Question

The potential energy of a particle varies with distance $x$ as $U=\frac{A{x}^{1/2}}{x^2+B}$, where A and B are constants. The dimensional formula for A x B is:

• A. $M^1{L}^{7/2}{T}^{-2}$
• B. $M^1{L}^{11/4}{T}^{-2}$
• C. $M^1{L}^{5/2}{T}^{-2}$
• D. $M^1{L}^{9/2}{T}^{-2}$

Answer 1

Answer: (B)

Given, $U=\frac{{Ax}^{1/2}}{x^2+B}$

$x^2+B$ should have some dimension $\Rightarrow$ $L^2=dim\left(B\right)$

$dim\left(A\right)=dim\left(\frac{U\times (x^2+B)}{x^{1/2}}\right)=\frac{\left[{ML}^2{T}^{-2}\right]\left[L^2\right]}{L^{1/2}}$

$={ML}^{7/2}{T}^{-2}$

$dim(A\times B)=\left[{ML}^{7/2}{T}^{-2}\right]\left[L^2\right]=\left[{ML}^{11/2}{T}^{-2}\right]$