JEE PYQ: Units & Measurements - Question ID f614c18dad6e (JEE Main 2023)

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), and time ($T$). The key formulas and concepts used here are:

• Torque ($\tau$): $\tau = \vec{r} \times \vec{F}$, where $\vec{r}$ is displacement and $\vec{F}$ is force. The dimensions of force are $MLT^{-2}$, and displacement is $L$, so torque has dimensions $ML^2T^{-2}$.
• Stress ($\sigma$): $\sigma = \frac{F}{A}$, where $F$ is force and $A$ is area. The dimensions are $\frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$.
• Pressure gradient ($\frac{dP}{dx}$): This is the rate of change of pressure with respect to distance. Pressure has dimensions $ML^{-1}T^{-2}$, and distance is $L$, so the gradient has dimensions $\frac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}$.
• Coefficient of viscosity ($\eta$): From Newton’s law of viscosity, $\eta = \frac{F}{A \cdot \frac{dv}{dx}}$, where $\frac{dv}{dx}$ is the velocity gradient. The dimensions are $\frac{MLT^{-2}}{L^2 \cdot \frac{LT^{-1}}{L}} = ML^{-1}T^{-1}$.

Step-by-Step Derivation:

We derive the dimensions for each quantity in List I and match them with List II.

1. Torque (A):

   Torque is given by $\tau = rF \sin \theta$. The dimensions of $r$ are $L$, and the dimensions of $F$ are $MLT^{-2}$. Thus: $[\tau] = L \cdot MLT^{-2} = ML^2T^{-2}.$ This matches II in List II.

2. Stress (B):

   Stress is force per unit area: $\sigma = \frac{F}{A}$. The dimensions of $F$ are $MLT^{-2}$, and the dimensions of $A$ are $L^2$. Thus: $[\sigma] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}.$ This matches IV in List II.

3. Pressure gradient (C):

   Pressure gradient is the change in pressure per unit distance: $\frac{dP}{dx}$. The dimensions of pressure ($P$) are $ML^{-1}T^{-2}$, and the dimensions of distance ($x$) are $L$. Thus: $\left[\frac{dP}{dx}\right] = \frac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}.$ This matches I in List II.

4. Coefficient of viscosity (D):

   The coefficient of viscosity is defined by $\eta = \frac{F}{A \cdot \frac{dv}{dx}}$. The dimensions of $F$ are $MLT^{-2}$, $A$ is $L^2$, and $\frac{dv}{dx}$ (velocity gradient) is $\frac{LT^{-1}}{L} = T^{-1}$. Thus: $[\eta] = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} = ML^{-1}T^{-1}.$ This matches III in List II.

The correct matching is:

• A → II
• B → IV
• C → I
• D → III

This corresponds to Option B.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Confusing torque and work: Both have dimensions $ML^2T^{-2}$, but they are different physical quantities. Torque is a vector, while work is a scalar.
• Misidentifying pressure gradient: Students sometimes forget to divide by $L$ when calculating the gradient, leading to incorrect dimensions $ML^{-1}T^{-2}$ (which is actually stress).
• Incorrect dimensions for viscosity: Students may forget that the velocity gradient has dimensions $T^{-1}$, leading to errors in the final dimensions of $\eta$.

Exam Tip: Always write down the defining formula for each quantity before deriving its dimensions. This avoids confusion between similar-looking dimensions.