UPSC NDA I 2022 General Ability Solved Paper

Concept:The gravitational constant $G$ is derived from Newton's law of gravitation: $F = \frac{G m_1 m_2}{r^2}$.Its dimension is found by rearranging the formula and substituting known dimensions of force, mass, and distance.Explanation:From Newton's law: $F = \frac{G m_1 m_2}{r^2}$.Rearrange to solve for $G$: $G = \frac{F r^2}{m_1 m_2}$.The dimension of force $F$ is $[MLT^{-2}]$ (mass $\times$ length $\times$ time$^{-2}$).The dimension of distance $r$ is $[L]$, so $r^2$ has dimension $[L^2]$.The dimension of mass $m_1$ and $m_2$ is $[M]$ each, so product $m_1 m_2$ has dimension $[M^2]$.Substitute: $[G] = \frac{[MLT^{-2}] \cdot [L^2]}{[M^2]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$.Answer:The dimension of gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$, which corresponds to option B.