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Units and Measurement question

2017 · Q164
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Units and Measurement question

2017 · Q164

NEETPhysicsUnits and MeasurementMCQ+4 / −1
A physical quantity of the dimensions of length that can be formed out of c, G and e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}}4πε0​e2​ is [c is velocity of light, G is the universal constant of gravitation and e is charge]
  1. A
    c2[G−e24πε0]1/2{c^2}{\left[ {G - {{{e^2}} \over {4\pi {\varepsilon _0}}}} \right]^{1/2}}c2[G−4πε0​e2​]1/2
  2. B
    1c2[e2G 4πε0]1/2{1 \over {{c^2}}}{\left[ {{{{e^2}} \over {G\,4\pi {\varepsilon _0}}}} \right]^{1/2}}c21​[G4πε0​e2​]1/2
  3. C
    1cGe2 4πε0{1 \over c}G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}c1​G4πε0​e2​
  4. D
    1c2[Ge2 4πε0]1/2{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}}c21​[G4πε0​e2​]1/2
View written solutionFree

Correct answer: D

Dimension of
e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}}4πε0​e2​ = [ F×\times× d2 ] = [ML3T-2]

Dimension of G = [M-1L3T-2],

Dimension of c = [LT-1]

Now assume dimension of length is related as,

L = [c]x[G]y[e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}}4πε0​e2​]z

∴\therefore∴ [L1] = [ML3T-2]z [M-1L3T-2]y [LT-1]x

Comparing both sides and solving we get,

x = -2, y = 12{1 \over 2}21​, z = 12{1 \over 2}21​

∴\therefore∴ L = 1c2[Ge2 4πε0]1/2{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}}c21​[G4πε0​e2​]1/2

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