JEE PYQ: Units & Measurements - Question ID b8d115a1f91d (JEE Main 2024)

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 key formulas and concepts for the given quantities are:

• Coefficient of viscosity ($\eta$): Defined via Newton’s law of viscosity: $F = \eta A \frac{dv}{dx}$. Here, $F$ is force, $A$ is area, and $\frac{dv}{dx}$ is the velocity gradient. The dimensions of force are $[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]$, area is $[\mathrm{L}^2]$, and velocity gradient is $[\mathrm{T}^{-1}]$. Thus, $\eta = \frac{F}{A \frac{dv}{dx}}$ has dimensions $[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-1}]$.
• Surface tension ($T$): Defined as force per unit length: $T = \frac{F}{L}$. Force has dimensions $[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]$, and length is $[\mathrm{L}]$. Thus, surface tension has dimensions $[\mathrm{M} \mathrm{T}^{-2}]$ or equivalently $[\mathrm{M} \mathrm{L}^0 \mathrm{T}^{-2}]$.
• Angular momentum ($L$): Given by $L = mvr$, where $m$ is mass, $v$ is velocity, and $r$ is radius. Dimensions of mass are $[\mathrm{M}]$, velocity is $[\mathrm{L} \mathrm{T}^{-1}]$, and radius is $[\mathrm{L}]$. Thus, angular momentum has dimensions $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}]$.
• Rotational kinetic energy ($K$): Given by $K = \frac{1}{2} I \omega^2$, where $I$ is moment of inertia and $\omega$ is angular velocity. Moment of inertia has dimensions $[\mathrm{M} \mathrm{L}^2]$, and angular velocity has dimensions $[\mathrm{T}^{-1}]$. Thus, rotational kinetic energy has dimensions $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]$.

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

1. (A) Coefficient of viscosity:

   From Newton’s law of viscosity:

   $F = \eta A \frac{dv}{dx}$

   Rearranging for $\eta$:

   $\eta = \frac{F}{A \frac{dv}{dx}}$

   Substitute dimensions:

   $$[\eta] = \frac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]}{[\mathrm{L}^2] \cdot [\mathrm{T}^{-1}]} = \frac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]}{[\mathrm{L}^2 \mathrm{T}^{-1}]} = [\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-1}]$$

   This matches option (III) in List II.

2. (B) Surface tension:

   Surface tension is force per unit length:

   $T = \frac{F}{L}$

   Substitute dimensions:

   $[T] = \frac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]}{[\mathrm{L}]} = [\mathrm{M} \mathrm{T}^{-2}] = [\mathrm{M} \mathrm{L}^0 \mathrm{T}^{-2}]$

   This matches option (IV) in List II.

3. (C) Angular momentum:

   Angular momentum is given by:

   $L = mvr$

   Substitute dimensions:

   $[L] = [\mathrm{M}] \cdot [\mathrm{L} \mathrm{T}^{-1}] \cdot [\mathrm{L}] = [\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}]$

   This matches option (II) in List II.

4. (D) Rotational kinetic energy:

   Rotational kinetic energy is given by:

   $K = \frac{1}{2} I \omega^2$

   Substitute dimensions (moment of inertia $I$ has dimensions $[\mathrm{M} \mathrm{L}^2]$, and angular velocity $\omega$ has dimensions $[\mathrm{T}^{-1}]$):

   $[K] = [\mathrm{M} \mathrm{L}^2] \cdot [\mathrm{T}^{-1}]^2 = [\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]$

   This matches option (I) in List II.

Thus, the correct matching is:

• (A) → (III)
• (B) → (IV)
• (C) → (II)
• (D) → (I)

This corresponds to Option D.

Students often confuse the dimensions of the following quantities:

• Coefficient of viscosity vs. surface tension: Both involve force, but viscosity is force per unit area per unit velocity gradient, while surface tension is force per unit length. This leads to different dimensions ($[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-1}]$ vs. $[\mathrm{M} \mathrm{T}^{-2}]$).
• Angular momentum vs. rotational kinetic energy: Angular momentum has dimensions $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}]$, while rotational kinetic energy has dimensions $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]$. The extra $\mathrm{T}^{-1}$ in angular momentum comes from velocity, while kinetic energy involves velocity squared.
• Surface tension vs. pressure: Both have dimensions involving $\mathrm{M} \mathrm{T}^{-2}$, but pressure is force per unit area ($[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}]$), while surface tension is force per unit length ($[\mathrm{M} \mathrm{T}^{-2}]$).

Exam Tip: Always derive dimensions from first principles (e.g., definitions or formulas) rather than memorizing them. This avoids confusion between similar-looking quantities.