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. $h G e^{-2} \mu^{0}$

Answer 1

Answer: (A)

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