Match List - I with List - II . List - I List - II (A) Mass density (I) [ML 2 T −3 ] (B) Impulse (II) [MLT −1 ] (C) Power (III) [ML 2 T 0 ] (D) Moment of inertia (IV) [ML −3 T 0 ] Choose the correct answer from the options given below :

(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Units and Measurements PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q56 with Solution

Match List - I with List - II.

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: D

Find dimensions of each quantity in List I

(A) Mass density $\text{Density} = \frac{\text{mass}}{\text{volume}}$ Mass has dimensions $[M]$ and volume has dimensions $[L^3]$ . So, $[\rho] = [ML^{-3}T^0]$ Hence, (A) $\to$ (IV).
Impulse

Impulse = change in momentum. Momentum = mass $\times$ velocity. $[\text{Impulse}] = [M]\,[LT^{-1}] = [MLT^{-1}]$ Hence, (B) $\to$ (II).
Power

Power = work done per unit time.

Work = force $\times$ displacement. Force has dimensions: $[F] = [MLT^{-2}]$ Therefore, $[\text{Work}] = [MLT^{-2}]\,[L] = [ML^2T^{-2}]$ Now, $[\text{Power}] = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$ Hence, (C) $\to$ (I).
Moment of inertia

$I = \sum mr^2$ So dimensions are: $[I] = [M][L^2] = [ML^2T^0]$ Hence, (D) $\to$ (III).
Final matching

$\boxed{(A)-(IV),\ (B)-(II),\ (C)-(I),\ (D)-(III)}$
Compare with options

This corresponds to Option D.
Comparison with stored answer

Stored correct answer = D.

Our derived answer also = D.

Therefore, the stored answer is correct.

Match List - I with List - II.

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: D

Find dimensions of each quantity in List I

(A) Mass density $\text{Density} = \frac{\text{mass}}{\text{volume}}$ Mass has dimensions $[M]$ and volume has dimensions $[L^3]$ . So, $[\rho] = [ML^{-3}T^0]$ Hence, (A) $\to$ (IV).
Impulse

Impulse = change in momentum. Momentum = mass $\times$ velocity. $[\text{Impulse}] = [M]\,[LT^{-1}] = [MLT^{-1}]$ Hence, (B) $\to$ (II).
Power

Power = work done per unit time.

Work = force $\times$ displacement. Force has dimensions: $[F] = [MLT^{-2}]$ Therefore, $[\text{Work}] = [MLT^{-2}]\,[L] = [ML^2T^{-2}]$ Now, $[\text{Power}] = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$ Hence, (C) $\to$ (I).
Moment of inertia

$I = \sum mr^2$ So dimensions are: $[I] = [M][L^2] = [ML^2T^0]$ Hence, (D) $\to$ (III).
Final matching

$\boxed{(A)-(IV),\ (B)-(II),\ (C)-(I),\ (D)-(III)}$
Compare with options

This corresponds to Option D.
Comparison with stored answer

Stored correct answer = D.

Our derived answer also = D.

Therefore, the stored answer is correct.