Previous year questions: System of Particles and Rotational Motion

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Write the axis and the relevant moment of inertia at the top of every solution before substituting numbers.

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part of mass M/8 is converted into a sphere of radius r, and the larger part is converted into a disc of thickness t and radius 2R. If I1 is the moment of inertia of the small sphere about an axis through its centre and I2 that of the disc about its diameter, the ratio I2 over I1 is

Answer: The small sphere has one eighth of the volume, so its radius is half of R. Its moment of inertia is two fifths of M/8 times R squared over four, which is M R squared over eighty. The disc has mass seven M over eight and radius 2R, so about a diameter its moment is one quarter of that mass times four R squared, which is seven M R squared over eight. The ratio is seven eighths divided by one eightieth, which is 70.

Hinglish: Chhoti sphere ka volume 1/8 hai, isliye radius R/2. I1 = 2/5 x (M/8) x (R^2/4) = MR^2/80. Disc ka mass 7M/8 aur radius 2R, diameter ke about I2 = 1/4 x (7M/8) x 4R^2 = 7MR^2/8. Ratio (7/8)/(1/80) = 70.

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A body of mass 0.1 kg has position r = (10 t squared i + 5 t cubed j) m. At t = 1 s, which of the following statements are correct? A. The linear momentum is (2i + 1.5j) kg m s^-1. B. The force is (2i + 3j) N. C. The angular momentum about the origin is 15k J s. D. The torque about the origin is 20k N m.

Answer: Differentiate the position: velocity is 20t i + 15 t squared j, and at t = 1 it is 20i + 15j, so the momentum is 0.1 times that, 2i + 1.5j, and A is correct. Differentiating again gives acceleration 20i + 30t j, which at t = 1 is 20i + 30j, so the force is 0.1 times that, 2i + 3j, and B is correct. The angular momentum is r cross p at t = 1, with r = 10i + 5j and p = 2i + 1.5j, giving (10 times 1.5 minus 5 times 2)k, which is 5k, so C is wrong. The torque is r cross F, giving (10 times 3 minus 5 times 2)k, which is 20k, so D is correct. The correct statements are A, B and D.

Hinglish: v = 20t i + 15t^2 j, t = 1 par 20i + 15j, isliye p = 2i + 1.5j, A sahi. a = 20i + 30j, F = 2i + 3j, B sahi. L = r x p = (10x1.5 - 5x2)k = 5k, C galat. tau = r x F = (10x3 - 5x2)k = 20k, D sahi. Answer A, B aur D.

Q3 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A solid sphere of radius 4 cm and mass 5 kg rotates about an axis through its centre at 1200 rpm. It is brought to rest in 10 s by a constant torque. The torque applied and the number of rotations made before it stops are

Answer: The moment of inertia is two fifths of 5 times 0.04 squared, which is 0.0032 kg metre squared. The initial angular speed is 1200 times 2 pi over 60, which is 40 pi radian per second, and the angular acceleration is minus 4 pi radian per second squared. The torque is the moment times the angular acceleration, 0.0032 times 4 pi, which is 0.0128 pi newton metre. The angle turned is omega nought squared over twice the angular acceleration, which is 1600 pi squared over 8 pi, that is 200 pi radian, and dividing by 2 pi gives 100 rotations. The option pairing 0.0128 pi newton metre with 100 rotations is correct.

Hinglish: I = 2/5 x 5 x 0.04^2 = 0.0032. omega0 = 40 pi rad/s, alpha = 4 pi. Torque tau = 0.0032 x 4 pi = 0.0128 pi N m. Angle theta = omega0^2/(2 alpha) = 1600 pi^2/(8 pi) = 200 pi rad, matlab 100 rotations. Option 0.0128 pi N m aur 100 rotations sahi hai.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it turns through an angle theta 1, and in the next 2 s it turns through theta 2. The ratio theta 2 over theta 1 is

Answer: With constant angular acceleration from rest, the angle in time t is half alpha t squared. For the first 2 s it is 2 alpha. For the first 4 s it is 8 alpha, so the angle in the second interval of 2 s is 8 alpha minus 2 alpha, which is 6 alpha. The ratio is 6 over 2, that is 3. The result is independent of the value of alpha, which is why it can be answered without any numbers.

Hinglish: Rest se constant alpha ke liye theta = 1/2 alpha t^2. Pehle 2 s mein 2 alpha, pehle 4 s mein 8 alpha, isliye agle 2 s mein 6 alpha. Ratio 6/2 = 3, alpha ke value par depend nahi karta.

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A solid cylinder of radius R slips on a rough horizontal plane. At t = 0 it has translational velocity v0 equal to 49 m per second and rotational velocity v0 over 4R. The time taken to start rolling is (coefficient of kinetic friction 0.25 and g = 9.8 m per second squared)

Answer: The friction force is mu m g, so the centre of mass decelerates at mu g, which is 2.45 m per second squared. The angular acceleration from the friction torque is mu m g R over half m R squared, which is 2 mu g over R. The velocity at time t is v0 minus mu g t, and the angular velocity is v0 over 4R plus 2 mu g t over R. Rolling begins when v equals omega R, so v0 minus mu g t equals v0 over 4 plus 2 mu g t. This gives three quarters of v0 equal to 3 mu g t, so t is v0 over 4 mu g, which is 49 over 9.8, that is 5 s.

Hinglish: Friction se a = mu g, aur torque se alpha = 2 mu g/R. Rolling tab shuru hota hai jab v = omega R. Equation se 3/4 v0 = 3 mu g t, isliye t = v0/(4 mu g) = 49/9.8 = 5 s.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A solid sphere A of mass 5m and a spherical shell B of mass m, both of radius R, rest on a rough surface. The same tangential force is applied at the highest point of each, and both roll without slipping. If their accelerations are aA and aB, the ratio aA over aB is

Answer: For a body rolling under a force applied at the top, the acceleration is the force divided by mass plus moment of inertia over radius squared. For the solid sphere the moment is two fifths of 5m times R squared, so the denominator is 5m plus 2m, which is 7m. For the shell the moment is two thirds of m R squared, so the denominator is m plus two thirds m, which is 5m over 3. The accelerations are F over 7m and 3F over 5m, and their ratio is 5 over 21.

Hinglish: Rolling body par top par force lagne se a = F/(m + I/R^2). Solid sphere: I = 2/5 x 5m R^2, denominator 7m. Shell: I = 2/3 mR^2, denominator 5m/3. aA = F/7m, aB = 3F/5m, ratio 5/21.

Q7 (NCERT Exemplar, 1 mark)

For which of the following does the centre of mass lie outside the body? (a) A pencil (b) A shotput (c) A dice (d) A bangle

Answer: The centre of mass of a body lies at the mass weighted average of its positions, and for a symmetric ring the average is the geometric centre, where there is no material. A bangle is such a ring, so its centre of mass lies in the empty space at its centre. A pencil, a shotput and a dice all have their mass filling the region around the centre, so their centre of mass lies inside the body. The answer is (d).

Hinglish: Bangle ek ring hai, iska CM centre par hota hai jahan material nahi hota, isliye body ke bahar padta hai. Pencil, shotput aur dice sab mein CM andar hota hai. Answer (d).

Q8 (NCERT Exemplar, 1 mark)

When a disc rotates with uniform angular velocity, which of the following is not true? (a) The sense of rotation remains the same (b) The orientation of the axis remains the same (c) The speed of rotation is non-zero and remains the same (d) The angular acceleration is non-zero and remains the same

Answer: Uniform angular velocity means the angular velocity vector is constant, so its direction, which is the axis orientation, stays fixed and the sense of rotation stays the same. The angular acceleration is the rate of change of angular velocity, so it is zero. Option (d) says it is non-zero and constant, which contradicts the uniform angular velocity. The answer is (d).

Hinglish: Uniform angular velocity ka matlab omega constant hai, isliye axis aur sense same rehte hain aur angular acceleration zero hota hai. Option (d) kehta hai angular acceleration non-zero hai, jo galat hai. Answer (d).

Q9 (NCERT Exemplar, 1 mark)

A uniform square plate has a small piece of irregular shape removed and glued to the centre of the plate. The moment of inertia about the axis perpendicular to the plate through its centre is then (a) increased (b) decreased (c) the same (d) changed in an unpredictable manner

Answer: The piece that is removed had some mass at a distance from the axis, so its contribution to the moment of inertia was positive. Glued to the centre, the same mass is at zero distance from the axis and contributes nothing. The total moment of inertia therefore decreases. The answer is (b).

Hinglish: Jo piece hataya gaya woh axis se kuch door tha, isliye uska contribution positive tha. Centre par chipkane se wahi mass axis par aa jaata hai aur contribution zero ho jaata hai, isliye total I kam ho jaata hai. Answer (b).

Q10 (NCERT Exemplar, 1 mark)

The density of a non-uniform rod of length 1 m is given by rho = a(1 + b x squared), where a and b are constants and x lies between 0 and 1. The centre of mass of the rod is at

Answer: The centre of mass of a rod with a variable linear density is the integral of x times the density divided by the integral of the density. With density proportional to 1 plus b x squared, the numerator is the integral of x plus b x cubed, which gives one half plus b over four, and the denominator is 1 plus b over three. Dividing gives three times (2 plus b) over four times (3 plus b), which is the first option.

Hinglish: CM integral x rho dx / integral rho dx hota hai. Numerator 1/2 + b/4, denominator 1 + b/3, divide karne par 3(2+b)/4(3+b) milta hai, jo pehla option hai.