The potential energy of a particle changes with distance x from a fixed origin as V = A√x / (x+B), where A and B are constant with appropriate dimensions. The dimensions of AB are ______.

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q27 – JEE-MAIN 2026

← Back

Home Physics Units & MeasurementsQuestion #571

The potential energy of a particle changes with distance

x

from a fixed origin as

V = \frac{A \sqrt{x}}{x + B}

, where

A

and

B

are constant with appropriate dimensions. The dimensions of

A B

are ______.

Clue 1 of 3: Principle of Homogeneity

For an equation to be dimensionally correct, the dimensions of all terms on both sides of the equation must be the same. Also, terms added or subtracted must have the same dimensions.

🥷

Ninja Strategy•Check Intermediate Dimensions and Basic Units

First, identify the dimensions of $B$ (length) and $A$ (derived from energy and length). Then, combine them. Options representing just $A$ or $V$ are common distractors, and option 2 has an incorrect mass dimension for an energy-related product.

🧠 Full Solution Path

Step 1: Determine the dimensions of B✦ Active

According to the principle of homogeneity of dimensions, terms added or subtracted must have the same dimensions. In the denominator $x + B$ , $x$ is distance, so its dimension is $[L^{1}]$ . Therefore, the dimension of $B$ must also be $[L^{1}]$ .

[B] = [x] = [L^{1}]

Step 2: Determine the dimensions of A○ Expand

The potential energy $V$ has dimensions of energy, which is $[M^{1} L^{2} T^{- 2}]$ . Using the given equation $V = \frac{A \sqrt{x}}{x + B}$ , we can write the dimensional equality:

[V] = \frac{[A] [\sqrt{x}]}{[x + B]} ⟹ [M^{1} L^{2} T^{- 2}] = \frac{[A] [L^{1 / 2}]}{[L^{1}]}

Solving for $[A]$ :

[A] = \frac{[M^{1} L^{2} T^{- 2}] [L^{1}]}{[L^{1 / 2}]} = [M^{1} L^{2 + 1 - 1 / 2} T^{- 2}] = [M^{1} L^{5 / 2} T^{- 2}]

Step 3: Calculate the dimensions of AB○ Expand

Now, multiply the dimensions of $A$ and $B$ to find the dimensions of $A B$ :

[A B] = [A] \times [B] = [M^{1} L^{5 / 2} T^{- 2}] \times [L^{1}] = [M^{1} L^{5 / 2 + 1} T^{- 2}] = [M^{1} L^{7 / 2} T^{- 2}]

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 27 – JEE-MAIN 2026

Clue 1 of 3: Principle of Homogeneity

🧠 Full Solution Path