Question

If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula \(P = E L^2 M^{-5} G^{-2}\) are :

• A. \([M^1 L^1 T^{-2}]\)
• B. \([M^{-1} L^{-1} T^2]\)
• C. \([M^0 L^1 T^0]\)
• D. \([M^0 L^0 T^0]\)

Hint:

Simplify power expressions using the law of exponents (\(a^m \cdot a^n = a^{m+n}\)) to avoid confusion during long dimensional substitutions.

Answer 1

The Correct Option is D

To determine the dimensions of \( P \) in the given equation \( P = E L^2 M^{-5} G^{-2} \), we need to analyze the dimensions of each component involved.

1. Energy (E): The dimensional formula for energy is \([M^1 L^2 T^{-2}]\).
2. Angular Momentum (L): The dimensional formula for angular momentum is \([M^1 L^2 T^{-1}]\).
3. Mass (M): The dimensional formula for mass is \([M^1]\).
4. Gravitational Constant (G): The dimensional formula for the gravitational constant is \([M^{-1} L^3 T^{-2}]\).

Substituting these into the formula \( P = E L^2 M^{-5} G^{-2} \):

• For \( E \), we have \([M^1 L^2 T^{-2}]\).
• For \( L^2 \), we need to square the dimensions and get \([M^2 L^4 T^{-2}]\).
• For \( M^{-5} \), the dimensions are \([M^{-5}]\).
• For \( G^{-2} \), we need to square and invert the dimensions and get \([M^2 L^{-6} T^4]\).

Now, combining these together:

P = E L^2 M^{-5} G^{-2} = [M^1 L^2 T^{-2}] \cdot [M^2 L^4 T^{-2}] \cdot [M^{-5}] \cdot [M^2 L^{-6} T^4]

By combining these dimensions based on the formula:

• For Mass \(([M]):\) \(1 + 2 - 5 + 2 = 0\)
• For Length \(([L]):\) \(2 + 4 - 6 = 0\)
• For Time \(([T]):\) \(-2 - 2 + 4 = 0\)

Therefore, the dimensions of \( P \) are \([M^0 L^0 T^0]\), which means it is dimensionless.

The correct answer is: \([M^0 L^0 T^0]\).