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.

Did you get this right?

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.

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

More JEE Main Mechanics questions

Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. ![JEE Main 2026 Physics, Gravitation — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Gravitation/6983c1888bfdbb24356395e8/question_1__q_6983c1888bfdbb24356395e8__cdn-question-pool.getmarks.app__22S1_q_34_1__f4f2b45079_final_ppt_sync.png)

2026medium

A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):

2026easy

Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:

2026medium

In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

2026medium

The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)

2026medium

View all Mechanics questions

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

JEE Main 2020 — Physics Mechanics

2020mcqeasy

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.

Did you get this right?

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.

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

JEE Main 2020 — Physics Mechanics

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

JEE Main 2020 — Physics Mechanics

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