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NEET 2017
Exam
NEET
Year
2017
Subject
Physics
Chapter
Units and Dimensions

NEET 2017 · Physics · Units and Dimensions

A physical quantity of the dimensions of length that can be formed...

A physical quantity of the dimensions of length that can be formed out of c,Gc, G and e24πε0\dfrac{{e}^{2}}{4\pi{ε}_{0}}, where [cc is velocity of light, GG is universal constant of gravitation, ee is charge & ε0{ε}_{0} is permittivity of free space]:

Options

  1. A\dfrac{1}{{c}^{2}}{\left[G\dfrac{{e}^{2}}{4\pi{ε}_{0}}\right]}^{\dfrac{1}{2}}
  2. B{c}^{2}{\left[G\dfrac{{e}^{2}}{4\pi{ε}_{0}}\right]}^{\dfrac{1}{2}}
  3. C\dfrac{1}{{c}^{2}}{\left[\dfrac{{e}^{2}}{G4\pi{ε}_{0}}\right]}^{\dfrac{1}{2}}
  4. D1cGe24πε0\dfrac{1}{c}G\dfrac{{e}^{2}}{4\pi{ε}_{0}}

Correct answer

A. \dfrac{1}{{c}^{2}}{\left[G\dfrac{{e}^{2}}{4\pi{ε}_{0}}\right]}^{\dfrac{1}{2}}

Step-by-step solution

Let e24πε0=A=ML3T−2\dfrac{{e}^{2}}{4\pi{ε}_{0}} = A = M{L}^{3}{T}^{- 2}

L=CxGy(A)zL = {C}^{x}{G}^{y}{\left(A\right)}^{z}

[L]=[LT−1]x[M−1L3T−2]y[ML3T−2]z[ L ] = {\left[L{T}^{- 1}\right]}^{x}{\left[{M}^{- 1}{L}^{3}{T}^{- 2}\right]}^{y}{\left[M{L}^{3}{T}^{- 2}\right]}^{z}

−y+z=0⇒y=z- y + z = 0 ⇒ y = z ......(i)

x+3y+3z=1x + 3y + 3z = 1 ......(ii)

−x−4z=0- x - 4z = 0 ......(iii)

From (i), (ii) and (iii)

z=y=12,x=−2z = y = \dfrac{1}{2}, x = - 2
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