{T}^{-2}}{M^{-1}{L}^3{T}^{-2}}$ = [M²] [L^-1] [T⁰]