A) frequency
B) velocity
C) angular momentum
D) time
Correct Answer: A
Solution :
\[E=hv\] \[\Rightarrow \] \[h=\]Plancks constant\[=\frac{E}{V}\] \[\therefore \] \[[h]=\frac{[E]}{[v]}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{T}^{-1}}]}\] \[=[M{{L}^{2}}{{T}^{-1}}]\] and \[I=\]moment of inertia\[=M{{R}^{2}}\] \[\Rightarrow \] \[[I]=[M][{{L}^{2}}]=[M{{L}^{2}}]\] Hence, \[\frac{[h]}{[I]}=\frac{[M{{L}^{2}}{{T}^{-1}}]}{[M{{L}^{2}}]}=[{{T}^{-1}}]\] \[=\frac{1}{[T]}=\] dimensions of frequency , Alternative: \[\frac{h}{I}=\frac{E\text{/}v}{I}\] \[=\frac{E\times T}{I}=\frac{(kg\text{-}{{m}^{2}}\text{/}{{s}^{2}})\times s}{(kg\text{-}{{m}^{2}})}\] \[=\frac{1}{s}=\frac{1}{\text{time}}=\text{frequency}\] Thus, dimensions of\[\frac{h}{I}\]is same as of frequency.
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