Question

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

• A. \(\left[ M^{\frac{3}{2}} L^{\frac{1}{2}} T^{-3} \right]\)
• B. \(\left[ M^{\frac{1}{2}} L T^{-3} \right]\)
• C. \(\left[ M^2 L^{\frac{1}{2}} T^{-4} \right]\)
• D. \(\left[ ML^2T^{-4} \right]\)

Answer 1

The Correct Option is C

Potential energy (V) has dimensions of [ML²T^-2].

The dimension of x is [L].

In the equation \(V = \frac{Ax^2}{\sqrt{x} + B}\), since B is added to \(\sqrt{x}\), the term \(\sqrt{x} + B\) must have the same dimensions as \(\sqrt{x}\).

Therefore,

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

\([ML^2T^{-2}] = [A][L]^{3/2}\)

\([A] = [ML^2T^{-2}L^{-3/2}] = [ML^{1/2}T^{-2}]\)

As \([V] = \frac{[A][L]^2}{[L]^{1/2}} = [A][L]^{3/2}\), it follows that [A] = [ML^1/2T^-2].

The dimensions of B are equivalent to \(\sqrt{x}\), thus [B] = [L]^1/2.

The dimensions of $\frac{A^2}{B}$ are calculated as follows:

\(\left[ \frac{A^2}{B} \right] = \frac{[ML^{1/2}T^{-2}]^2}{[L]^{1/2}} = \frac{[M^2L^1T^{-4}]}{[L]^{1/2}} = [M^2L^{1/2}T^{-4}]\)