Physical World Units and Measurements - Result Question 8
8. A physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^{2}}{4 \pi \varepsilon_0}$ is [c is velocity of light, $G$ is universal constant of gravitation and $e$ is charge]
[2017]
(a) $c^{2}[G \frac{e^{2}}{4 \pi \varepsilon_0}]^{1 / 2}$
(b) $\frac{1}{c^{2}}[\frac{e^{2}}{G 4 \pi \varepsilon_0}]^{1 / 2}$
(c) $\frac{1}{c} G \frac{e^{2}}{4 \pi \varepsilon_0}$
(d) $\frac{1}{c^{2}}[G \frac{e^{2}}{4 \pi \varepsilon_0}]^{1 / 2}$
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Answer:
Correct Answer: 8. (d)
Solution:
- (d) Let dimensions of length is related as,
$[L]=[c]^{x}[G]^{y}[\frac{e^{2}}{4 \pi \varepsilon_0}]^{z}$ $\Rightarrow \frac{e^{2}}{4 \pi \varepsilon_0}=[ML^{3} T^{-2}]$
$[L]=[LT^{-1}]^{x}[M^{-1} L^{3} T^{-2}}^{y}[ML^{3} T^{-2}]^{z}$
$[L]=[L^{x+3 y+3 z} M^{-y+z} T^{-x-2 y-2 z}]$
Comparing both sides
$-y+z=0 \Rightarrow y=z$
$x+3 y+3 z=1$
$-x-4 z=0 \quad(\because y=z)$
From (i), (ii) and (iii)
$z=y=\frac{1}{2}, x=-2$
Hence, $[L]=c^{-2}[G \cdot \frac{e^{2}}{4 \pi \varepsilon_0}]^{1 / 2}$
