NEET Solved Paper 2017 Question 18

Question: 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]

Options:

A) $ \frac{1}{c}G\frac{e^{2}}{4\pi {\varepsilon_0}} $

B) $ \frac{1}{c^{2}}{{[ G\frac{e^{2}}{4\pi {\varepsilon_0}} ]}^{\frac{1}{2}}} $

C) $ c^{2}{{[ G\frac{e^{2}}{4\pi {\varepsilon_0}} ]}^{\frac{1}{2}}} $

D) $ \frac{1}{c^{2}}{{[ \frac{e^{2}}{G4\pi {\varepsilon_0}} ]}^{\frac{1}{2}}} $

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Answer:

Correct Answer: B

Solution:

Let $ \frac{e^{2}}{4\pi {\varepsilon_0}}=A=ML^{3}{T^{-2}} $ $ l=C^{x}G^{y}{{(A)}^{z}} $

$ L={{[L/{T^{-1}}]}^{x}}{{[{M^{-1}}L^{3}{T^{-2}}]}^{y}}{{[ML^{3}{T^{-2}}]}^{z}} $ $ -y+z=0\Rightarrow y=z $ …(i)

$ x+3y+3z=1 $ …(ii)

$ -x-4z=0 $ …(iii)

From (i), (ii) & (iii) 

$ z=y=\frac{1}{2},x=-2 $