JEE Physics 2025 Question with Solution

Given: Match the physical quantities in List-I with their dimensional formulae in List-II.

Find: The correct matching option.

For each quantity:

1. Mass density is mass per unit volume.

$$[\text{Density}] = \frac{M}{L^3} = [ML^{-3}T^0]$$

So, (A) (\to) (IV).

2. Impulse has the dimensions of momentum.

$$[\text{Impulse}] = MLT^{-1}$$

So, (B) (\to) (II).

3. Power is work done per unit time.

$$[\text{Power}] = \frac{ML^2T^{-2}}{T} = [ML^2T^{-3}]$$

So, (C) (\to) (I).

4. Moment of inertia is given by $I = mr^2$, hence

$$[\text{Moment of Inertia}] = ML^2$$

So, (D) (\to) (III).

Therefore, the correct matching is (A)-(IV), (B)-(II), (C)-(I), (D)-(III). The correct option is C.

Given:

• (A) Mass density
• (B) Impulse
• (C) Power
• (D) Moment of inertia

Find: Which option gives the correct dimensional matching.

Step 1: Determine the dimensional formula for each quantity in List-I.

(A) Mass density: Mass density $\rho$ is defined as mass per unit volume.

$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{L^3} = ML^{-3}T^{0}$$

So, (A) matches with (IV).

(B) Impulse: Impulse $J$ is defined as the change in momentum or the product of force and time.

$$J = \Delta p = m \Delta v = M \cdot LT^{-1} = MLT^{-1}$$

Alternatively,

$$J = F \cdot t = (ma) \cdot t = (MLT^{-2}) \cdot T = MLT^{-1}$$

So, (B) matches with (II).

(C) Power: Power $P$ is defined as the rate of doing work or the product of force and velocity.

$$P = \frac{\text{Work}}{\text{Time}} = \frac{F \cdot d}{t} = \frac{(MLT^{-2}) \cdot L}{T} = ML^2T^{-3}$$

Alternatively,

$$P = F \cdot v = (MLT^{-2}) \cdot (LT^{-1}) = ML^2T^{-3}$$

So, (C) matches with (I).

(D) Moment of inertia: Moment of inertia $I$ of a particle is given by $mr^2$.

$$I = M \cdot L^2 = ML^2T^{0}$$

So, (D) matches with (III).

Step 2: Collect the matches.

• (A) (\to) (IV)
• (B) (\to) (II)
• (C) (\to) (I)
• (D) (\to) (III)

Step 3: Choose the correct option. This corresponds to option (3).

Therefore, the correct option is C.