Question

If E and G respectively denote energy and gravitational constant, then \(\frac{E}{G}\) has the dimensions

• A. [M²] [L⁻²] [T⁻¹]
• B. [M²] [L⁻¹] [T⁰]
• C. [M] [L⁻¹] [T⁻¹]
• D. [M] [L⁰] [T⁰]

Answer 1

The Correct Option is B

To solve the problem of determining the dimensions of \(\frac{E}{G}\), we need to understand the dimensional formulas for energy and the gravitational constant first.

1. Energy (E):
   The dimensional formula for energy is derived from its basic definition. It can be expressed in terms of work done, which is force times distance.
   • Force (F) has the dimensional formula: \([M][L][T^{-2}]\)
   • Distance (d) has the dimensional formula: \([L]\)
   Therefore, Energy (E) is: \([M][L][T^{-2}] \times [L] = [M][L^2][T^{-2}]\).
2. Gravitational Constant (G):
   The gravitational constant G relates force between two masses (\(F\)) in Newton's Law of Gravitation: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] Rearranging this, we find G: \[ G = \frac{F \cdot r^2}{m_1 \cdot m_2} \] Substituting the dimensional formulae, since \(F = [M][L][T^{-2}]\) and \(r = [L]\), we get: \[ G = \frac{[M][L][T^{-2}] \cdot [L]^2}{[M]^2} = [M^{-1}][L^3][T^{-2}] \]
3. Combine the two formulas to find \(\frac{E}{G}\):
   • The dimensional formula for Energy (E) is: \([M][L^2][T^{-2}]\)
   • The dimensional formula for Gravitational Constant (G) is: \([M^{-1}][L^3][T^{-2}]\)
4. Compute the dimensions of \(\frac{E}{G}\): \[ \frac{E}{G} = \frac{[M][L^2][T^{-2}]}{[M^{-1}][L^3][T^{-2}]} = [M]^{1-(-1)}[L]^{2-3}[T]^{0} \] Simplifying the exponents, we get: \[ \frac{E}{G} = [M]^2[L]^{-1}[T]^0 \]
5. Conclusion: The dimensions of \(\frac{E}{G}\) are \([M^2][L^{-1}][T^0]\), which matches the given correct option: \([M^2][L^{-1}][T^0]\).

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