The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of

Solution :
$$\eqalign{ & {\text{Energy carried by photon is given by }}E = hv \cr & \Rightarrow h = {\text{Planc's constant}} = \frac{E}{v} \cr & \therefore \left[ h \right] = \frac{{\left[ {M{L^2}\;{T^{ - 2}}} \right]}}{{\left[ {{T^{ - 1}}} \right]}} = \left[ {M{L^2}\;{T^{ - 1}}} \right] \cr & {\text{and}}\,I = {\text{moment of inertia }} = M{R^2} \cr & \Rightarrow \left[ I \right] = \left[ {M{L^2}} \right] \cr & {\text{Hence,}}\,\frac{{\left[ h \right]}}{{\left[ I \right]}} = \frac{{\left[ {M{L^2}\;{T^{ - 1}}} \right]}}{{\left[ {M{L^2}} \right]}} = \left[ {{T^{ - 1}}} \right] \cr & = \frac{1}{{[\;T]}} = {\text{dimension of frequency}} \cr} $$
Alternative
$$\eqalign{ & \frac{h}{I} = \frac{{\frac{E}{v}}}{I} \cr & = \frac{{E \times T}}{I} = \frac{{\left( {\frac{{kg - {m^2}}}{{{s^2}}}} \right) \times s}}{{\left( {kg - {m^2}} \right)}} \cr & = \frac{1}{{\;s}} = \frac{1}{{{\text{ time }}}} = {\text{frequency}} \cr} $$
Thus, dimensions of $$\frac{h}{I}$$ is same as that of frequency.