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 276 questions; sign in to attempt the remaining 272 in the exam simulator.

1. Q1JEE Main 2026 (28 Jan, Shift 2)Angular momentum and Angular impulse
   When the position vector $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ changes sign as $-\vec{r}$, which one of the following vector will not flip under sign change ?
   1. A.Angular momentum
   2. B.Velocity
   3. C.Acceleration
   4. D.Linear momentum
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   Answer: (A)

   The transformation $\vec{r} \to -\vec{r}$ is known as a parity transformation (inversion through the origin). Vectors that change sign under parity are called polar vectors, while those that do not change sign are called axial vectors (or pseudovectors). Velocity is defined as $\vec{v} = \frac{d\vec{r}}{dt}$. Since $\vec{r} \to -\vec{r}$, we have $\vec{v} \to -\vec{v}$. Acceleration is defined as $\vec{a} = \frac{d\vec{v}}{dt}$. Since $\vec{v} \to -\vec{v}$, we have $\vec{a} \to -\vec{a}$. Linear momentum is defined as $\vec{p} = m\vec{v}$. Since $\vec{v} \to -\vec{v}$, we have $\vec{p} \to -\vec{p}$. Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. Under the transformation $\vec{r} \to -\vec{r}$ and $\vec{p} \to -\vec{p}$, the cross product becomes: $\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$. Thus, angular momentum does not flip its sign under the transformation $\vec{r} \to -\vec{r}$.
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.$\frac{\lambda L^3}{8 \pi^2}$
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   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 (04 Apr, Shift 1)Rolling without Slipping
   A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v$. The sphere and the cylinder reaches upto maximum heights $h_1$ and $h_2$, respectively, above the initial level. The ratio $h_1: h_2$ is $\frac{n}{10}$. The value of $n$ is______.
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   Answer: 7

   Gain in P.E. = Loss in K.E. $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right)$ $\mathrm{h} \propto 1+\frac{\mathrm{K}^2}{\mathrm{R}^2}$ $\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{1+\frac{2}{5}}{1+1}=\frac{7}{5 \times 2}=\frac{7}{10}$ $\mathrm{n}=7$
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 ______ $m{s}^{-1}$.
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   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, $2240=\dfrac{1}{2}2(v{)}^{2}+\dfrac{1}{2}\left[\dfrac{2}{5}(2){R}^{2}\right]{\left(\dfrac{v}{R}\right)}^{2}$ $\Rightarrow 2240={v}^{2}+\dfrac{2}{5}{v}^{2}$ $\Rightarrow v=40m{s}^{-1}$.