Question
A physical quantity of the dimensions of length that can be formed out of $$c, G$$ and $$\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}$$ is [$$c$$ is velocity of light, $$G$$ is universal constant of gravitation and $$e$$ is charge]
A.
$$\frac{1}{{{c^2}}}{\left[ {G\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\frac{1}{2}}}$$
B.
$${c^2}{\left[ {G\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\frac{1}{2}}}$$
C.
$$\frac{1}{{{c^2}}}{\left[ {\frac{{{e^2}}}{{G\,4\pi {\varepsilon _0}}}} \right]^{\frac{1}{2}}}$$
D.
$$\frac{1}{c}G\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}$$
Answer :
$$\frac{1}{{{c^2}}}{\left[ {G\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\frac{1}{2}}}$$
Solution :
$$\eqalign{
& {\text{As force }}F = \frac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}} \Rightarrow \frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = {r^2} \cdot F \cr
& {\text{Putting dimensions of }}r{\text{ and }}F,{\text{ we get,}} \cr
& \Rightarrow \left[ {\frac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right] = \left[ {M{L^3}\;{T^{ - 2}}} \right]\,......\left( {\text{i}} \right) \cr
& {\text{Also, force,}}\,F = \frac{{G{m^2}}}{{{r^2}}} \cr
& \Rightarrow \left[ G \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {{M^2}} \right]}} \cr
& \Rightarrow \left[ G \right] = \left[ {{{\text{M}}^{ - 1}}\;{{\text{L}}^3}\;{{\text{T}}^{ - 2}}} \right]\,......\left( {{\text{ii}}} \right) \cr
& {\text{and}}\,\left[ {\frac{1}{{{c^2}}}} \right] = \frac{1}{{\left[ {\;{L^2}\;{T^{ - 2}}} \right]}} = \left[ {{L^{ - 2}}\;{T^2}} \right]\,......\left( {{\text{iii}}} \right) \cr
& {\text{Now, checking optionwise,}} \cr
& = \frac{1}{{{c^2}}}{\left( {\frac{{G{e^2}}}{{4\pi {\varepsilon _0}}}} \right)^{\frac{1}{2}}} \cr
& = \left[ {{L^{ - 2}}\;{T^2}} \right]{\left[ {{L^6}\;{T^{ - 4}}} \right]^{\frac{1}{2}}} = \left[ L \right] \cr} $$