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NEET 2017 · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from NEET 2017

Q65NEET 2017Medium

Applications of Dimensional Analysis

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 the universal constant of gravitation and e is charge]

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Answer: (d)

$\dfrac{e^2}{4\pi\varepsilon_0} = F r^2$ , so its dimension is $ML^3T^{-2}$ .
$G\dfrac{e^2}{4\pi\varepsilon_0}$ : $M^{-1}L^3T^{-2}\cdot ML^3T^{-2} = L^6T^{-4}$ ; square root gives $L^3T^{-2}$ .
Dividing by $c^2 = L^2T^{-2}$ gives $L$ .
So the answer is (d).

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