The quantities x= 1 √μ _0 ∈… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020mcqeasy

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

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

A
x, y and z have the same dimension.
B
Only x and z have the same dimension
C
Only x and y have the same dimension
D
Only y and z have the same dimension.

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Solution

(a) Dimensions of quantity are given bellow

${\mu }_{0}=[Tm{A}^{-1}]=[\mathrm{Wb}{m}^{-1}{A}^{-1}]=[Vs{m}^{-1}{C}^{-1}s]=[N{C}^{-2}{s}^{2}] {\epsilon }_{0}=[{C}^{2}{N}^{-1}{m}^{-2}]$

Now the dimension of $x$ is given by

$[x]=\frac{1}{[\sqrt{{\mu }_{0}{\in }_{0}}]}=\sqrt{\frac{1}{[N{C}^{-2}{s}^{2}][{C}^{2}{N}^{-1}{m}^{-2}]}} [x]=m{s}^{-1}$

(b) The dimension of electric and magnetic field is given by,

$B=[T]=[\mathrm{Wb}{m}^{-2}]=[Vs{m}^{-2}] E=[V{m}^{-1}]$

Now the dimension of $y$ is given by,

$[y]=\frac{[E]}{[B]}=\frac{[V{m}^{-1}]}{[Vs{m}^{-2}]}=m{s}^{-1}$

$C=[C{V}^{-1}] R=[V{C}^{-1}s]$

Now the dimension of $z$ is given by

$[z]=\frac{[l]}{[CR]}=\frac{[m]}{[C{V}^{-1}][V{C}^{-1}s]}=m{s}^{-1}$

All quantities are having the dimensions of velocity.

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Solution

(a) Dimensions of quantity are given bellow

${\mu }_{0}=[Tm{A}^{-1}]=[\mathrm{Wb}{m}^{-1}{A}^{-1}]=[Vs{m}^{-1}{C}^{-1}s]=[N{C}^{-2}{s}^{2}] {\epsilon }_{0}=[{C}^{2}{N}^{-1}{m}^{-2}]$

Now the dimension of $x$ is given by

$[x]=\frac{1}{[\sqrt{{\mu }_{0}{\in }_{0}}]}=\sqrt{\frac{1}{[N{C}^{-2}{s}^{2}][{C}^{2}{N}^{-1}{m}^{-2}]}} [x]=m{s}^{-1}$

(b) The dimension of electric and magnetic field is given by,

$B=[T]=[\mathrm{Wb}{m}^{-2}]=[Vs{m}^{-2}] E=[V{m}^{-1}]$

Now the dimension of $y$ is given by,

$[y]=\frac{[E]}{[B]}=\frac{[V{m}^{-1}]}{[Vs{m}^{-2}]}=m{s}^{-1}$

$C=[C{V}^{-1}] R=[V{C}^{-1}s]$

Now the dimension of $z$ is given by

$[z]=\frac{[l]}{[CR]}=\frac{[m]}{[C{V}^{-1}][V{C}^{-1}s]}=m{s}^{-1}$

All quantities are having the dimensions of velocity.

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics