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

The correct option is [M²] [L⁻¹] [T⁰].

To find the dimensions of the ratio of energy to the gravitational constant, we determine the individual dimensional formulas of both quantities.

1. Dimensional Formula of Energy (E):
Energy is defined as work done, which is the product of force and displacement.

$E=\text{Force}\times \text{Displacement}$

The dimensional formula of Force is $\left[ML{T}^{-2}\right]$, and for Displacement it is $\left[L\right]$.
Therefore, the dimensional formula of energy $E$ is:

$\left[E\right]=\left[ML{T}^{-2}\right]\times \left[L\right]=\left[M{L}^2{T}^{-2}\right]$

2. Dimensional Formula of the Gravitational Constant (G):
According to Newton's Law of Gravitation, the force ($F$) between two masses $m_1$ and $m_2$ separated by distance $r$ is given by:

$F=G\frac{m_1{m}_2}{r^2}$

Rearranging the formula to solve for $G$:

$G=\frac{F{r}^2}{m_1{m}_2}$

Substituting the dimensional formulas for Force $\left[ML{T}^{-2}\right]$, distance $\left[L\right]$, and mass $\left[M\right]$:

$\left[G\right]=\frac{\left[ML{T}^{-2}\right]\times \left[L^2\right]}{\left[M\right]\times \left[M\right]}=\left[M^{-1}{L}^3{T}^{-2}\right]$

3. Dimensions of the Ratio E/G:
Dividing the dimensional formula of energy by the dimensional formula of the gravitational constant:

$\left[\frac{E}{G}\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[M^{-1}{L}^3{T}^{-2}\right]}$

Applying the laws of exponents:

$\left[\frac{E}{G}\right]=\left[M^{1-(-1)}{L}^{2-3}{T}^{-2-(-2)}\right]$

$\left[\frac{E}{G}\right]=\left[M^2{L}^{-1}{T}^0\right]$