A) \[[M{{L}^{2}}{{T}^{-1}}]\]
B) \[[M{{L}^{2}}{{T}^{-3}}]\]
C) \[[ML{{T}^{-1}}]\]
D) \[[M{{L}^{3}}{{T}^{-3}}]\]
Correct Answer: A
Solution :
Energy (E) =Plancks constant (h)\[\times \]Frequency(v) \[\Rightarrow \] \[E=hv\]\[\Rightarrow \]\[[E]=[h][v]\] \[\Rightarrow \] \[[M{{L}^{2}}{{T}^{-2}}]=[h]\left[ \frac{1}{T} \right]\] \[\Rightarrow \] \[[h]=[M{{L}^{2}}{{T}^{-1}}]\] \[\therefore \] Dimension of\[h=[h]=[M{{L}^{2}}{{T}^{-1}}]\]
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