The ratio of the dimensions of Planck's constant and that of the...

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: $\frac{h}{I}=\frac{E / \mu}{I}$ $$\begin{aligned} & =\frac{E \times T}{I}=\frac{\left(\mathrm{kg}-\mathrm{m}^2 / \mathrm{s}^2\right) \times \mathrm{s}}{\left(\mathrm{kg}-\mathrm{m}^2\right)} \\ & =\frac{1}{s}=\frac{1}{\text{ time }}=\text{ frequency }\end{aligned}$$ Thus, dimensions of $\frac{h}{I}$ is same as of frequency.