A) \[{{c}^{2}}\left[ G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]\]
B) \[\frac{1}{{{c}^{2}}}{{\left[ \frac{{{e}^{2}}}{G4\pi {{\varepsilon }_{0}}} \right]}^{1/2}}\]
C) \[\frac{1}{c}G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\]
D) \[\frac{1}{{{c}^{2}}}{{\left[ G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]}^{1/2}}\]
Correct Answer: D
Solution :
[d] Let dimensions of length is related as, \[L={{[c]}^{x}}{{[G]}^{y}}{{\left[ \frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]}^{z}}\Rightarrow \frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}=M{{L}^{3}}{{T}^{-2}}\] \[L={{[L{{T}^{-1}}]}^{x}}{{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]}^{y}}{{[M{{L}^{3}}{{T}^{-2}}]}^{z}}\] \[[L]=[{{L}^{x+3y+3z}}{{M}^{-y+z}}{{T}^{-x-2y-2z}}]\] Comparing both sides \[z=y=\frac{1}{2},x=-2\] Hence, L= \[{{c}^{-2}}{{\left[ G.\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]}^{1/2}}\]
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