Question

A physical quantity of the dimensions of length that can be formed out of $c,G\text{and}\frac{e^2}{4\pi {ϵ}_0}$ is [$c$ is velocity of light, $G$ is universal constant of gravitation and $e$ is charge] :

• A. $[c{]}^2{\left[G\frac{e^2}{4\pi {ϵ}_0}\right]}^{\frac{1}{2}}$
• B. $\frac{1}{c^2}{\left[\frac{e^2}{G4\pi {ϵ}_0}\right]}^{\frac{1}{2}}$
• C. $\frac{1}{c}G\frac{e^2}{4\pi {ϵ}_0}$
• D. $\frac{1}{c^2}{\left[G\frac{e^2}{4\pi {ϵ}_0}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is D $\frac{1}{c^2}{\left[G\frac{e^2}{4\pi {ϵ}_0}\right]}^{\frac{1}{2}}$

Given: $c,G,$ $\frac{e^2}{4\pi {ϵ}_0}$
So, $\left[L\right]=\left[c{]}^x\right[G{]}^y{\left[\frac{e^2}{4\pi {ϵ}_0}\right]}^z$
$\left[L\right]=\left[L{T}^{-1}{]}^x\right[M^{-1}{L}^3{T}^{-2}{]}^y[M{L}^3{T}^{-2}{]}^z$
$\left[L\right]=L^{x+3y+3z}{M}^{-y+z}{T}^{-x-2y-2z}$
on comparing we get,
$x+3y+3z=1$
$-y+z=0$
$-x-2y-2z=0$
on solving we get
$x=-2,y=\frac{1}{2},z=\frac{1}{2}$
Thus, $L=\frac{1}{c^2}\sqrt{\frac{G{e}^2}{4\pi {ϵ}_0}}$
$L=\frac{1}{c^2}{\left[\frac{G{e}^2}{4\pi {ϵ}_0}\right]}^{\frac{1}{2}}$