Question

If $E$ and $G$ respectively denote energy and gravitational constant, then $\frac{ E }{ G }$ has the dimensions of:

• A. $\left[ M ^{2}\right]\left[ L ^{-1}\right]\left[ T ^{0}\right]$
• B. $[ M ]\left[ L ^{-1}\right]\left[ T ^{-1}\right]$
• C. $[ M ]\left[ L ^{0}\right]\left[ T ^{0}\right]$
• D. $\left[ M ^{2}\right]\left[ L ^{-2}\right]\left[ T ^{-1}\right]$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the dimensions of the expression \(\frac{E}{G}\), where \(E\) is energy and \(G\) is the gravitational constant. Let's analyze their dimensional formulas step-by-step.

1. The dimensional formula for energy \((E)\) is: \([M^1 L^2 T^{-2}]\). This is derived from the formula for kinetic energy or potential energy, which is expressed as: \( E = \frac{1}{2} mv^2 \) or \( mgh \).
2. The dimensional formula for the gravitational constant \((G)\) is: \([M^{-1} L^3 T^{-2}]\). This is derived from Newton's law of gravitational force: \( F = G\frac{m_1 m_2}{r^2} \).
3. To find the dimensions of \(\frac{E}{G}\), we divide the dimensions of energy by those of the gravitational constant:

\[ \frac{[M^1 L^2 T^{-2}]}{[M^{-1} L^3 T^{-2}]} \]

This simplifies to:

\[ [M^1 L^2 T^{-2}] \times [M^1 L^{-3} T^2] \]

1. Performing the multiplication and simplification, we get:

\[ [M^{1+1} L^{2-3} T^{-2+2}] \]

This results in:

\[ [M^2 L^{-1} T^{0}] \]

Thus, the dimensional formula for \(\frac{E}{G}\) is \([ M^2 ][ L^{-1} ][ T^0 ]\).

Therefore, the correct answer is: \(\left[ M ^{2}\right]\left[ L ^{-1}\right]\left[ T ^{0}\right]\).