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 $A B$ 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 $A=\left[\frac{U x^{2}}{\sqrt{x}}\right]$
$=\frac{\left[ ML ^{2} T ^{-2}\right]\left[ L ^{2}\right]}{\left[ L ^{1 / 2}\right]}$
$=\left[ ML ^{7 / 2} T ^{-2}\right]$
$\therefore $ The dimensional formula for $A B$
$=\left[ ML ^{7 / 2} T ^{-2}\right]\left[ L ^{2}\right]=\left[ M ^{1} L ^{11 / 2} T ^{-2}\right]$