Rotational Motion JEE Main previous year questions with solutions

4 solved JEE Main questions on Rotational Motion, free to read — no sign-in needed. The full chapter has 272 questions; sign in to attempt the remaining 268 in the exam simulator.

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1. Q1JEE Main 2026 (23 Jan, Shift 1)Angular momentum and Angular impulse
   Two small balls with masses $m$ and 2 m are attached to both ends of a rigid rod of length $d$ and negligible mass. If angular momentum of this system is $L$ about an axis ($A$) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about $A$ is :
   1. A.$\frac{2 L}{5 m d^{2}}$
   2. B.$\frac{4}{3} \frac{L}{m d^{2}}$
   3. C.$\frac{3}{2} \frac{L}{m d^{2}}$
   4. D.$\frac{2 L}{m d^{2}}$
   Show answer & solution

   Answer: (C)

   CM from mass $m$: $x_{cm} = \frac{2md}{3m} = \frac{2d}{3}$. Distances: $m$ is $2d/3$ from CM, $2m$ is $d/3$ from CM. $I_{cm} = m(2d/3)^2 + 2m(d/3)^2 = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{2md^2}{3}$. $L = I\omega \Rightarrow \omega = \frac{L}{I} = \frac{3L}{2md^2}$.
2. Q2JEE Main 2025 (08 Apr, Shift 2)Moment of inertia of rigid bodies
   A rod of linear mass density ' $\lambda$ ' and length ' $L$ ' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is :
   1. A.$\frac{\lambda \mathrm{L}^3}{16 \pi^2}$
   2. B.$\frac{\lambda \mathrm{L}^3}{12}$
   3. C.$\frac{\lambda \mathrm{L}^3}{4 \pi^2}$
   4. D.$\frac{\lambda L^3}{8 \pi^2}$
   Show answer & solution

   Answer: (D)

   $L=2 \pi R$ $\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}$
3. Q3JEE Main 2024 (31 Jan, Shift 1)Energy
   A solid circular disc of mass $50kg$ rolls along a horizontal floor so that its center of mass has a speed of $0.4m{s}^{-1}$. The absolute value of work done on the disc to stop it is ______ $J$.
   Show answer & solution

   Answer: 6

   Using work-energy theorem, we get $W=∆KE$ $\Rightarrow W=0-\left(\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2}\right)$ $\Rightarrow W=0-\dfrac{1}{2}m{v}^{2}\left(1+\dfrac{{K}^{2}}{{R}^{2}}\right)$ $=-\dfrac{1}{2}\times 50\times 0.{4}^{2}\left(1+\dfrac{1}{2}\right)=-6J$ Absolute work $=+6J$ Therefore, required value of $W=6J$
4. Q4JEE Main 2023 (29 Jan, Shift 1)Combination of translation and rotation
   A solid sphere of mass $2kg$ is making pure rolling on a horizontal surface with kinetic energy $2240J$. The velocity of centre of mass of the sphere will be ______ $m{s}^{-1}$.
   Show answer & solution

   Answer: 40

   Let velocity of centre of mass of sphere be $v$. Total kinetic energy of solid sphere is $KE=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2}$, where, $I$ is moment of inertia of the solid sphere about the axis passing through it's centre of mass and $\omega =\dfrac{v}{R}\left(\text{pure rolling condition}\right)$. Therefore, $2240=\dfrac{1}{2}2(v{)}^{2}+\dfrac{1}{2}\left[\dfrac{2}{5}(2){R}^{2}\right]{\left(\dfrac{v}{R}\right)}^{2}$ $\Rightarrow 2240={v}^{2}+\dfrac{2}{5}{v}^{2}$ $\Rightarrow v=40m{s}^{-1}$.