Circular Motion JEE Main previous year questions with solutions

5 solved JEE Main questions on Circular Motion, free to read — no sign-in needed. The full chapter has 80 questions; sign in to attempt the remaining 75 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 2)Frictional force
   A $0.5$ kg mass is in contact against the inner wall of a cylindrical drum of radius $4$ m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is $5$ rad/s. The coefficient of friction between the drum's inner wall surface and mass is _______. (Take $g = 10$ m/s$^2$)
   1. A.$0.1$
   2. B.$0.5$
   3. C.$0.7$
   4. D.$0.3$
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   Answer: (A)

   The normal force $N$ provides the necessary centripetal force for the mass to move in a circle: $N = m \omega^2 R$ For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force: $f \ge mg$ Since the maximum static friction is $f_{max} = \mu N$, we have: $\mu N \ge mg$ $\mu (m \omega^2 R) \ge mg$ $\mu \ge \dfrac{g}{\omega^2 R}$ Substituting the given values $g = 10$ m/s$^2$, $\omega = 5$ rad/s, and $R = 4$ m: $\mu \ge \dfrac{10}{5^2 \times 4}$ $\mu \ge \dfrac{10}{100}$ $\mu \ge 0.1$ The minimum coefficient of friction is $0.1$. Answer: $0.1$
2. Q2JEE Main 2025 (04 Apr, Shift 1)Angular momentum and Angular impulse
   If $\overrightarrow{\mathrm{L}}$ and $\overrightarrow{\mathrm{P}}$ represent the angular momentum and linear momentum respectively of a particle of mass ' m ' having position vector $\overrightarrow{\mathrm{r}}=\mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t}+\hat{\mathrm{j}} \sin \omega \mathrm{t})$. The direction of force is
   1. A.Opposite to the direction of $\vec{r}$
   2. B.Opposite to the direction of $\overrightarrow{\mathrm{L}}$
   3. C.Opposite to the direction of $\vec{P}$
   4. D.Opposite to the direction of $\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{P}}$
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   Answer: (A)

   $\overrightarrow{\mathrm{a}}=-\omega^2 \overrightarrow{\mathrm{r}}$ $\therefore \overrightarrow{\mathrm{F}}$ opposite to $\overrightarrow{\mathrm{r}}-$
3. Q3JEE Main 2024 (30 Jan, Shift 2)Electric Field and Electric Field Lines
   A particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then time period will be given as: (Consider $k$ as Coulomb's constant)
   1. A.${T}^{2}=\dfrac{4{\pi }^{2}m}{2k\lambda q}{r}^{3}$
   2. B.$T=2\pi r\sqrt{\dfrac{m}{2k\lambda q}}$
   3. C.$T=\dfrac{1}{2\pi r}\sqrt{\dfrac{m}{2k\lambda q}}$
   4. D.$T=\dfrac{1}{2\pi }\sqrt{\dfrac{2k\lambda q}{m}}$
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   Answer: (B)

   Attractive electrostatic force due to line charge on the charged particle will provide the required centripetal force. Therefore, $q\left(\dfrac{2k\lambda }{r}\right)=m{\omega }^{2}r$ $\Rightarrow {\omega }^{2}=\dfrac{2k\lambda q}{m{r}^{2}}$ $\Rightarrow {\left(\dfrac{2\pi }{T}\right)}^{2}=\dfrac{2k\lambda q}{m{r}^{2}}$ $\Rightarrow T=2\pi r\sqrt{\dfrac{m}{2k\lambda q}}$
4. Q4JEE Main 2023 (29 Jan, Shift 1)Non-uniform Circular Motion
   A car is moving on a horizontal curved road with radius $50m$. The approximate maximum speed of car will be, if friction between tyres and road is $0.34$. [Take $g=10m{s}^{-2}$]
   1. A.$3.4m{s}^{-1}$
   2. B.$22.4m{s}^{-1}$
   3. C.$13m{s}^{-1}$
   4. D.$17m{s}^{-1}$
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   Answer: (C)

   For a car moving with uniform speed along a circular track, the centripetal acceleration is provided by the frictional force acting radially inwards. Therefore, applying Newton's second law along radial direction, $\mu mg=\dfrac{m{v}^{2}}{r}$ $\Rightarrow v=\sqrt{\mu rg}$ $=\sqrt{0.34\times 50\times 10}$ $\approx 13m{s}^{-1}$
5. Q5JEE Main 2022 (24 Jun, Shift 1)Uniform Circular Motion
   A boy ties a stone of mass $100g$ to the end of a $2m$ long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of $80N$. If the maximum speed with which the stone can revolve is $\dfrac{K}{\pi }rev{\min }^{-1}$. The value of $K$ is : (Assume the string is massless and un-stretchable)
   1. A.$400$
   2. B.$300$
   3. C.$600$
   4. D.$800$
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   Answer: (C)

   The maximum angular velocity of the stone is $\omega =rpm\times \dfrac{2\pi }{60}=\dfrac{K}{30}rad{s}^{-1}$ The tension in the string will provide the required centripetal force. Therefore, $T=m{\omega }^{2}l \Rightarrow 80=0.1\times {\left(\dfrac{K}{30}\right)}^{2}\times 2 \Rightarrow {K}^{2}=360000\Rightarrow K=600$