Dimension of c = $[M^{0} L^{1} T^{- 1}]$
Dimensions of G = $[M^{- 1} L^{3} T^{- 2}]$
Dimensions of $\frac{e^{2}}{4 {πε}_{o}},$
We know, $F$ $=$ $\frac{q^{2}}{4 {πε}_{o}} \frac{1}{r^{2}}$
So, $\frac{e^{2}}{4 {πε}_{o}}$ $=$ $F \times r^{2}$
Here, dimensions of $\frac{e^{2}}{4 {πε}_{o}}$ $=$ $[M^{1} L^{3} T^{- 2}]$
Now, let dimensions of length in terms of given quantities is given by:
L = $c^{x}$ $G^{y}$ ${(\frac{e^{2}}{4 {πε}_{o}})}^{z}$
$L$ $=$ ${[M^{0} L^{1} T^{- 1}]}^{x}$ ${[M^{- 1} L^{3} T^{- 2}]}^{y}$ ${[M^{1} L^{3} T^{- 2}]}^{z}$
$[M^{0} L^{1} T^{0}]$ $=$ $[M^{- y + z} L^{x + 3 y + 3 z} T^{- x - 2 y - 2 z}]$
By comparing, we get,

$y = z$ $. . . . . . . . . . . . (i)$
$- x - 2 y - 2 z = 0 \Rightarrow x = - 4 y$ $. . . . . . . . . . . . . (ii)$
$And$ $x + 3 y + 3 z$ $= 1 \Rightarrow - 4 y + 3 y + 3 y$ $=$ $1$
$So,$ $y = \frac{1}{2}$ $,$ $x = - 2,$ $z = \frac{1}{2}$

Finally the dimensions of $L = c^{- 2}$ $G^{\frac{1}{2}} {(\frac{e^{2}}{4 {πε}_{o}})}^{\frac{1}{2}}$

NEET 2017 Exam Paper