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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2014
  • Hard
  • MCQ

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is

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Answer & Solution

Correct Answer

(B) \(\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{l}
The\,{\rm{Dimensional}}\,{\rm{formulae}}\,{\rm{of}}\\
{\rm{e}}\,{\rm{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} \right]\\
{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{-3}}{T^4}{A^2}} \right]\\
G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]\\
and\,{m_e} = \left[ {{M^1}{L^0}{T^0}} \right]\\
Now,\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}
\end{array}\)

\(\begin{array}{l}
 = \frac{{{{\left[ {{M^0}{L^0}{T^1}{A^1}} \right]}^2}}}{{2\pi \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]{{\left[ {{M^1}{L^0}{T^o}} \right]}^2}}}\\
 = \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^{ - 1 - 1 + 2}}{L^{ - 3 + 3}}{T^{4 - 2}}{A^2}} \right]}}\\
 = \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^0}{L^0}{T^2}{A^2}} \right]}} = \frac{1}{{2\pi }}
\end{array}\)

\(\begin{array}{l}
\frac{1}{{2\pi }}\,is\,{\rm{Dimensionl}}ess\,thus\,the\,combination\\
\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\end{array}\)

would d have the same value in diffierent systems of units 

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