JEE Physics 2023 Question with Solution

Given: A matching question on dimensional units for surface tension, pressure, viscosity, and impulse.

Find: The correct correspondence between List I and List II.

Use the units of each physical quantity:

For surface tension,

$$\text{Surface tension} = \frac{F}{\ell}$$

Since

$$[F] = MLT^{-2}, \quad [\ell] = L$$

therefore

$$\left[\frac{F}{\ell}\right] = \frac{MLT^{-2}}{L} = MT^{-2}$$

So the SI unit is $\text{kg} \, \text{s}^{-2}$, which corresponds to (IV).

For pressure,

$$\text{Pressure} = \frac{F}{A}$$

Since

$$[A] = L^2$$

therefore

$$\left[\frac{F}{A}\right] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$$

So the SI unit is $\text{kg} \, \text{m}^{-1} \, \text{s}^{-2}$, which corresponds to (III).

For viscosity,

$$\text{Viscosity} = \frac{F}{A\left(\dfrac{dv}{dz}\right)}$$

Now,

$$[A] = L^2, \quad \left[\frac{dv}{dz}\right] = \frac{LT^{-1}}{L} = T^{-1}$$

Therefore,

$$\left[\text{Viscosity}\right] = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} = ML^{-1}T^{-1}$$

So the SI unit is $\text{kg} \, \text{m}^{-1} \, \text{s}^{-1}$, which corresponds to (I).

For impulse,

$$\text{Impulse} = \int F \, dt$$

Hence,

$$[\text{Impulse}] = [F][t] = MLT^{-2} \cdot T = MLT^{-1}$$

So the SI unit is $\text{kg} \, \text{m} \, \text{s}^{-1}$, which corresponds to (II).

Thus,

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

Therefore, the matching obtained from the working is A-IV, B-III, C-I, D-II.

However, the solution explicitly states The Correct Option is C, while this matching text corresponds to option B in the listed options. Following the solution as primary source, the correct option is C.