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Units and Measurements question

2019 · 12 Apr · Shift 1 · Q56
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Units and Measurements question

2019 · 12 Apr · Shift 1 · Q56

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
Which of the following combinations has the dimension of electrical resistance (∈\in∈ 0 is the permittivity of vacuum and μ\muμ 0 is the permeability of vacuum)?
  1. A
    ∈0μ0\sqrt {{{{ \in _0}} \over {{\mu _0}}}}μ0​∈0​​​
  2. B
    ∈0μ0{{{{ \in _0}} \over {{\mu _0}}}}μ0​∈0​​
  3. C
    μ0∈0\sqrt {{{{\mu _0}} \over {{ \in _0}}}}∈0​μ0​​​
  4. D
    μ0∈0{{{{\mu _0}} \over {{ \in _0}}}}∈0​μ0​​
View written solutionFree

Correct answer: C

  1. Use the known relation between μ0, ε0\mu_0,\ \varepsilon_0μ0​, ε0​ and speed of light

We know that

μ0ε0=1c2\mu_0 \varepsilon_0 = \frac{1}{c^2}μ0​ε0​=c21​

Also, the intrinsic impedance of free space is

Z0=μ0ε0Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}Z0​=ε0​μ0​​​

which has the unit of electrical resistance (ohm).

So from standard electromagnetic dimensional relations,

μ0ε0\sqrt{\frac{\mu_0}{\varepsilon_0}}ε0​μ0​​​

has the dimension of resistance.

  1. Verify by dimensional analysis

Electrical resistance is

R=VIR = \frac{V}{I}R=IV​

so

[R]=[V][I]=ML2T−3I−1I=ML2T−3I−2[R] = \frac{[V]}{[I]} = \frac{ML^2T^{-3}I^{-1}}{I} = ML^2T^{-3}I^{-2}[R]=[I][V]​=IML2T−3I−1​=ML2T−3I−2

Now,

  • Permittivity:
[ε0]=Q2FL2[\varepsilon_0] = \frac{Q^2}{F L^2}[ε0​]=FL2Q2​

But more directly from Coulomb's law,

14πε0⋅q2r2=F\frac{1}{4\pi\varepsilon_0}\cdot \frac{q^2}{r^2} = F4πε0​1​⋅r2q2​=F

so

[ε0]=Q2FL2[\varepsilon_0] = \frac{Q^2}{F L^2}[ε0​]=FL2Q2​

Using [Q]=IT[Q]=IT[Q]=IT and [F]=MLT−2[F]=MLT^{-2}[F]=MLT−2,

[ε0]=(IT)2(MLT−2)L2=I2T2ML3T−2=M−1L−3T4I2[\varepsilon_0] = \frac{(IT)^2}{(MLT^{-2})L^2} = \frac{I^2T^2}{ML^3T^{-2}} = M^{-1}L^{-3}T^4I^2[ε0​]=(MLT−2)L2(IT)2​=ML3T−2I2T2​=M−1L−3T4I2
  • Permeability: Using
c2=1μ0ε0c^2 = \frac{1}{\mu_0\varepsilon_0}c2=μ0​ε0​1​

we get

[μ0]=1[ε0][c2][\mu_0] = \frac{1}{[\varepsilon_0][c^2]}[μ0​]=[ε0​][c2]1​

Since [c2]=L2T−2[c^2]=L^2T^{-2}[c2]=L2T−2,

[μ0]=1(M−1L−3T4I2)(L2T−2)=MLT−2I−2[\mu_0] = \frac{1}{\left(M^{-1}L^{-3}T^4I^2\right)\left(L^2T^{-2}\right)} = MLT^{-2}I^{-2}[μ0​]=(M−1L−3T4I2)(L2T−2)1​=MLT−2I−2

Now,

[μ0ε0]=MLT−2I−2M−1L−3T4I2=M2L4T−6I−4\left[\frac{\mu_0}{\varepsilon_0}\right] = \frac{MLT^{-2}I^{-2}}{M^{-1}L^{-3}T^4I^2} = M^2L^4T^{-6}I^{-4}[ε0​μ0​​]=M−1L−3T4I2MLT−2I−2​=M2L4T−6I−4

Therefore,

[μ0ε0]=ML2T−3I−2\left[\sqrt{\frac{\mu_0}{\varepsilon_0}}\right] = ML^2T^{-3}I^{-2}[ε0​μ0​​​]=ML2T−3I−2

which is exactly the dimension of resistance.

  1. Check options
  • A: ε0μ0\sqrt{\dfrac{\varepsilon_0}{\mu_0}}μ0​ε0​​​ → inverse of resistance
  • B: ε0μ0\dfrac{\varepsilon_0}{\mu_0}μ0​ε0​​ → inverse square of resistance
  • C: μ0ε0\sqrt{\dfrac{\mu_0}{\varepsilon_0}}ε0​μ0​​​ → resistance ✅
  • D: μ0ε0\dfrac{\mu_0}{\varepsilon_0}ε0​μ0​​ → square of resistance

Hence, the correct option is

C\boxed{\text{C}}C​
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