JEE Physics 2026 Question with Solution

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

Find: The correct matching among the given options.

Coefficient of viscosity From Newton's law of viscosity,

$$\eta = \frac{\text{shear stress}}{\text{velocity gradient}}$$

Since shear stress has dimensions

$$[ML^{-1}T^{-2}]$$

and velocity gradient has dimensions

$$[T^{-1}]$$

therefore,

$$[\eta] = \frac{[ML^{-1}T^{-2}]}{[T^{-1}]} = [ML^{-1}T^{-1}]$$

So, A (\rightarrow) IV.

Surface tension Surface tension is force per unit length,

$$S = \frac{F}{l}$$

Hence,

$$[S] = \frac{[MLT^{-2}]}{[L]} = [ML^{0}T^{-2}]$$

So, B (\rightarrow) III.

Pressure Pressure is force per unit area,

$$P = \frac{F}{A}$$

Hence,

$$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$

So, C (\rightarrow) I.

Surface energy Surface energy is energy per unit area,

$$E_s = \frac{\text{energy}}{\text{area}}$$

Hence,

$$[E_s] = \frac{[ML^{2}T^{-2}]}{[L^{2}]} = [ML^{-2}T^{-2}]$$

So, D (\rightarrow) II.

Therefore, the final matching is A–IV, B–III, C–I, D–II.

The correct option is C.

Given: Physical quantities involving force, area, length, and energy.

Find: A fast way to identify the correct matching.

Reduce each quantity directly to standard forms:

• Coefficient of viscosity = stress divided by rate of strain
• Surface tension = force per unit length
• Pressure = force per unit area
• Surface energy = energy per unit area

Then use

$$[\text{force}] = [MLT^{-2}], \qquad [\text{energy}] = [ML^{2}T^{-2}]$$

So,

$$[\eta] = [ML^{-1}T^{-1}], \qquad [S] = [ML^{0}T^{-2}],$$ $$[P] = [ML^{-1}T^{-2}], \qquad [E_s] = [ML^{-2}T^{-2}]$$

Thus the matching is A–IV, B–III, C–I, D–II.

This works because reducing everything to force, area, length, and time avoids confusing closely related quantities such as pressure and surface tension.

The correct option is C.