What is the dimensional formula of gravitational constant G ?

The correct option is [M⁻¹L³T⁻²].

To find the dimensional formula of the universal gravitational constant ($G$), we start with Newton's Law of Gravitation, which describes the gravitational force between two masses:

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

Where:
• $F$ is the gravitational force between the masses,
• $m_1$ and $m_2$ are the masses of the two objects,
• $r$ is the distance between the centers of the two masses,
• $G$ is the universal gravitational constant.

Rearranging the formula to solve for $G$:

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

Now, let's write down the dimensional formulas of the individual physical quantities involved:
• Force ($F$) = Mass × Acceleration = $\left[{\mathrm{MLT}}^{-2}\right]$
• Distance ($r$) = $\left[L\right]$ (so, $r^2=\left[L^2\right]$)
• Mass ($m_1$ or $m_2$) = $\left[M\right]$ (so, $m_1{m}_2=\left[M^2\right]$)

Substitute these dimensional formulas into the rearranged equation for $G$:

$\left[G\right]=\frac{\left[{\mathrm{MLT}}^{-2}\right]·\left[L^2\right]}{\left[M\right]·\left[M\right]}$

Simplify the expression in the numerator first:

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

Now divide by the denominator:

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

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

Thus, the dimensional formula of the gravitational constant $G$ is [M⁻¹L³T⁻²].