JEE PYQ: Units & Measurements - Question ID 0eebf50eb27f (JEE Main 2023)

In dimensional analysis, every physical quantity is expressed in terms of the fundamental dimensions: mass ($\mathrm{M}$), length ($\mathrm{L}$), and time ($\mathrm{T}$). The goal is to find the dimensional formula of each quantity in List I and match it with the corresponding entry in List II.

Key formulas and definitions:

• Young’s Modulus ($Y$): Defined as stress divided by strain. Stress is force per unit area, and strain is dimensionless. Thus, $Y = \frac{\text{Force}}{\text{Area}} = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}]} = [\mathrm{ML^{-1}T^{-2}}]$
• Coefficient of Viscosity ($\eta$): Given by the formula $\eta = \frac{F}{A \cdot \frac{dv}{dx}}$, where $F$ is force, $A$ is area, and $\frac{dv}{dx}$ is the velocity gradient. Dimensional analysis yields $[\eta] = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}] \cdot [\mathrm{T^{-1}}]} = [\mathrm{ML^{-1}T^{-1}}]$
• Planck’s Constant ($h$): Relates energy of a photon to its frequency via $E = h \nu$. Since energy has dimensions $[\mathrm{ML^2T^{-2}}]$ and frequency has $[\mathrm{T^{-1}}]$, $h = \frac{E}{\nu} = [\mathrm{ML^2T^{-1}}]$
• Work Function ($\varphi$): The minimum energy required to remove an electron from a metal surface. Energy has dimensions $[\mathrm{ML^2T^{-2}}]$

Step 1: Derive the dimensional formula for Young’s Modulus ($Y$)
Stress = Force / Area = $\frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}]} = [\mathrm{ML^{-1}T^{-2}}]$
Strain is dimensionless, so $Y = \frac{\text{Stress}}{\text{Strain}} = [\mathrm{ML^{-1}T^{-2}}]$
This matches III in List II.

Step 2: Derive the dimensional formula for Coefficient of Viscosity ($\eta$)
$\eta = \frac{F}{A \cdot \frac{dv}{dx}}$
Force ($F$) = $[\mathrm{MLT^{-2}}]$
Area ($A$) = $[\mathrm{L^2}]$
Velocity gradient ($\frac{dv}{dx}$) = $\frac{[\mathrm{LT^{-1}}]}{[\mathrm{L}]} = [\mathrm{T^{-1}}]$
Thus, $\eta = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}] \cdot [\mathrm{T^{-1}}]} = [\mathrm{ML^{-1}T^{-1}}]$
This matches I in List II.

Step 3: Derive the dimensional formula for Planck’s Constant ($h$)
Energy ($E$) = $[\mathrm{ML^2T^{-2}}]$
Frequency ($\nu$) = $[\mathrm{T^{-1}}]$
Thus, $h = \frac{E}{\nu} = \frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{T^{-1}}]} = [\mathrm{ML^2T^{-1}}]$
This matches II in List II.

Step 4: Derive the dimensional formula for Work Function ($\varphi$)
Work function is an energy term, so its dimensions are the same as energy: $[\mathrm{ML^2T^{-2}}]$
This matches IV in List II.

Step 5: Match List I with List II
A (Young’s Modulus) → III
B (Coefficient of Viscosity) → I
C (Planck’s Constant) → II
D (Work Function) → IV
This corresponds to Option D.

• Confusing Young’s Modulus with Pressure: Both have the same dimensions ($[\mathrm{ML^{-1}T^{-2}}]$), but students sometimes misassign them due to oversight.
• Planck’s Constant vs. Angular Momentum: Both have dimensions $[\mathrm{ML^2T^{-1}}]$, but the question specifically asks for Planck’s constant, so the match is unambiguous.
• Work Function as Energy: Some students mistakenly associate it with power ($[\mathrm{ML^2T^{-3}}]$) instead of energy.
• Exam Tip: Always derive dimensions from first principles rather than relying on memory. Cross-verify each step to avoid dimensional mismatches.