JEE PYQ: Units & Measurements - Question ID ccb29f5e3b19 (JEE Main 2021)

ID: ccb29f5e3b19JEE Main 2021Single Correct MCQ
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 = EL2M-5G-2 are :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass (MM), length (LL), and time (TT). The given problem involves determining the dimensions of a quantity PP defined by the formula: P=EL2M5G2P = E L^2 M^{-5} G^{-2} where:

  • EE is energy,
  • LL is angular momentum,
  • MM is mass,
  • GG is the gravitational constant.
To find the dimensions of PP, we first express each of these quantities in terms of MM, LL, and TT.

Key dimensional formulas:

  • Energy (EE): [E]=[ML2T2][E] = [M L^2 T^{-2}]
  • Angular momentum (LL): [L]=[ML2T1][L] = [M L^2 T^{-1}]
  • Mass (MM): [M]=[M][M] = [M]
  • Gravitational constant (GG): [G]=[M1L3T2][G] = [M^{-1} L^3 T^{-2}]

Step-by-Step Derivation:

Step 1: Write the dimensional formula for each quantity in the expression for PP.
Given: P=EL2M5G2P = E \cdot L^2 \cdot M^{-5} \cdot G^{-2} Substitute the dimensions: [P]=[E][L]2[M]5[G]2[P] = [E] \cdot [L]^2 \cdot [M]^{-5} \cdot [G]^{-2} Step 2: Substitute the known dimensions. [P]=[ML2T2][ML2T1]2[M]5[M1L3T2]2[P] = [M L^2 T^{-2}] \cdot [M L^2 T^{-1}]^2 \cdot [M]^{-5} \cdot [M^{-1} L^3 T^{-2}]^{-2} Step 3: Expand the exponents. [P]=[ML2T2][M2L4T2][M]5[M2L6T4][P] = [M L^2 T^{-2}] \cdot [M^2 L^4 T^{-2}] \cdot [M]^{-5} \cdot [M^{2} L^{-6} T^{4}] Note: [G]2=[M1L3T2]2=[M2L6T4][G]^{-2} = [M^{-1} L^3 T^{-2}]^{-2} = [M^{2} L^{-6} T^{4}] Step 4: Combine the dimensions by adding exponents for each fundamental dimension (MM, LL, TT).

  • For mass (MM): 1+25+2=01 + 2 - 5 + 2 = 0
  • For length (LL): 2+4+(6)=02 + 4 + (-6) = 0
  • For time (TT): 22+4=0-2 - 2 + 4 = 0
Step 5: Write the final dimensional formula for PP. [P]=[M0L0T0][P] = [M^0 L^0 T^0] Conclusion: The dimensions of PP are [M0L0T0][M^0 L^0 T^0], which corresponds to a dimensionless quantity. This matches option D. Common Traps & Exam Tip:

Students often make the following mistakes in such problems:

  1. Incorrect dimensional formulas: Confusing the dimensions of angular momentum (LL) with linear momentum (MLT1M L T^{-1}). Angular momentum has dimensions [ML2T1][M L^2 T^{-1}]. Always verify the dimensions of each quantity before substitution.
  2. Sign errors in exponents: Forgetting that negative exponents in the formula (e.g., M5M^{-5} or G2G^{-2}) reverse the sign of the dimensions. For example, G2G^{-2} becomes [M2L6T4][M^2 L^{-6} T^4], not [M2L6T4][M^{-2} L^6 T^{-4}].
  3. Arithmetic errors in combining exponents: Adding or subtracting exponents incorrectly when combining dimensions. Double-check each step to avoid simple calculation mistakes.
  4. Misinterpreting dimensionless quantities: Assuming that a quantity with dimensions [M0L0T0][M^0 L^0 T^0] is invalid or trivial. In reality, such quantities are dimensionless and can represent ratios or pure numbers (e.g., fine-structure constant).

Exam Tip: Always write down the dimensions of each quantity explicitly before substituting. This minimizes errors and makes the derivation clearer. If the final dimensions simplify to [M0L0T0][M^0 L^0 T^0], the quantity is dimensionless, which is a valid and important result.

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