RE-NEET 2024 Physics Dimensional Formulae MCQ Question with Solution

The potential energy of a particle moving along x-direction varies as V = $\frac{Ax^2}{\sqrt{x} + B}$ . The dimensions of $\frac{A^2}{B}$ are:

A

$M^{3/2} L^{1/2} T^{-3}$

B

$M^{1/2} L T^{-3}$

C

$M^2 L^{1/2} T^{-4}$

D

$ML^{2}T^{-4}$

Option C

Topic: Dimensional Analysis Difficulty: 🟡 Medium

Given:

V=\frac{Ax^2}{\sqrt{x}+B}

Since $V$ is potential energy:

[V]=ML^2T^{-2}

By principle of homogeneity:

[B]=[\sqrt{x}]=L^{1/2}

Now,

ML^2T^{-2} =\frac{[A]L^2}{L^{1/2}}

[A]=ML^{1/2}T^{-2}

Therefore,

\left[\frac{A^2}{B}\right] =\frac{(ML^{1/2}T^{-2})^2}{L^{1/2}}

=\boxed{M^2L^{1/2}T^{-4}}

📌 NCERT: Terms added/subtracted must have the same dimensions. (questions.collegedunia.com)

🧠 NEET Trick: First find $B$ from the denominator, then find $A$ using the dimensions of energy.

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RE-NEET 2024 Physics Dimensional Formulae MCQ Question