Question

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

Answer 1

$x=$ distance from a fixed origin

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

unit of $B$ is same as $x^2$. Unit of $x^2=\left[L^2\right]$

$B=\left[L^2\right]$

$u=\frac{\left[x^{1/2}\right]A}{x^2+B}$

$A=\frac{u(x^2+B)}{\sqrt{x}}=\frac{{kgm}^2}{s^2}\times \frac{m^2}{m^{1/2}}$

$A=\frac{kg{m}^{7/2}}{s^2}$

$A=\left[{ML}^{7/2}{T}^{-2}\right]$

Dimensions of $AB=\left[{ML}^{7/2}{T}^{-2}\right]\left[L^2\right]$

Dimensions of $AB=\left[{ML}^{11/2}{T}^{-2}\right]$