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QNEET-2017-Ph-974: NEET 2017 Physics - Class 11 NEET-2017-Ph-974 Solution: A physical quantity of the dimensions of length that can be…

A physical quantity of the dimensions of length that can be formed out of c G and e24πϵ0\frac{e^2}{4\pi\epsilon_0} is [c is velocity of light G is the universal constant of gravitation and e is charge] c2[Ge24πϵ0]1/2c^2[G\frac{e^2}{4\pi\epsilon_0}]^{1/2} 1c2[e2G4πϵ0]1/2\frac{1}{c^2}[\frac{e^2}{G4\pi\epsilon_0}]^{1/2} 1c[Ge24πϵ0]1/2\frac{1}{c}[G\frac{e^2}{4\pi\epsilon_0}]^{1/2} 1c2[Ge24πϵ0]1/2\frac{1}{c^2}[G\frac{e^2}{4\pi\epsilon_0}]^{1/2}

Multiple Choice Options

  • A c2[G−e24πε0]1/2{c^2}{\left[ {G - {{{e^2}} \over {4\pi {\varepsilon _0}}}} \right]^{1/2}} c 2 [ G _ 4 � _ 0 _ e 2 _ ] 1/2
  • B 1c2[e2G 4πε0]1/2{1 \over {{c^2}}}{\left[ {{{{e^2}} \over {G\,4\pi {\varepsilon _0}}}} \right]^{1/2}} c 2 1 _ [ G 4 � _ 0 _ e 2 _ ] 1/2
  • C 1cGe2 4πε0{1 \over c}G{{{e^2}} \over {\,4\pi {\varepsilon _0}}} c 1 _ G 4 � _ 0 _ e 2 _
  • D 1c2[Ge2 4πε0]1/2{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}} c 2 1 _ [ G 4 � _ 0 _ e 2 _ ] 1/2

Step-by-Step Verified Solution

Dimension of e 2 4 � _ 0 = [ F _ d 2 ] = [ML 3 T -2 ] Dimension of G = [M -1 L 3 T -2 ], Dimension of c = [LT -1 ] Now assume dimension of length is related as, L = [c] x [G] y [ e 2 4 � _ 0 ] z _ [L 1 ] = [ML 3 T -2 ] z [M -1 L 3 T -2 ] y [LT -1 ] x Comparing both sides and solving we get, x = -2, y = 1 2 , z = 1 2 _ L = 1 c 2 [ G e 2 4 � _ 0 ] 1 / 2

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