Kerala KEAM 5 June 2024 Solved Paper

The gravitational constant $G$ appears in Newton's law of gravitation:$$F = \frac{G m_1 m_2}{r^2}$$where:- $F$ is the force with dimensional formula $[M L T^{-2}]$- $m_1$ and $m_2$ are masses with dimensional formula $[M]$- $r$ is the distance with dimensional formula $[L]$The dimensional formula of $G$ can be derived as follows:$$[F] = \frac{[G] [M]^2}{[L]^2}$$Substituting the dimensional formulas:$$[M L T^{-2}] = \frac{[G] M^2}{L^2}$$Solving for $[G]$:$$[G] = \frac{M L^3 T^{-2}}{M^2} = M^{-1} L^3 T^{-2}$$Thus, the dimensional formula of $G$ is:$$[G] = M^{-1} L^3 T^{-2}$$Therefore, comparing this with $M^a L^b T^c$, we get:$$a = -1, \quad b = 3, \quad c = -2$$Thus, the correct values of $a$, $b$, and $c$ are $-1, 3, -2$.$$\boxed{(-1, 3, -2)}$$ Quick Tip: To find the dimensional formula of physical constants, use the fundamental equations where the constant is present, and balance the dimensions of all quantities involved.