Question

The potential energy $U$ of a particle varies with distance $x$ from a fixed origin as $U=\frac{A\sqrt{x}}{x^2+B}$ where $A$ and $B$ are dimensional constants. The dimensional formula for $AB$ is

• A. $\left[M^1{L}^{7/2}{T}^{-2}\right]$
• B. $\left[M^1{L}^{11/2}{T}^{-2}\right]$
• C. $\left[M^1{L}^{5/2}{T}^{-2}\right]$
• D. $\left[M^1{L}^{9/2}{T}^{-2}\right]$

Answer 1

Answer: (B)

In the expression, $U=\frac{A\sqrt{x}}{x^2+B}$
$B$ must have the dimensions of $x^2$ i.e., $\left[L^2\right]$
Dimensions of $A=\left[\frac{U{x}^2}{\sqrt{x}}\right]$
$=\frac{\left[M{L}^2{T}^{-2}\right]\left[L^2\right]}{\left[L^{1/2}\right]}$
$=\left[M{L}^{7/2}{T}^{-2}\right]$
$\therefore$ The dimensional formula for $AB$
$=\left[M{L}^{7/2}{T}^{-2}\right]\left[L^2\right]=\left[M^1{L}^{11/2}{T}^{-2}\right]$