System of Particles and Rotational Motion NEET PYQs (1988–2026)

Solved System of Particles and Rotational Motion PYQs

3 recent questions with answers. The full set of 117 is free to practise.

1. NEET 2025MediumImpulse & Conservation of Angular Momentum
   Q1

   The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

   • A115 days
   • B108 days
   • C100 days
   • D105 days

   Show answer & solution

   Correct answer: (B)

   When considering the Sun as a solid sphere, the formula for the moment of inertia is given by:
   
   $$I=\frac{2}{5}m{R}^2$$
   
   Here, $m$ is the mass and $R$ is the radius of the Sun.
   To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum. The principle states that angular momentum before and after the expansion must be equal:
   
   $$l^{\prime }{\omega }^{\prime }=l\omega$$
   
   Substituting the expression for angular momentum $(I\times \omega )$ for both initial and expanded states, we get:
   
   $$\frac{2}{5}m(2R{)}^2\times \frac{2\pi }{T^{\prime }}=\frac{2}{5}m{R}^2\times \frac{2\pi }{T}$$
   
   This simplifies to:
   
   $$\Rightarrow 4m{R}^2\times \frac{2\pi }{T^{\prime }}=m{R}^2\times \frac{2\pi }{T}$$
   
   Cancelling out common terms, we find:
   
   $$\Rightarrow 4\times \frac{1}{T^{\prime }}=\frac{1}{T}$$
   
   Thus:
   
   $$\Rightarrow T^{\prime }=4T=4\times 27=108\text{ days }$$
   
   Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.
2. NEET 2024EasyCentre of Mass
   Q2

   Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity $v_1$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1:v_2$ is
   [NEET 2024]

   • A$1:2$
   • B$2:1$
   • C$4:1$
   • D$1:4$

   Show answer & solution

   Correct answer: (B)

   Before collision $\Rightarrow (A)\to v_1(B)$ rest
   It undergoes completely inelastic collision
   Using conservation of linear momentum
   Initial momentum = Final momentum
   $$\Rightarrow m{v}_1=m{v}_2+m{v}_2\Rightarrow m{v}_1=2m{v}_2\Rightarrow \frac{v_1}{v_2}=\frac{2}{1}$$
3. NEET 2024EasyMoment of Inertia
   Q3

   The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 gcm 2 . The length of the $400g$ rod is nearly:
   [NEET 2024]

   • A8.5 cm
   • B17.5 cm
   • C20.7 cm
   • D72.0 cm

   Show answer & solution

   Correct answer: (A)

   Moment of inertia of rod $=I=\frac{m{l}^2}{12}$
   
   $$\begin{array}{cc}& \Rightarrow 2400=400\frac{l^2}{12}\\ & \Rightarrow 72=l^2\\ & \Rightarrow l=\sqrt{72}=8.48cm=8.5cm\end{array}$$

Topic-wise question count

Where NEET has asked from System of Particles and Rotational Motion, most frequent first.

• Moment of Inertia27
• Rotational Kinematics & Dynamics26
• Rolling Motion18
• Impulse & Conservation of Angular Momentum16
• Centre of Mass16
• Torque & Angular Momentum13
• Centre of Mass & Torque1

Year-wise System of Particles and Rotational Motion questions in NEET

Number of questions from this chapter in each year's paper(s).