Physics - Unit & Dimension Question with Solution | TestHub

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

Where U = Potential energy

$x=$ distance

Dimensional formula of potential energy, $U=\left[{\mathrm{M}}^1\mathrm{ }{\mathrm{L}}^2\mathrm{ }{\mathrm{T}}^{-2}\right]$

Dimensional formula of distance, $x=\left[{\mathrm{M}}^0\mathrm{ }{\mathrm{L}}^1\mathrm{ }{\mathrm{T}}^0\right]$

From equation (i), we can write

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

$\left[B\right]= {\left[{\mathrm{M}}^0\mathrm{ }{\mathrm{L}}^1\mathrm{ }{\mathrm{T}}^0\right]}^2$

$\left[B\right]=\left[{\mathrm{M}}^0\mathrm{ }{\mathrm{L}}^2\mathrm{ }{\mathrm{T}}^0\right]$

Dimensional formula of $B=\left[{\mathrm{M}}^0\mathrm{ }{\mathrm{L}}^2\mathrm{ }{\mathrm{T}}^0\right]$

We can write equation (i) as,

$U=\frac{A\sqrt{x}}{x^2}$

$U=\frac{A}{x^{\frac{3}{2}}}$

$A=U{x}^{\frac{3}{2}}$

$A=\left[{\mathrm{M}}^1\mathrm{ }{\mathrm{L}}^2\mathrm{ }{\mathrm{T}}^{-2}\right]\times {\left[{\mathrm{M}}^0\mathrm{ }{\mathrm{L}}^1\mathrm{ }{\mathrm{T}}^0\right]}^{\frac{3}{2}}$

$A=\left[{\mathrm{M}}^1\mathrm{ }{\mathrm{L}}^{\frac{7}{2}}\mathrm{ }{\mathrm{T}}^{-2}\right]$