JEE PYQ: Units & Measurements - Question ID 90e99e22a630 (JEE Main 2004)

The coefficient of viscosity (denoted by $\eta$) is a measure of a fluid’s resistance to flow. It quantifies the internal frictional force that arises when adjacent layers of the fluid move at different velocities.

The fundamental physics relation governing viscosity is Newton’s law of viscosity: $F = \eta \, A \, \frac{dv}{dx}$ where

• $F$ is the viscous force (dimensions of force),
• $A$ is the area of contact between the layers (dimensions of area),
• $\frac{dv}{dx}$ is the velocity gradient perpendicular to the flow (dimensions of velocity per unit length).

Step 1: Write the dimensions of each quantity in Newton’s law.
Force $F$ has dimensions $\mathrm{MLT}^{-2}$.
Area $A$ has dimensions $\mathrm{L}^{2}$.
Velocity gradient $\frac{dv}{dx}$ has dimensions $\frac{\mathrm{LT}^{-1}}{\mathrm{L}} = \mathrm{T}^{-1}$.

Step 2: Substitute these dimensions into Newton’s law: $\mathrm{MLT}^{-2} = \eta \cdot \mathrm{L}^{2} \cdot \mathrm{T}^{-1}$

Step 3: Solve for $\eta$ by isolating it on one side: $\eta = \frac{\mathrm{MLT}^{-2}}{\mathrm{L}^{2} \cdot \mathrm{T}^{-1}}$ Simplify the powers of $\mathrm{L}$ and $\mathrm{T}$: $\eta = \mathrm{ML}^{-1}\mathrm{T}^{-1}$

Step 4: Compare the derived dimensions with the given options.
Option A is $\mathrm{ML}^{-1}\mathrm{T}^{-1}$, which matches exactly.

Students often confuse the dimensions of viscosity with those of force or pressure. A frequent mistake is to forget that the velocity gradient $\frac{dv}{dx}$ carries dimensions of $\mathrm{T}^{-1}$, not $\mathrm{LT}^{-1}$. This oversight leads to incorrect cancellation of powers of $\mathrm{L}$ and $\mathrm{T}$, resulting in wrong options like $\mathrm{MLT}^{-1}$ (Option B) or $\mathrm{ML}^{-1}\mathrm{T}^{-2}$ (Option C).

Exam Tip: Always write down the defining formula and substitute dimensions systematically. Double-check each step to ensure correct cancellation of units.