1.Units, Dimensions and Measurement
hard

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

A

$\frac{{ch}}{{2\pi \varepsilon _0^2}}$ 

B

$\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}$

C

$\frac{{{\mu _0}{\varepsilon _0}G}}{{{c^2}h{e^2}}}$

D

$\frac{{2\pi \sqrt {{\mu _0}{\varepsilon _0}} h}}{{c{e^2}G}}$

(JEE MAIN-2014)

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 

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.