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

Match the quantities in LIST-I with their corresponding dimensional formulas in LIST-II. Proficiency in quantities and their dimensional formulas is crucial for solving such problems.

1. Gravitational Constant (A):

The gravitational constant \(G\) is a component of Newton's law of universal gravitation: \(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\), where \(F\) represents gravitational force. Rearranging the formula to solve for \(G\) yields: \(G = \frac{F \cdot r^2}{m_1 \cdot m_2}\). Consequently, its dimensional formula is \([M^{-1}L^3T^{-2}]\).

1. Gravitational Potential Energy (B):

This energy arises from an object's position within a gravitational field and is calculated using the formula \(U = m \cdot g \cdot h\). In this equation, \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is height. Therefore, its dimensional formula is \([ML^2T^{-2}]\).

1. Gravitational Potential (C):

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

1. Acceleration due to Gravity (D):

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

The correct answer, by matching descriptions with dimensional formulas, is: A-IV, B-III, C-II, D-I.