Question

Which of the following product of $e,h,\mu ,G$ (where $\mu$ is the permeability) be taken so that the dimensions of the product are same as that of the speed of light?

• A. $h{e}^{-2}{\mu }^{-1}{G}^0$
• B. $h^{2}e{G}^0\mu$
• C. $h^0{e}^2{G}^{-1}\mu$
• D. $hG{e}^{-2}{\mu }^0$

Answer 1

Answer: (A)

Here $v=e^a{h}^b{\mu }^c{G}^d$.
Taking the dimensions, $M^{0}L{T}^{-1}{A}^0$
$={\left[A{T}^1\right]}^a{\left[M{L}^2{T}^{-1}\right]}^b{\left[ML{T}^{-2}{A}^{-2}\right]}^c{\left[M^{-1}{L}^3{T}^{-2}\right]}^d$
There will be four simultaneous equations by equating the dimensions of $M,L,T$, and $A$.
These are $a-2c=0,a-b-2c-2d=-1,b+c-d=0$ and $2b+c+3d=1$.
Solving for $a,b,c$ and $d$, we get
$a=-2,b=1,c=-1,d=0$
Thus, $v=e^{-2}h{\mu }^{-1}{G}^0$