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

Let's match the quantities in LIST-I with their corresponding dimensional formulas in LIST-II. Understanding the quantities and their respective dimensional formulas is essential in solving this type of question.

1. Gravitational Constant (A):

The gravitational constant \(G\) is used in Newton's law of universal gravitation: \(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\), where \(F\) is the gravitational force. The dimensional formula for \(G\) can be derived by rearranging the formula: \(G = \frac{F \cdot r^2}{m_1 \cdot m_2}\). Thus, its dimensional formula is \([M^{-1}L^3T^{-2}]\).

1. Gravitational Potential Energy (B):

This is the energy due to position in a gravitational field, given by \(U = m \cdot g \cdot h\). Here, \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height. So, its dimensional formula is \([ML^2T^{-2}]\).

1. Gravitational Potential (C):

The gravitational potential at a point is defined as the work done per unit mass to bring a mass from infinity to that point: \(V = \frac{U}{m} = \frac{m \cdot g \cdot h}{m} = g \cdot h\). Its dimensional formula can be derived as \([L^2T^{-2}]\).

1. Acceleration due to Gravity (D):

The acceleration due to gravity \(g\) is the acceleration experienced by an object due to the gravitational force. Its dimensional formula is derived from \(F = m \cdot g\), leading to \(g = \frac{F}{m}\). Hence, its dimensional formula is \([LT^{-2}]\).

By matching the descriptions and dimensional formulas, we get the correct answer: A-IV, B-III, C-II, D-I.

Answer 2

(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)