The potential energy of a particle moving along x -direction varies as V= A x^2 x +B . The dimensions of (A^2)/(B) are:

Correct answer: C. [M^2 ~L^ 1 / 2 ~T^ -4 ] V= A x^2 x +B As per homogeneous rule B= L ML^2 ~T^ -2 = A ~L^2 ~L^ 1 / 2 A=ML^ 1 / 2 ~T^ -2 (A^2)/(B)= M^2 L T^ -4 ~L^ 1 / 2 =M^2 ~L^ 1 / 2 ~T^ -4

NEET 2024

Exam: NEET
Year: 2024
Subject: Physics
Chapter: Units and Dimensions

NEET 2024 · Physics · Units and Dimensions

The potential energy of a particle moving along x -direction varies...

The potential energy of a particle moving along

x

-direction varies as

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

. The dimensions of

\frac{A^2}{B}

are:

Options

A $\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{1 / 2} \mathrm{~T}^{-3}\right]$
B $\left[M^{1 / 2} L^{-3}\right]$
C $\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$
D $\left[\mathrm{ML}^2 \mathrm{~T}^{-4}\right]$

Correct answer

C. $\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$

Step-by-step solution

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

As per homogeneous rule

B=\sqrt{L}

\mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{A \mathrm{~L}^2}{\mathrm{~L}^{1 / 2}}

A=\mathrm{ML}^{1 / 2} \mathrm{~T}^{-2}

\frac{A^2}{B}=\frac{\mathrm{M}^2 L T^{-4}}{\mathrm{~L}^{1 / 2}}=\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}

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