Question

Match the LIST-I with LIST-II

\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]

Choose the correct answer from the options given below:

• A. A-IV, B-III, C-II, D-I
• B. A-III, B-II, C-I, D-IV
• C. A-II, B-IV, C-III, D-I
• D. A-I, B-III, C-IV, D-II

Hint:

Use the formulas for gravitational constant, gravitational potential energy, gravitational potential, and acceleration due to gravity to derive their dimensional formulas.

Answer 1

The Correct Option is A

(A) \( G = \frac{Fr^2}{m^2} \) \( [G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}] \) (IV) 

(B) P.E. = mgh = \( [MLT^{-2}L] = [ML^2T^{-2}] \) (III) 

(C) Gravitational Potential = \( \frac{GM}{r} \) \( [M^{-1}L^3T^{-2}] \frac{[M]}{[L]} = [M^0L^2T^{-2}] = [L^2T^{-2}] \) (II) 

(D) Acceleration due to gravity = \( \frac{GM}{r^2} \) \( [M^{-1}L^3T^{-2}] \frac{[M]}{[L^2]} = [M^0LT^{-2}] = [LT^{-2}] \) (I)