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Physics Question on Units and measurement

The potential energy of a particle moving along the x-direction varies as:
V=Ax2x+BV = \frac{Ax^2}{\sqrt{x} + B}
The dimensions of A2B\frac{A^2}{B} are:

A

[M32L12T−3]\left[ M^{\frac{3}{2}} L^{\frac{1}{2}} T^{-3} \right]

B

[M12LT−3]\left[ M^{\frac{1}{2}} L T^{-3} \right]

C

[M2L12T−4]\left[ M^2 L^{\frac{1}{2}} T^{-4} \right]

D

[ML2T−4]\left[ ML^2T^{-4} \right]

Answer

[M2L12T−4]\left[ M^2 L^{\frac{1}{2}} T^{-4} \right]

Explanation

Solution

The dimensions of potential energy (V) are [ML2T-2].

The dimensions of xare [L].

In the equation V=Ax2x+BV = \frac{Ax^2}{\sqrt{x} + B}, the term x+B\sqrt{x} + B must have the same dimensions as x\sqrt{x} because B is added to it.

Thus,

[V]=[A][x]2[x]1/2[V] = \frac{[A][x]^2}{[x]^{1/2}}

[ML2T−2]=[A][L]3/2[ML^2T^{-2}] = [A][L]^{3/2}

[A]=[ML2T−2L−3/2]=[ML1/2T−2][A] = [ML^2T^{-2}L^{-3/2}] = [ML^{1/2}T^{-2}]

Also, [V]=[A][L]2[L]1/2=[A][L]3/2[V] = \frac{[A][L]^2}{[L]^{1/2}} = [A][L]^{3/2}, thus, [A] = [ML1/2T-2]

The dimensions of B are same as x\sqrt{x}, thus [B] = [L]1/2

Then, the dimensions of A2B\frac{A^2}{B} are:

[A2B]=[ML1/2T−2]2[L]1/2=[M2L1T−4][L]1/2=[M2L1/2T−4]\left[ \frac{A^2}{B} \right] = \frac{[ML^{1/2}T^{-2}]^2}{[L]^{1/2}} = \frac{[M^2L^1T^{-4}]}{[L]^{1/2}} = [M^2L^{1/2}T^{-4}]