JEE PYQ: Units & Measurements - Question ID b13e94996690 (JEE Main 2026)

In physics, the dimensional consistency principle states that every term in a physically meaningful equation must have the same dimensions. Potential energy \( V \) has the dimensions of energy, which is: \[ [V] = \left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}\right] \] Here, \( \mathrm{M} \), \( \mathrm{L} \), and \( \mathrm{T} \) represent the dimensions of mass, length, and time, respectively.

Given the expression for potential energy: \[ V = \frac{A \sqrt{x}}{x + B} \] we must ensure that the dimensions of each term in the numerator and denominator are consistent. Specifically:

• The term \( \sqrt{x} \) has dimensions \( \left[\mathrm{L}^{1/2}\right] \).
• The term \( x \) in the denominator has dimensions \( \left[\mathrm{L}^1\right] \).
• The constant \( B \) must have the same dimensions as \( x \) to ensure dimensional homogeneity in the denominator \( (x + B) \). Thus, \( [B] = \left[\mathrm{L}^1\right] \).

Step 1: Determine the dimensions of \( A \).
The given potential energy expression is: \[ V = \frac{A \sqrt{x}}{x + B} \] Since \( [V] = \left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}\right] \), we equate the dimensions of both sides: \[ \left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}\right] = \frac{[A] \cdot \left[\mathrm{L}^{1/2}\right]}{\left[\mathrm{L}^1\right]} \] Simplify the right-hand side: \[ \left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}\right] = \frac{[A] \cdot \mathrm{L}^{1/2}}{\mathrm{L}^1} = [A] \cdot \mathrm{L}^{-1/2} \] Now, solve for \( [A] \): \[ [A] = \left[\mathrm{M}^1 \mathrm{L}^2 \mathrm{T}^{-2}\right] \cdot \mathrm{L}^{1/2} = \left[\mathrm{M}^1 \mathrm{L}^{5/2} \mathrm{T}^{-2}\right] \] Step 2: Determine the dimensions of \( B \).
As established earlier, \( B \) must have the same dimensions as \( x \), so: \[ [B] = \left[\mathrm{L}^1\right] \] Step 3: Compute the dimensions of \( AB \).
Multiply the dimensions of \( A \) and \( B \): \[ [AB] = [A] \cdot [B] = \left[\mathrm{M}^1 \mathrm{L}^{5/2} \mathrm{T}^{-2}\right] \cdot \left[\mathrm{L}^1\right] = \left[\mathrm{M}^1 \mathrm{L}^{7/2} \mathrm{T}^{-2}\right] \] Step 4: Match with the given options.
The derived dimensions \( \left[\mathrm{M}^1 \mathrm{L}^{7/2} \mathrm{T}^{-2}\right] \) correspond to Option D.

Trap 1: Ignoring dimensional homogeneity in the denominator.
Students often forget that \( x \) and \( B \) must have the same dimensions for the expression \( (x + B) \) to be valid. If \( B \) is incorrectly assumed to be dimensionless, the final dimensions of \( AB \) will be wrong. Trap 2: Misapplying the square root dimension.
The term \( \sqrt{x} \) has dimensions \( \left[\mathrm{L}^{1/2}\right] \), not \( \left[\mathrm{L}^1\right] \). Confusing this leads to incorrect exponents in the final answer. Trap 3: Arithmetic errors in combining exponents.
When multiplying \( [A] \) and \( [B] \), students may incorrectly add the exponents of \( \mathrm{L} \). For example, \( \mathrm{L}^{5/2} \cdot \mathrm{L}^1 = \mathrm{L}^{7/2} \), not \( \mathrm{L}^{3/2} \). Exam Tip:
Always verify that the dimensions of each term in an equation are consistent. For expressions involving addition or subtraction (like \( x + B \)), ensure the terms have identical dimensions. This is a quick sanity check to avoid errors.