A physical quantity of the dimensions of length that can be formed out of c, G and $\frac{\mathrm{e}^{2}}{4…

A physical quantity of the dimensions of length that can be formed out of c, G and $\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}$ is [c is velocity of light, G is universal constant of gravitation and e is charge]
  1. $\mathrm{c}^{2}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}$
  2. $\frac{1}{\mathrm{c}^{2}}\left[\frac{\mathrm{e}^{2}}{\mathrm{G} 4 \pi \varepsilon_{0}}\right]^{1 / 2}$
  3. $\frac{1}{c} \mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}$
  4. $\frac{1}{\mathrm{c}^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}$

Solution

Let dimensions of length is related as,
$L=[c]^{x}[G]^{y}\left[\frac{e^{2}}{4 \pi \varepsilon_{0}}\right]^{z}$
$\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}=\mathrm{ML}^{3} \mathrm{~T}^{-2}$
$\mathrm{L}=\left[\mathrm{LT}^{-1}\right]^{\mathrm{x}}\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]^{\mathrm{y}}\left[\mathrm{ML}^{3} \mathrm{~T}^{-2}\right]^{2}$
$[L]=\left[L^{x+3 y+3 z} M^{-y+z} T^{-x-2 y-2 z}\right]$
Comparing both sides
$-y+z=0 \Rightarrow y=z$.....(i)
$x+3 y+3 z=1$.....(ii)
$-x-4 z=0 \quad(\because y=z)$.....(iii)
From (i), (ii) and (iii) $z=y=\frac{1}{2}, x=-2$
Hence, $L=c^{-2}\left[G \cdot \frac{e^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}$ ,

Asked in: JEE Mains - Units and Dimensions - Test 2

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