Question

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

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

Answer 1

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

Let the physical quantity be represented by $l$.
Then,
$l\propto c^x{G}^y(\frac{e^2}{4\pi {ϵ}_0}{)}^z$
$\begin{array}{}\left[l\right]=\left[c{]}^x\right[G{]}^y{\left[\frac{e^2}{4\pi {ϵ}_0}\right]}^z& \text{(1)}\end{array}$
as we know the dimension of these physical quantities are
$l=\left[M^0{L}^1{T}^0\right]$
$c=\left[M^0{L}^0{T}^{-1}\right]$
$G=\left[M^{-1}{L}^3{T}^{-2}\right]$
$\frac{e^2}{4\pi {\epsilon }_0}=\left[M{L}^3{T}^{-2}\right]$
Putting these values in equation (1) we get,
$$L={\left[L{T}^{-1}\right]}^x{\left[M^{-1}{L}^3{T}^{-2}\right]}^y{\left[M{L}^3{T}^{-2}\right]}^z$$$$-y+z=0\Rightarrow y=z$$$$x+3y+3z=1$$$$-x-4z=0$$Solving the above three equations, we get
$z=y=\frac{1}{2},x=-2$
Thus, $l$ will be
$$\frac{1}{c^2}{\left[G\frac{e^2}{4\pi {\epsilon }_0}\right]}^{\frac{1}{2}}$$