A) \[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
B) \[[M{{L}^{0}}{{T}^{0}}]\]
C) \[[{{M}^{0}}L{{T}^{0}}]\]
D) \[[{{M}^{0}}{{L}^{0}}T]\]
Correct Answer: A
Solution :
\[[e]=AT,\,\,{{\varepsilon }_{0}}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\] \[[h]=[M{{L}^{2}}{{T}^{-1}}]\]and\[[c]=[L{{T}^{-1}}]\] \[\therefore \]\[\left[ \frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}hc} \right]=\left[ \frac{{{A}^{2}}{{T}^{2}}}{{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}\times M{{L}^{2}}{{T}^{-1}}\times L{{T}^{-1}}} \right]\] \[=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
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