Question

The potential energy of a particle varies with distance $x$ from fixed a origin as $U=\left(\frac{A\sqrt{x}}{x+B}\right)$; where, $A$ and $B$ are constants. The dimensions of $AB$ are

• A. $\left[M{L}^{5/2}{T}^{-2}\right]$
• B. $\left[M{L}^2{T}^{-2}\right]$
• C. $\left[M^{3/2}{L}^{3/2}{T}^{-2}\right]$
• D. $\left[M{L}^{7/2}{T}^{-2}\right]$

Answer 1

The correct option is D $\left[M{L}^{7/2}{T}^{-2}\right]$

Since $x$ and $B$ can be added together, their dimensions must be the same. Thus, $\left[x\right]=\left[B\right]=[x+B]=\left[L\right]$

Now, $\left[U\right]=\frac{\left[A\right]\left[\sqrt{x}\right]}{[x+B]}$

$\Rightarrow \left[U\right][x+B]=\left[A\right]\left[\sqrt{x}\right]$

$\Rightarrow \left[A\right]=\frac{\left[U\right][x+B]}{\left[\sqrt{x}\right]}$

$\Rightarrow \left[A\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L^{0.5}\right]}=\left[M{L}^{2.5}{T}^{-2}\right]$

Therefore, $\left[AB\right]=\left[A\right]\left[B\right]=\left[M{L}^{7/2}{T}^{-2}\right]$