JEE PYQ: Units & Measurements - Question ID cb970cbe3654 (JEE Main 2022)

Core Formula & Concept:

The question asks for the dimensional formula of a physical quantity whose SI unit is pascal-second (Pa·s).

• Pascal (Pa): The SI unit of pressure (or stress). Pressure is defined as force per unit area: $\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}$ The dimensional formula of force is $[MLT^{-2}]$, and area is $[L^2]$. Thus, the dimensional formula of pressure is: $[Pa] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$
• Second (s): The SI unit of time, with dimensional formula $[T]$.
• Pascal-second (Pa·s): This is the unit of dynamic viscosity ($\eta$), a measure of a fluid's resistance to flow. It is defined via Newton's law of viscosity: $F = \eta A \frac{dv}{dz}$ where $F$ is the viscous force, $A$ is the area, and $\frac{dv}{dz}$ is the velocity gradient. Rearranging for $\eta$: $\eta = \frac{F}{A \frac{dv}{dz}}$ Since $\frac{dv}{dz}$ has dimensions of $[T^{-1}]$, the dimensional formula of $\eta$ is: $[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]$

Step-by-Step Derivation:

We derive the dimensional formula of pascal-second (Pa·s) as follows:

1. Express pascal (Pa) in terms of base units: $1 \, \text{Pa} = 1 \, \text{N/m}^2 = 1 \, \frac{\text{kg} \cdot \text{m/s}^2}{\text{m}^2} = 1 \, \text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}$ Thus, the dimensional formula of pascal is: $[Pa] = [ML^{-1}T^{-2}]$
2. Multiply by the dimensional formula of second ($[T]$): $[Pa \cdot s] = [Pa] \cdot [s] = [ML^{-1}T^{-2}] \cdot [T] = [ML^{-1}T^{-1}]$
3. Compare with the given options:
   • A: $[ML^{-1}T^{-1}]$ (Matches our derivation)
   • B: $[ML^{-1}T^{-2}]$ (Incorrect, this is the dimensional formula of pressure, not Pa·s)
   • C: $[ML^{2}T^{-1}]$ (Incorrect, this is the dimensional formula of angular momentum or torque)
   • D: $[M^{-1}L^{3}T^{0}]$ (Incorrect, this is the dimensional formula of volume per unit mass)

Common Traps & Exam Tip:

Students often confuse the dimensional formula of pascal-second with that of pascal alone. Key mistakes include:

• Ignoring the "second" in Pa·s: Some students stop at $[ML^{-1}T^{-2}]$ (Option B), forgetting to multiply by $[T]$.
• Misidentifying the quantity: Pa·s is the unit of dynamic viscosity, not pressure or any other quantity. Confusing it with other quantities (e.g., momentum or energy) leads to incorrect options like C or D.
• Sign errors in exponents: Students may incorrectly write $[MLT^{-1}]$ (missing the negative sign for $L$) or $[M^{-1}LT^{-1}]$ (wrong mass exponent).

Exam Tip: Always break down compound units into their base units (kg, m, s) and derive the dimensions step-by-step. For Pa·s, remember: $\text{Pa·s} = \text{Pressure} \times \text{Time} = \frac{\text{Force}}{\text{Area}} \times \text{Time} = \frac{[MLT^{-2}]}{[L^2]} \times [T] = [ML^{-1}T^{-1}]$