JEE PYQ: Units & Measurements - Question ID d464f1fc2789 (JEE Main 2025)

To determine which quantity has the dimensions of momentum, we recall the following fundamental concepts and formulas:

• Momentum ($p$): The dimension of momentum is mass × length / time, denoted as $[M L T^{-1}]$.
• Charge ($q$): The SI unit of charge is the coulomb (C), and its dimension is $[I T]$, where $I$ is current and $T$ is time.
• Current ($I$): The dimension of current is simply $[I]$.
• Permeability of vacuum ($\mu_0$): The dimension of $\mu_0$ is derived from the force between two current-carrying wires. The formula for the magnetic force per unit length between two parallel wires is: $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$ Rearranging for $\mu_0$, we get: $\mu_0 = \frac{2 \pi d F}{l I_1 I_2}$ The dimension of force ($F$) is $[M L T^{-2}]$, length ($l$ or $d$) is $[L]$, and current ($I$) is $[I]$. Thus, the dimension of $\mu_0$ is: $[\mu_0] = \frac{[L] \cdot [M L T^{-2}]}{[L] \cdot [I]^2} = [M L T^{-2} I^{-2}]$

Our goal is to check the dimensions of each option and compare them with the dimension of momentum, $[M L T^{-1}]$.

We analyze each option one by one by substituting the dimensions of $q$, $I$, and $\mu_0$.

1. Option A: $\frac{q I}{\mu_0}$
   Substitute the dimensions: $\left[ \frac{q I}{\mu_0} \right] = \frac{[I T] \cdot [I]}{[M L T^{-2} I^{-2}]} = \frac{[I^2 T]}{[M L T^{-2} I^{-2}]} = [I^2 T] \cdot [M^{-1} L^{-1} T^{2} I^{2}]$ Simplify: $= [M^{-1} L^{-1} T^{3} I^{4}]$ This does not match the dimension of momentum, $[M L T^{-1}]$.


2. Option B: $q^2 \mu_0 I$
   Substitute the dimensions: $[q^2 \mu_0 I] = [I T]^2 \cdot [M L T^{-2} I^{-2}] \cdot [I] = [I^2 T^2] \cdot [M L T^{-2} I^{-2}] \cdot [I]$ Simplify: $= [M L T^{0} I^{1}]$ This does not match the dimension of momentum, $[M L T^{-1}]$.


3. Option C: $\frac{q \mu_0}{I}$
   Substitute the dimensions: $\left[ \frac{q \mu_0}{I} \right] = \frac{[I T] \cdot [M L T^{-2} I^{-2}]}{[I]} = [I T] \cdot [M L T^{-2} I^{-2}] \cdot [I^{-1}]$ Simplify: $= [M L T^{-1} I^{-1}]$ This does not match the dimension of momentum, $[M L T^{-1}]$, because of the extra $[I^{-1}]$ term.


4. Option D: $q \mu_0 I$
   Substitute the dimensions: $[q \mu_0 I] = [I T] \cdot [M L T^{-2} I^{-2}] \cdot [I] = [I T] \cdot [M L T^{-2} I^{-2}] \cdot [I]$ Simplify: $= [M L T^{-1} I^{0}] = [M L T^{-1}]$ This matches the dimension of momentum, $[M L T^{-1}]$.

Thus, the correct option is D.

Students often make the following mistakes in such dimensional analysis questions:

• Incorrect dimension of $\mu_0$: Many students mistakenly assume $\mu_0$ has dimensions of $[M L T^{-2} I^{-1}]$ or $[M L^2 T^{-2} I^{-2}]$. It is crucial to derive $\mu_0$'s dimension correctly from the force between current-carrying wires.
• Ignoring current ($I$) in charge dimension: Charge $q$ has dimension $[I T]$, not just $[T]$. Forgetting the current dimension leads to incorrect simplification.
• Algebraic errors in simplification: When multiplying or dividing dimensions, students sometimes misapply exponent rules (e.g., $I^2 \cdot I^{-2} = I^0 = 1$, not $I^4$). Careful step-by-step simplification is essential.
• Overlooking the final comparison: After deriving the dimension of an option, students may forget to compare it with the target dimension ($[M L T^{-1}]$). Always cross-check the final result.

Exam Tip: For dimensional analysis questions, always:

1. Write down the dimensions of all given quantities clearly.
2. Substitute and simplify step-by-step, keeping track of exponents.
3. Compare the final dimension with the target dimension (here, momentum).
4. Double-check calculations to avoid sign or exponent errors.