Question

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

Answer 1

Step1: Potential Energy and dimensional constants.

1. Potential Energy: The energy possessed by an object due to its position.
2. Dimensional Constants: A physical quantity with dimensions and a fixed value is referred to as a dimensional constant.

Step 2: Given parameters

1. $U=\frac{\left(A\sqrt{x}\right)}{(x^2+B)}$

where, $x$ is the distance from the fixed origin.

Step 3: Calculations

The potential energy of a particle varies with distance $x$ from a fixed origin as,

$U=\frac{\left(A\sqrt{x}\right)}{(x^2+B)}$

From the principle of homogeneity,

$$\begin{gathered} \left[x^2\right]=\left[B\right] \\ \therefore \left[B\right]=\left[{\mathrm{L}}^2\right] \end{gathered}$$

$\mathrm{Unit}\mathrm{of}B=\mathrm{unit}\mathrm{of}{x}^2=\left[{\mathrm{L}}^2\right]$

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

$$\begin{gathered} \left[M{L}^2{T}^{-2}\right]=\frac{A\left[L^{1/2}\right]}{\left[L^2\right]} \\ \Rightarrow A=\left[{\mathrm{ML}}^{7/2}{\mathrm{T}}^{-2}\right] \\ \Rightarrow AB=\left[{\mathrm{ML}}^{7/2}{\mathrm{T}}^{-2}\right]\left[{\mathrm{L}}^2\right] \\ \Rightarrow AB=\left[{\mathrm{ML}}^{11/2}{\mathrm{T}}^{-2}\right] \end{gathered}$$

Hence, the dimensional formula for $AB$ is $\left[{\mathrm{ML}}^{11/2}{\mathrm{T}}^{-2}\right]$.