System of Particles and Rotational Motion NEET PYQs (2005–2026) with Solutions

10 solved System of Particles and Rotational Motion PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyRotational Kinematics

   The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:

   1. AOption A: 900
   2. BOption B: 600
   3. COption C: 150
   4. DOption D: 300

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   Correct answer: (C) 150

   Explanation

   Convert angular speeds into SI units: $$\omega_1 = 600\times\frac{2\pi}{60}=20\pi\,\text{rad/s}$$ $$\omega_2 = 1200\times\frac{2\pi}{60}=40\pi\,\text{rad/s}$$ Angular acceleration: $$\alpha=\frac{\omega_2-\omega_1}{t} =\frac{40\pi-20\pi}{10} =2\pi\,\text{rad/s}^2$$ Using rotational equation of motion: $$\omega_2^2=\omega_1^2+2\alpha\theta$$ Substitute values: $$(40\pi)^2=(20\pi)^2+2(2\pi)\theta$$ $$1600\pi^2=400\pi^2+4\pi\theta$$ $$1200\pi^2=4\pi\theta$$ $$\theta=300\pi\,\text{rad}$$ Number of revolutions: $$N=\frac{\theta}{2\pi} =\frac{300\pi}{2\pi} =150$$ Therefore, the correct answer is: $$\boxed{150}$$

2. Question 2 (NEET 2026)

   NEET 2026HardLinear momentum of a system of particles

   A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point $A\left(\theta = \frac{\pi}{2}\right)$ with identical uniform angular speeds in opposite directions, and meet again at point $B\left(\theta = -\frac{\pi}{2}\right)$. During this time, which of the following figures schematically represent the magnitude of the total linear momentum $P$ of the system, as a function of $\theta$?

   1. AOption A: Graph A
   2. BOption B: Graph B
   3. COption C: Graph C
   4. DOption D: Graph D

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   Correct answer: (C) Graph C

   Explanation

   The graphs ($P$ against $\theta$ from $\pi/2$ to $-\pi/2$): (A) $P$ constant; (B) $P$ falls from a maximum to zero midway and rises again; (C) $P$ rises from zero to a maximum midway and falls back to zero (dome shape); (D) $P$ decreases linearly to zero. Take one particle at angle $\theta$ (moving clockwise) and the other at the mirror position $\pi - \theta$ (moving anticlockwise), each with speed $\omega$ (unit radius, unit mass). Particle 1: $\vec{v}_1 = \omega(\sin\theta, -\cos\theta)$; particle 2: $\vec{v}_2 = \omega(-\sin\theta, -\cos\theta)$. The horizontal components cancel and the vertical components add. $|\vec{P}| = 2\omega|\cos\theta|$. At $\theta = \frac{\pi}{2}$ (start) and $\theta = -\frac{\pi}{2}$ (meeting point) $P = 0$; at $\theta = 0$ it is maximum, $2\omega$. So $P$ rises from zero to a maximum and falls back to zero, the dome-shaped graph of option (C).

3. Question 3 (NEET 2026)

   NEET 2026MediumParallel axis theorem

   A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is:

   1. AOption A: $(M - m)(R + r)^2$
   2. BOption B: $(m - M)(R + r)^2$
   3. COption C: $(m - M)(R - r)^2$
   4. DOption D: $0$

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   Correct answer: (B) $(m - M)(R + r)^2$

   Explanation

   The spheres touch at a point, so the distance between their centres is $d = R + r$. Using the parallel axis theorem: $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R + r)^2$ $I_B = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + M(R + r)^2$ $I_A - I_B = (m - M)(R + r)^2$ (negative, since $m < M$), option (B).

4. Question 4 (NEET 2026)

   NEET 2026MediumAngular momentum about different points

   A thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$. Its angular momentum is $L_A$ and $L_B$ computed about points $A$ and $B$, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is:

   1. AOption A: $\frac{1}{4}$
   2. BOption B: $\frac{1}{2}$
   3. COption C: $1$
   4. DOption D: $2$

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   Correct answer: (C) $1$

   Explanation

   Angular momentum about any point $P$: $\vec{L}_P = \vec{L}_{cm} + \vec{r}_{cm} \times M\vec{v}_{cm}$, where $\vec{r}_{cm}$ is the position of the centre of mass relative to $P$. The disc rotates about its fixed centre, so its centre of mass is at rest: $\vec{v}_{cm} = 0$ (total linear momentum is zero). Hence the angular momentum is the same about every point: $L_A = L_B = I\omega$, whatever the distances $OA$ and $OB$. $\frac{L_A}{L_B} = 1$, option (C).

5. Question 5 (NEET 2026)

   NEET 2026MediumMoment of Inertia

   A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre C as shown in figure. The moment of inertia of the ring about an axis yy′ will be:

   1. AOption A: 3mL³ / 8π
   2. BOption B: 3mL³ / 8π²
   3. COption C: 3mL² / 8π
   4. DOption D: 3mL² / 8π²

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   Correct answer: (B) 3mL³ / 8π²

   Explanation

   Mass of ring M=mL. Radius r=L/(2π). Using parallel axis theorem: Iyy′=ICM+Mr²=(Mr²/2)+Mr²=(3/2)Mr²=(3/2)(mL)(L/2π)²=3mL³/(8π²). Hence option (B) is correct.

6. Question 6 (NEET 2025)

   NEET 2025HardMoment of Inertia

   A sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is:

   1. AOption A: 7/8
   2. BOption B: 7/40
   3. COption C: 7/57
   4. DOption D: 7/64

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   Correct answer: (C) 7/57

   Explanation

   Let the density of the material be $\\rho$. Mass of larger sphere: $$M = \\rho \\cdot \\frac{4}{3}\\pi (2R)^3 = \\frac{32}{3}\\pi \\rho R^3$$ Mass of smaller sphere: $$m = \\rho \\cdot \\frac{4}{3}\\pi R^3$$ Hence, $$M = 8m$$ Moment of inertia of the smaller sphere about the given $Y$-axis: Using parallel axis theorem,$$I_s = \\frac{2}{5}mR^2 + mR^2 = \\frac{7}{5}mR^2$$Moment of inertia of the larger sphere about the same axis:$$I_L = \\frac{2}{5}(8m)(2R)^2 = \\frac{64}{5}mR^2$$Moment of inertia of the remaining portion:$$I_r = I_L - I_s = \\frac{64}{5}mR^2 - \\frac{7}{5}mR^2 = \\frac{57}{5}mR^2$$Therefore,$$\\frac{I_s}{I_r} = \\frac{\\frac{7}{5}mR^2}{\\frac{57}{5}mR^2} = \\frac{7}{57}$$ Hence, option (C) is correct.

7. Question 7 (NEET 2024)

   NEET 2024EasyRotational Dynamics

   A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?

   1. AOption A: Both the points P and Q move with equal speed.
   2. BOption B: Point P has zero speed.
   3. COption C: Point P moves slower than point Q.
   4. DOption D: Point P moves faster than point Q.

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   Correct answer: (D) Point P moves faster than point Q.

   Explanation

   In pure rolling motion, the velocity of the topmost point of the wheel is 2v, while the velocity of the bottommost point is zero. Therefore, point P moves faster than point Q. Hence, option (D) is correct.

8. Question 8 (NEET 2024)

   NEET 2024EasyMoment of Inertia

   The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400\,\text{g cm}^2$. The length of the $400\,\text{g}$ rod is nearly:

   1. AOption A: $20.7\,\text{cm}$
   2. BOption B: $72.0\,\text{cm}$
   3. COption C: $8.5\,\text{cm}$
   4. DOption D: $17.5\,\text{cm}$

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   Correct answer: (C) $8.5\,\text{cm}$

   Explanation

   For a thin rod about an axis through its centre and perpendicular to its length: Given: $$I=2400\,\text{g cm}^2,\quad M=400\,\text{g}$$ So, $$2400=\frac{1}{12}(400)L^2$$ $$L^2=72$$ $$L=\sqrt{72}\approx 8.5\,\text{cm}$$ Hence the correct answer is: **(3) $8.5\,\text{cm}$**

9. Question 9 (NEET 2022)

   NEET 2022EasyCentre of Mass

   Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the centre of mass of the system from the 10 kg mass is:

   1. AOption A: 5 m
   2. BOption B: 10/3 m
   3. COption C: 20/3 m
   4. DOption D: 10 m

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   Correct answer: (C) 20/3 m

   Explanation

   Position of centre of mass: x = (m₂L)/(m₁ + m₂) = (20 × 10)/(10 + 20) = 200/30 = 20/3 m Hence, option (C) is correct.

10. Question 10 (NEET 2021)

    NEET 2021HardEquilibrium of Rigid Bodies

    A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm mark and another unknown mass m is suspended from the rod at 160 cm mark. Find the value of m such that the rod is in equilibrium. (g = 10 m/s²)

    1. AOption A: 1/12 kg
    2. BOption B: 1/2 kg
    3. COption C: 1/3 kg
    4. DOption D: 1/6 kg

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    Correct answer: (A) 1/12 kg

    Explanation

    Taking moments about 40 cm mark: Torque due to 2 kg mass = 2×10×0.2 = 4 Nm. Torque due to rod's weight = 0.5×10×0.6 = 3 Nm. Torque due to mass m at 160 cm = m×10×1.2 = 12m Nm. Equilibrium: 4 = 3 + 12m ⇒ m = 1/12 kg.