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

Step 1: Understanding the Concept:
Dimensional analysis involves substituting the base dimensions (\(M, L, T\)) for each physical quantity in a formula to determine the overall dimensions of the resultant.
Step 2: Key Formula or Approach:
We find the dimensions of each variable first:
1. Energy (\(E\)): \([M L^2 T^{-2}]\)
2. Angular Momentum (\(L\)): \([M L^2 T^{-1}]\)
3. Mass (\(M\)): \([M]\)
4. Gravitational Constant (\(G\)): \([M^{-1} L^3 T^{-2}]\)
Step 3: Detailed Explanation:
Substitute the dimensions into the formula \(P = \frac{E L^2}{M^5 G^2}\):
\[ [P] = \frac{[M L^2 T^{-2}] [M L^2 T^{-1}]^2}{[M]^5 [M^{-1} L^3 T^{-2}]^2} \]
\[ [P] = \frac{[M L^2 T^{-2}] [M^2 L^4 T^{-2}]}{M^5 [M^{-2} L^6 T^{-4}]} \]
\[ [P] = \frac{M^3 L^6 T^{-4}}{M^{5-2} L^6 T^{-4}} = \frac{M^3 L^6 T^{-4}}{M^3 L^6 T^{-4}} \]
\[ [P] = [M^0 L^0 T^0] \]
Step 4: Final Answer:
The dimensions of P are \([M^0 L^0 T^0]\), meaning it is a dimensionless quantity.