- Home
- Standard 11
- Physics
ભૌતિક અચળાંકોના નીચે દર્શાવેલા સમીકરણો માથી (તેમના સામાન્ય ચિન્હોથી દર્શાવેલા) કયું એકમાત્ર સમીકરણ કે જે અલગ અલગ માપન પદ્ધતિમાં સમાન મૂલ્ય આપે?
$\frac{{ch}}{{2\pi \varepsilon _0^2}}$
$\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}$
$\frac{{{\mu _0}{\varepsilon _0}G}}{{{c^2}h{e^2}}}$
$\frac{{2\pi \sqrt {{\mu _0}{\varepsilon _0}} h}}{{c{e^2}G}}$
Solution
$\begin{array}{l}
The\,{\rm{Dimensional}}\,{\rm{formulae}}\,{\rm{of}}\\
{\rm{e}}\,{\rm{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} \right]\\
{\varepsilon _0} = \left[ {{M^{ – 1}}{L^3}{T^4}{A^2}} \right]\\
G = \left[ {{M^{ – 1}}{L^{-3}}{T^{ – 2}}} \right]\\
and\,{m_e} = \left[ {{M^1}{L^0}{T^0}} \right]\\
Now,\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}
\end{array}$
$\begin{array}{l}
= \frac{{{{\left[ {{M^0}{L^0}{T^1}{A^1}} \right]}^2}}}{{2\pi \left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]{{\left[ {{M^1}{L^0}{T^o}} \right]}^2}}}\\
= \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^{ – 1 – 1 + 2}}{L^{ – 3 + 3}}{T^{4 – 2}}{A^2}} \right]}}\\
= \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^0}{L^0}{T^2}{A^2}} \right]}} = \frac{1}{{2\pi }}
\end{array}$
$\begin{array}{l}
\frac{1}{{2\pi }}\,is\,{\rm{Dimensionl}}ess\,thus\,the\,combination\\
\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\end{array}$
would d have the same value in diffierent systems of units
