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

The correct answer is [M²] [L^-1] [T⁰].

To find the dimensional formula of the ratio of Energy (E) to the Gravitational constant (G), we first need to determine the dimensional formula for each quantity separately.

Step 1: Dimensional formula of Energy (E)
Energy is dimensionally equivalent to work, which is defined as force multiplied by displacement.

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

The dimensional formula for Force is [M L T^-2] and for Displacement is [L]. Multiplying these gives the dimensions of Energy:

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

Step 2: Dimensional formula of the Gravitational constant (G)
According to Newton's law of universal gravitation, the force between two masses is given by:

$F=\frac{G{m}_1{m}_2}{r^2}$

Rearranging this formula to solve for the gravitational constant G gives:

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

Substituting the dimensions for force [M L T^-2], distance squared [L²], and the product of the two masses [M²]:

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

Step 3: Calculate the dimensions of E/G
Now, we divide the dimensional formula of E by the dimensional formula of G:

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

Using the laws of exponents to subtract the exponents of the denominator from the numerator for each base:

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

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

Thus, the final dimensional formula for E/G evaluates to [M²] [L^-1] [T⁰].