Question

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

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

Answer 1

The correct option is B $M^1{L}^{\frac{11}{2}}{T}^{-2}$

Lets find out dimensions of $B$ first:

Now in the equation given, we have $x^2+B$ in the denominator. So dimensions of $B$ must be equal to dimensions of $x^2$.
$\therefore \left[B\right]=\left[L^2\right]$

Now we will find dimensions of $A$:
$U=\frac{A{x}^{1/2}}{x^2+B}$
$\left[M{L}^2{T}^{-2}\right]=\frac{\left[A\right]\left[L^{1/2}\right]}{\left[L^2\right]}$

$\left[A\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[L^2\right]}{\left[L^{1/2}\right]}$

So, the dimensions of $A\times B$ are:
$\left[A\right]\times \left[B\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[L^2\right]}{\left[L^{1/2}\right]}.\left[L^2\right]$