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Units and Measurements question

2013 · Shift 0 · Q75
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Units and Measurements question

2013 · Shift 0 · Q75

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Let [ε0{\varepsilon _0}ε0​] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
  1. A
    ε0=[M−1L−3T2A]{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]ε0​=[M−1L−3T2A]
  2. B
    ε0={\varepsilon _0} =ε0​= [M−1L−3T4A2]\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right][M−1L−3T4A2]
  3. C
    ε0=[M1L2T1A2]{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]ε0​=[M1L2T1A2]
  4. D
    ε0=[M1L2T1A]{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]ε0​=[M1L2T1A]
View written solutionFree

Correct answer: B

  1. We use Coulomb's law:

F=14πε0q1q2r2F = \frac{1}{4\pi \varepsilon_0}\frac{q_1 q_2}{r^2}F=4πε0​1​r2q1​q2​​

So,

ε0=q2Fr2\varepsilon_0 = \frac{q^2}{F r^2}ε0​=Fr2q2​

  1. Write dimensions of each quantity:
  • Charge: [q]=[AT][q] = [AT][q]=[AT]
  • Force: [F]=[MLT−2][F] = [MLT^{-2}][F]=[MLT−2]
  • Distance: [r]=[L][r] = [L][r]=[L]

Thus,

[ε0]=[q]2[F][r]2[\varepsilon_0] = \frac{[q]^2}{[F][r]^2}[ε0​]=[F][r]2[q]2​

  1. Substitute dimensions:

[ε0]=(AT)2(MLT−2)(L2)[\varepsilon_0] = \frac{(AT)^2}{(MLT^{-2})(L^2)}[ε0​]=(MLT−2)(L2)(AT)2​

=A2T2ML3T−2= \frac{A^2 T^2}{M L^3 T^{-2}}=ML3T−2A2T2​

=M−1L−3T4A2= M^{-1} L^{-3} T^{4} A^{2}=M−1L−3T4A2

  1. Therefore,

[ε0]=[M−1L−3T4A2][\varepsilon_0] = [M^{-1}L^{-3}T^4A^2][ε0​]=[M−1L−3T4A2]

  1. Compare with options:
  • A: [M−1L−3T2A][M^{-1}L^{-3}T^2A][M−1L−3T2A] — incorrect
  • B: [M−1L−3T4A2][M^{-1}L^{-3}T^4A^2][M−1L−3T4A2] — correct
  • C: [M1L2T1A2][M^1L^2T^1A^2][M1L2T1A2] — incorrect
  • D: [M1L2T1A][M^1L^2T^1A][M1L2T1A] — incorrect

Hence, the correct option is B.

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