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Units and Measurements question

2020 · 5 Sep · Shift 2 · Q51
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Units and Measurements question

2020 · 5 Sep · Shift 2 · Q51

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The quantities x = 1μ0ε0{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}μ0​ε0​​1​, y = EB{E \over B}BE​ and z =lCR{l \over {CR}}CRl​ are defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and ε0{{\varepsilon _0}}ε0​, μ0{{\mu _0}}μ0​, - free space permittivity and permeability respectively. Then :
  1. A
    Only y and z have the same dimension
  2. B
    x, y and z have the same dimension
  3. C
    Only x and y have the same dimension
  4. D
    Only x and z have the same dimension
View written solutionFree

Correct answer: B

  1. We compare the dimensions of x=1μ0ε0,y=EB,z=lCR.x=\frac{1}{\sqrt{\mu_0\varepsilon_0}},\qquad y=\frac{E}{B},\qquad z=\frac{l}{CR}.x=μ0​ε0​​1​,y=BE​,z=CRl​.

  2. Dimension of xxx

    From electromagnetism, the speed of light is c=1μ0ε0.c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}.c=μ0​ε0​​1​. Therefore, [x]=[c]=LT−1.[x]=[c]=LT^{-1}.[x]=[c]=LT−1.

  3. Dimension of yyy

    We use the relation for electromagnetic waves: E=cB⇒EB=c.E=cB \quad \Rightarrow \quad \frac{E}{B}=c.E=cB⇒BE​=c. Hence, [y]=LT−1.[y]=LT^{-1}.[y]=LT−1.

    Alternatively, [E]=Fq=MLT−2IT=MLT−3I−1,[E]=\frac{F}{q}=\frac{MLT^{-2}}{IT}=MLT^{-3}I^{-1},[E]=qF​=ITMLT−2​=MLT−3I−1, and [B]=Fqv=MLT−2(IT)(LT−1)=MT−2I−1.[B]=\frac{F}{qv}=\frac{MLT^{-2}}{(IT)(LT^{-1})}=MT^{-2}I^{-1}.[B]=qvF​=(IT)(LT−1)MLT−2​=MT−2I−1. So, [EB]=MLT−3I−1MT−2I−1=LT−1.\left[\frac{E}{B}\right]=\frac{MLT^{-3}I^{-1}}{MT^{-2}I^{-1}}=LT^{-1}.[BE​]=MT−2I−1MLT−3I−1​=LT−1.

  4. Dimension of zzz

    We need the dimension of CRCRCR.

    Since RC=time constant,RC=\text{time constant},RC=time constant, we have [RC]=T.[RC]=T.[RC]=T. Therefore, [z]=LT=LT−1.[z]=\frac{L}{T}=LT^{-1}.[z]=TL​=LT−1.

    For completeness: [R]=VI=ML2T−3I−2,[R]=\frac{V}{I}=ML^2T^{-3}I^{-2},[R]=IV​=ML2T−3I−2, [C]=QV=ITML2T−3I−1=M−1L−2T4I2,[C]=\frac{Q}{V}=\frac{IT}{ML^2T^{-3}I^{-1}}=M^{-1}L^{-2}T^4I^2,[C]=VQ​=ML2T−3I−1IT​=M−1L−2T4I2, hence [CR]=T.[CR]=T.[CR]=T.

  5. Comparison

    We found: [x]=[y]=[z]=LT−1.[x]=[y]=[z]=LT^{-1}.[x]=[y]=[z]=LT−1.

  6. Correct option

    Therefore, all three quantities have the same dimension.

    Option B\boxed{\text{Option B}}Option B​

  7. Comparison with stored correct answer

    Stored correct answer: B

    Our derived answer: B

    So they agree.

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