Question

The dimensional formula for the gravitational constant is:

• A. $\text{ }\\!\\![\\!\\!\text{ }{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }$
• B. $\text{ }\\!\\![\\!\\!\text{ ML}{{\text{T}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }$
• C. $\text{ }\\!\\![\\!\\!\text{ M}{{\text{L}}^{2}}{{\text{T}}^{-2}}\text{ }\\!\\!]\\!\\!\text{ }$
• D. $\text{ }\\!\\![\\!\\!\text{ }{{\text{M}}^{-1}}{{\text{L}}^{3}}\text{T }\\!\\!]\\!\\!\text{ }$

Answer 1

Answer: (A)

$\text{ }\\!\\![\\!\\!\text{ }{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }$

Answer 2

Gravitational constant is equal in magnitude to that force of attraction which acts between two particles each of unit mass separated by a unit distance apart.
$\therefore$ $G=\frac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$
(Newtons law of gravitation) where
${{m}_{1}}$ and ${{m}_{2}}$ are masses, r is the distance between them, and F is force.
$\therefore$ Dimensions of gravitational constant
$\text{=}\,\,\frac{\text{dimensions}\,\text{of}\,\text{force}\,\,\text{ }\\!\\!\times\\!\\!\text{ }\,{{\text{(length)}}^{\text{2}}}}{{{\text{(dimensions}\,\text{of}\,\text{mass)}}^{\text{2}}}}$
$=\frac{[ML{{T}^{-2}}][{{L}^{2}}]}{[{{M}^{2}}]}$
$=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]$