JEE PYQ: Units & Measurements - Question ID 69de455d11e2 (JEE Main 2025)

In dimensional analysis, every physical quantity is expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), and time ($T$). The dimension of a derived quantity is written as a product of powers of these fundamental dimensions, e.g., $[Q] = M^a L^b T^c$.

Key formulas used in this question:

• Mass density ($\rho$): $\rho = \frac{\text{mass}}{\text{volume}} \implies [\rho] = \frac{M}{L^3} = ML^{-3}T^0$.
• Impulse ($J$): $J = \text{force} \times \text{time} = ma \cdot t \implies [J] = MLT^{-2} \cdot T = MLT^{-1}$.
• Power ($P$): $P = \frac{\text{work}}{\text{time}} = \frac{F \cdot s}{t} \implies [P] = \frac{MLT^{-2} \cdot L}{T} = ML^2T^{-3}$.
• Moment of inertia ($I$): $I = \text{mass} \times (\text{distance})^2 \implies [I] = M \cdot L^2 = ML^2T^0$.

1. Mass density (A):

   $\rho = \frac{\text{mass}}{\text{volume}} = \frac{M}{L^3} = ML^{-3}T^0$.

   This matches (IV) in List-II.



2. Impulse (B):

   $J = \text{force} \times \text{time} = ma \cdot t = (MLT^{-2}) \cdot T = MLT^{-1}$.

   This matches (II) in List-II.



3. Power (C):

   $P = \frac{\text{work}}{\text{time}} = \frac{F \cdot s}{t} = \frac{(MLT^{-2}) \cdot L}{T} = ML^2T^{-3}$.

   This matches (I) in List-II.



4. Moment of inertia (D):

   $I = \text{mass} \times (\text{distance})^2 = M \cdot L^2 = ML^2T^0$.

   This matches (III) in List-II.

Thus, the correct matching is:

• (A) → (IV)
• (B) → (II)
• (C) → (I)
• (D) → (III)

This corresponds to Option D.

Students often confuse the dimensions of impulse and momentum. Both have the same dimension ($MLT^{-1}$), but impulse is force × time, while momentum is mass × velocity. Another common mistake is misidentifying the dimension of power as $ML^2T^{-2}$ (which is work/energy) instead of $ML^2T^{-3}$.

Exam Tip: Always verify dimensions by breaking down the formula into fundamental quantities (mass, length, time). Cross-check each derived dimension with the given options to avoid mismatches.