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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
The\,{\rm{Dimensional}}\,{\rm{formulae}}\,{\rm{of}}\\
{\rm{e}}\,{\rm{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} \right]\\
{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{-3}}{T^4}{A^2}} \right]\\
G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]\\
and\,{m_e} = \left[ {{M^1}{L^0}{T^0}} \right]\\
Now,\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}
\end{array}\)
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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