JEE PYQ: Units & Measurements - Question ID 9d3ce8efbf66 (JEE Main 2022)

Core Formula & Concept:

The coefficient of viscosity ($\eta$) is defined through Newton’s law of viscous force: $F = \eta \, A \,\frac{dv}{dx}$ where

• $F$ is the viscous force,
• $A$ is the area of contact,
• $\frac{dv}{dx}$ is the velocity gradient perpendicular to the flow.

Rearranging gives $\eta = \frac{F}{A \,\frac{dv}{dx}}.$

In the usual mass–length–time (MLT) system, the dimensions are:

• Force $F \to [M\,L\,T^{-2}]$,
• Area $A \to [L^2]$,
• Velocity gradient $\frac{dv}{dx} \to [T^{-1}]$.

Hence $\eta \to \frac{[M\,L\,T^{-2}]}{[L^2]\cdot[T^{-1}]} = [M\,L^{-1}\,T^{-1}].$

In the new fundamental system, we take momentum $P$, area $A$, and time $T$ as base quantities. We must express the MLT dimensions of $\eta$ in terms of $P$, $A$, and $T$.

Step-by-Step Derivation:

Step 1: Express momentum in MLT
Momentum $P = m\,v \to [M\,L\,T^{-1}]$.
Therefore $[M] = \frac{[P]}{[L\,T^{-1}]} = [P\,L^{-1}\,T].$

Step 2: Express length in terms of area and momentum
Area $A \to [L^2]$, so $[L] = [A^{1/2}].$ Substitute into the expression for $[M]$: $[M] = [P\,L^{-1}\,T] = [P\,A^{-1/2}\,T].$

Step 3: Rewrite $\eta$ in the new base
Recall $\eta \to [M\,L^{-1}\,T^{-1}]$. Substitute $[M]$ and $[L]$: $\eta \to \bigl[P\,A^{-1/2}\,T\bigr]\cdot\bigl[A^{1/2}\bigr]^{-1}\cdot[T^{-1}] = [P\,A^{-1/2}\,T]\cdot[A^{-1/2}]\cdot[T^{-1}] = [P\,A^{-1}\,T^{0}].$

Step 4: Match with the given options
The derived formula is $[\mathrm{P}\,\mathrm{A}^{-1}\,\mathrm{T}^{0}]$, which corresponds exactly to option A.

Common Traps & Exam Tip:

Students often confuse the exponent of $A$ or $T$ when switching bases. A frequent mistake is to forget that momentum already carries a time dimension, leading to incorrect cancellation of $T$. Always write each step explicitly in terms of the new base quantities to avoid sign errors.