Laws of Motion JEE Main previous year questions with solutions

5 solved JEE Main questions on Laws of Motion, free to read — no sign-in needed. The full chapter has 187 questions; sign in to attempt the remaining 182 in the exam simulator.

1. Q1JEE Main 2026 (24 Jan, Shift 2)Equilibrium of Forces
   A flexible chain of mass $m$ hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is $30^{\circ}$. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is $\_\_\_\_$.
   1. A.$\frac{\sqrt{3}}{2} \mathrm{mg}$
   2. B.$\mathrm{mg}$
   3. C.$\frac{1}{2} m \mathrm{~g}$
   4. D.$\sqrt{3} \mathrm{mg}$
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   Answer: (A)

   A chain of mass $m$ hangs between two fixed points at the same level with inclination $30°$ at support points. Consider equilibrium of one half of the chain (mass $m/2$): Horizontal equilibrium: $T_0 = T_1\cos(30°) = T_1 \cdot \frac{\sqrt{3}}{2}$ Vertical equilibrium: $T_1\sin(30°) = \frac{m}{2}g$ From vertical equation: $T_1 \cdot \frac{1}{2} = \frac{mg}{2}$, giving $T_1 = mg$ Substituting into horizontal equation: $T_0 = mg \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}mg$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Newton's laws of motion
   An object with mass 500 g moves along x -axis with speed $v=4 \sqrt{x} \mathrm{~m} / \mathrm{s}$. The force acting on the object is :
   1. A.8 N
   2. B.5 N
   3. C.6 N
   4. D.4 N
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   Answer: (D)

   $\begin{aligned} & F=M \times a \\ & v=4 \sqrt{x} \\ & v^2=16 x \\ & 2 v \frac{d v}{d x}=16 \\ & \frac{v d v}{d x}=\frac{16}{2}=8 \\ & F=0.5 \times 8=4 N\end{aligned}$
3. Q3JEE Main 2024 (30 Jan, Shift 2)Frictional force
   A block of mass $m$ is placed on a surface having vertical cross section given by $y=\dfrac{{x}^{2}}{4}$. If coefficient of friction is $0.5$, the maximum height above the ground at which block can be placed without slipping is:
   1. A.$\dfrac{1}{4}m$
   2. B.$\dfrac{1}{2}m$
   3. C.$\dfrac{1}{6}m$
   4. D.$\dfrac{1}{3}m$
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   Answer: (A)

   At equilibrium, $\tan \theta =\mu$. Now, $\tan \theta$ is slope of the curve. Therefore, $\tan \theta =\dfrac{dy}{dx}=\dfrac{x}{2}$ Now, $\dfrac{x}{2}=0.5\Rightarrow x=1$ Required height $y=\dfrac{{x}^{2}}{4}=\dfrac{1}{4}m$.
4. Q4JEE Main 2023 (12 Apr, Shift 1)Addition and Subtraction of Vectors
   Three forces ${F}_{1}=10N,{F}_{2}=8N,{F}_{3}=6N$ are acting on a particle of mass $5kg$. The forces ${F}_{2}$ and ${F}_{3}$ are applied perpendicularly so that particle remains at rest. If the force ${F}_{1}$ is removed, then the acceleration of the particle is
   1. A.$7m{s}^{–2}$
   2. B.$0.5m{s}^{–2}$
   3. C.$4.8m{s}^{–2}$
   4. D.$2m{s}^{–2}$
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   Answer: (D)

   When the force ${F}_{1}$ is removed, the particle starts accelerating because of the resultant force between ${F}_{2}$ and ${F}_{3}$. The resultant force can be calculated as follows: $\begin{matrix}F & = & \sqrt{{{F}_{2}}^{2}+{{F}_{3}}^{2}} \\ & = & 10N\end{matrix}$ Hence, the acceleration of the particle is given by $\begin{matrix}a & = & \dfrac{10N}{5kg} \\ & = & 2m{s}^{–2}\end{matrix}$
5. Q5JEE Main 2022 (27 Jun, Shift 2)Spring force
   One end of a massless spring of spring constant $k$ and natural length ${l}_{0}$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing trough fixed end, then the elongation of the spring will be
   1. A.$\dfrac{k-m{\omega }^{2}{l}_{0}}{m{\omega }^{2}}$
   2. B.$\dfrac{m{\omega }^{2}{l}_{0}}{k+m{\omega }^{2}}$
   3. C.$\dfrac{m{\omega }^{2}{l}_{0}}{k-m{\omega }^{2}}$
   4. D.$\dfrac{k+m{\omega }^{2}{l}_{0}}{m{\omega }^{2}}$
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   Answer: (C)

   The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude $\dfrac{{v}^{2}}{r}$. The horizontal force on the particle is due to the spring and is given by $kx$, where $x$ is the elongation and $k$ is the spring constant. $kx=\dfrac{m{v}^{2}}{r}=m{\omega }^{2}r=m{\omega }^{2}\left({l}_{0}+x\right)$ $\Rightarrow \left(k-m{\omega }^{2}\right)x=m{\omega }^{2}{l}_{0}$ $\Rightarrow x=\dfrac{m{\omega }^{2}{l}_{0}}{k-m{\omega }^{2}}$