JEE PYQ: Units & Measurements - Question ID b90d0a7129c6 (JEE Main 2025)

Core Formula & Concept:

In dimensional analysis, every physical quantity is expressed in terms of the fundamental dimensions: mass ($\mathrm{M}$), length ($\mathrm{L}$), time ($\mathrm{T}$), and electric current ($\mathrm{A}$). The key formulas and concepts used here are:

• Permeability of free space ($\mu_0$): From Ampere’s law, the magnetic field $B$ around a long straight wire carrying current $I$ is given by: $B = \frac{\mu_0 I}{2 \pi r}$ Rearranging for $\mu_0$, we get: $\mu_0 = \frac{2 \pi r B}{I}$
• Magnetic field ($B$): The force on a charge $q$ moving with velocity $v$ in a magnetic field is: $F = q v B \sin \theta$ Solving for $B$, we have: $B = \frac{F}{q v \sin \theta}$
• Magnetic moment ($\vec{m}$): The torque $\vec{\tau}$ on a current loop of area $A$ carrying current $I$ in a magnetic field $B$ is: $\vec{\tau} = \vec{m} \times \vec{B}$ where $m = I A$. Thus: $m = \frac{\tau}{B}$
• Torsional constant ($k$): The restoring torque in a torsional pendulum is proportional to the angular displacement $\theta$: $\tau = k \theta$ Solving for $k$, we get: $k = \frac{\tau}{\theta}$ Since $\theta$ is dimensionless, the dimensions of $k$ are the same as torque.

Step-by-Step Derivation:

Step 1: Dimensional formula for Permeability of free space ($\mu_0$)

From the formula: $\mu_0 = \frac{2 \pi r B}{I}$ The dimensions of each term are:

• $r$: $\left[\mathrm{L}\right]$
• $B$: Derived below as $\left[\mathrm{M} \mathrm{T}^{-2} \mathrm{A}^{-1}\right]$
• $I$: $\left[\mathrm{A}\right]$

Thus: $$\left[\mu_0\right] = \frac{\left[\mathrm{L}\right] \left[\mathrm{M} \mathrm{T}^{-2} \mathrm{A}^{-1}\right]}{\left[\mathrm{A}\right]} = \left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$$ This matches List-II (III).

Step 2: Dimensional formula for Magnetic field ($B$)

From the formula: $B = \frac{F}{q v \sin \theta}$ The dimensions of each term are:

• $F$: $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$
• $q$: $\left[\mathrm{A} \mathrm{T}\right]$ (since current $I = q/t$)
• $v$: $\left[\mathrm{L} \mathrm{T}^{-1}\right]$
• $\sin \theta$: Dimensionless

Thus: $$\left[B\right] = \frac{\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]}{\left[\mathrm{A} \mathrm{T}\right] \left[\mathrm{L} \mathrm{T}^{-1}\right]} = \left[\mathrm{M} \mathrm{T}^{-2} \mathrm{A}^{-1}\right]$$ This matches List-II (II).

Step 3: Dimensional formula for Magnetic moment ($m$)

From the formula: $m = I A$ The dimensions of each term are:

• $I$: $\left[\mathrm{A}\right]$
• $A$: $\left[\mathrm{L}^2\right]$ (area)

Thus: $\left[m\right] = \left[\mathrm{A}\right] \left[\mathrm{L}^2\right] = \left[\mathrm{L}^2 \mathrm{A}\right]$ This matches List-II (IV).

Step 4: Dimensional formula for Torsional constant ($k$)

From the formula: $\tau = k \theta$ Since $\theta$ is dimensionless, the dimensions of $k$ are the same as torque: $\left[\tau\right] = \left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$ This matches List-II (I).

Matching List-I to List-II:

• (A) Permeability of free space → (III)
• (B) Magnetic field → (II)
• (C) Magnetic moment → (IV)
• (D) Torsional constant → (I)

This corresponds to Option A.

Common Traps & Exam Tip:

Students often confuse the dimensions of:

• Magnetic moment ($m$): Some mistakenly associate it with torque dimensions ($\left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$) instead of $\left[\mathrm{L}^2 \mathrm{A}\right]$. Remember, magnetic moment is current times area.
• Permeability ($\mu_0$): It is easy to forget the $\mathrm{A}^{-2}$ term. Always derive it from Ampere’s law.
• Torsional constant ($k$): Since $\theta$ is dimensionless, $k$ has the same dimensions as torque. Do not introduce extra dimensions.

Exam Tip: Always derive dimensions from fundamental formulas rather than memorizing them. This reduces errors and builds conceptual clarity.