Units & Dimensions

JEE Main level, single correct, medium

112 of 149

Question 112 of 149

The quantities x=1μ0∈0,y=EBx=\dfrac{1}{\sqrt{{\mu }_{0}{\in }_{0}}},y=\dfrac{E}{B} and z=lCRz=\dfrac{l}{CR} are defined where C-capacitance, R-Resistance, ℓ−\ell -length, E-Electric field, B-magnetic field and∈0,μ0,−\in 0,\mu 0,-free space permittivity and permeability respectively. Then:

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Answer: (A) x, y and z have the same dimension.

(a) Dimensions of quantity are given bellow
μ0=[TmA−1]=[Wbm−1A−1]=[Vsm−1C−1s]=[NC−2s2]ϵ0=[C2N−1m−2]{\mu }_{0}=\left[Tm{A}^{-1}\right]=\left[Wb{m}^{-1}{A}^{-1}\right]=\left[Vs{m}^{-1}{C}^{-1}s\right]=\left[N{C}^{-2}{s}^{2}\right] {\epsilon }_{0}=\left[{C}^{2}{N}^{-1}{m}^{-2}\right]
Now the dimension of xx is given by
[x]=1[μ0∈0]=1[NC−2s2][C2N−1m−2][x]=ms−1\left[x\right]=\dfrac{1}{\left[\sqrt{{\mu }_{0}{\in }_{0}}\right]}=\sqrt{\dfrac{1}{\left[N{C}^{-2}{s}^{2}\right]\left[{C}^{2}{N}^{-1}{m}^{-2}\right]}} \left[x\right]=m{s}^{-1}
(b) The dimension of electric and magnetic field is given by,
B=[T]=[Wbm−2]=[Vsm−2]E=[Vm−1]B=\left[T\right]=\left[Wb{m}^{-2}\right]=\left[Vs{m}^{-2}\right] E=\left[V{m}^{-1}\right]
Now the dimension of yy is given by,
[y]=[E][B]=[Vm−1][Vsm−2]=ms−1\left[y\right]=\dfrac{\left[E\right]}{\left[B\right]}=\dfrac{\left[V{m}^{-1}\right]}{\left[Vs{m}^{-2}\right]}=m{s}^{-1}
(c) The dimension of capacitance and resistance is given by,
C=[CV−1]R=[VC−1s]C=\left[C{V}^{-1}\right] R=\left[V{C}^{-1}s\right]
Now the dimension of zz is given by
[z]=[l][CR]=[m][CV−1][VC−1s]=ms−1\left[z\right]=\dfrac{\left[l\right]}{\left[CR\right]}=\dfrac{\left[m\right]}{\left[C{V}^{-1}\right]\left[V{C}^{-1}s\right]}=m{s}^{-1}
All quantities are having the dimensions of velocity.

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