A) \[\frac{1}{c}G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\]
B) \[\frac{1}{{{c}^{2}}}{{\left[ G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]}^{\frac{1}{2}}}\]
C) \[{{c}^{2}}{{\left[ G\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right]}^{\frac{1}{2}}}\]
D) \[\frac{1}{{{c}^{2}}}{{\left[ \frac{{{e}^{2}}}{G4\pi {{\varepsilon }_{0}}} \right]}^{\frac{1}{2}}}\]
Correct Answer: B
Solution :
| [b] Let \[\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}=A=M{{L}^{3}}{{T}^{-2}}\] |
| \[l={{C}^{x}}{{G}^{y}}{{(A)}^{z}}\] |
| \[L={{[L/{{T}^{-1}}]}^{x}}{{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]}^{y}}{{[M{{L}^{3}}{{T}^{-2}}]}^{z}}\] |
| \[-y+z=0\Rightarrow y=z\] (i) |
| \[x+3y+3z=1\] (ii) |
| \[-x-4z=0\] (iii) |
| From (i), (ii) & (iii) |
| \[z=y=\frac{1}{2},\,x=-2\] |
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