Physics - Unit & Dimension Question with Solution | TestHub

$E=h\nu$ $\Rightarrow h=$ Planck's constant $=\frac{E}{\mu }$ $\therefore [h]=\frac{[E]}{[\mu ]}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{T}}^{-1}\right]}$ $=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]$ and $I=$ moment of inertia $=M{R}^2$ Hence, $\frac{[h]}{[I]}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}{\left[{\mathrm{ML}}^2\right]}=\left[{\mathrm{T}}^{-1}\right]$ $=\frac{1}{[\mathrm{T}]}=$ dimensions of frequency Alternative: $\frac{h}{I}=\frac{E/\mu }{I}$ $\begin{array}{rl}& =\frac{E\times T}{I}=\frac{\left(\mathrm{kg}-{\mathrm{m}}^2/{\mathrm{s}}^2\right)\times \mathrm{s}}{\left(\mathrm{kg}-{\mathrm{m}}^2\right)}\\ & =\frac{1}{s}=\frac{1}{\text{ time }}=\text{ frequency }\end{array}$ Thus, dimensions of $\frac{h}{I}$ is same as of frequency.