Physics · System of particles and rigid body · chapter 6 of 14

Class 11 Physics Chapter 6: Systems of Particles and Rotational Motion MCQs with answers

10 multiple-choice questions on Systems of Particles and Rotational Motion, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
A rigid disc rotates about a fixed axis through its centre. Point X is 0.1 m from the axis and point Y is 0.3 m from it. The ratio vY/vX of their linear speeds is
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Answer: (c) 3

Every particle has the same ω, and v = ωr, so vY/vX = 0.3/0.1 = 3. Taking 1 forgets that v grows with r; 9 squares the ratio.

Question 2
Two small balls P and Q are joined by a light rod 60 cm long. The centre of mass of the pair is 15 cm from P. The ratio of the mass of P to the mass of Q is
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Answer: (b) 3

The centre of mass is nearer the heavier ball, and mP × 15 = mQ × 45, so mP/mQ = 45/15 = 3. The answer 1/3 puts the centre of mass nearer the lighter ball.

Question 3
A uniform wire 2.0 m long is bent at its midpoint so that its two halves are at right angles. The distance of the centre of mass of the bent wire from the bend is about
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Answer: (a) 0.35 m

Take the bend as origin and the halves along x and y. Each 1.0 m half has its centre of mass at its midpoint: (0.5, 0) and (0, 0.5) m, with equal masses. So the centre of mass is at (0.25, 0.25) m, a distance 0.25√2 ≈ 0.35 m from the bend; it lies outside the wire.

Question 4
Masses of 1 kg, 2 kg and 1 kg are at (2, 0) m, (0, 1) m and (−1, −1) m in the x-y plane. A fourth mass of 2 kg is to be placed so that the centre of mass of all four is at the origin. Its x-coordinate must be
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Answer: (d) −0.5 m

For X = 0 we need Σmᵢxᵢ = 0: 1(2) + 2(0) + 1(−1) + 2x = 0, so x = −0.5 m. Dividing by the total mass 6 kg or 4 kg instead gives the wrong distractors.

Question 5
The centre of gravity of a body coincides with its centre of mass when
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Answer: (b) the acceleration due to gravity is the same at every point of the body

With g the same for all particles it comes out of the sum Σrᵢ × mᵢg = 0, giving Σmᵢrᵢ = 0 about the CG, which is the condition for the centre of mass. Symmetry, rest or rigidity is not what makes the two points coincide.

Question 6
A shell fired from a gun on level ground would have landed 240 m away. At the highest point it explodes into two fragments of equal mass, which reach the ground at the same instant. One fragment lands 180 m from the gun, on the line of fire. Neglecting air resistance, the other lands at a distance from the gun of
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Answer: (a) 300 m

The explosion forces are internal, so the centre of mass follows the original parabola and lands at 240 m. Both fragments land together, so 240 = (180 + x)/2 and x = 300 m.

Question 7
A girl stands at rest on smooth, frictionless ice and throws a ball horizontally forward. Just after the throw, the horizontal velocity of the centre of mass of the girl and ball together is
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Answer: (c) zero

There is no external horizontal force (the ice is frictionless), so the horizontal momentum of the system stays zero; the girl moves back as the ball moves forward and their centre of mass has no horizontal velocity.

Question 8
A nucleus at rest breaks into two fragments whose masses are in the ratio 2 : 3. External forces are negligible. The ratio of the kinetic energy of the lighter fragment to that of the heavier fragment is
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Answer: (b) 3/2

Total momentum stays zero, so the fragments have equal and opposite momenta p. K = p²/2m, so K₁/K₂ = m₂/m₁ = 3/2. The answer 1 assumes equal energies; 9/4 squares the mass ratio.

Question 9
Vectors a and b have magnitudes 4 and 5 units and the angle between them is 30°. The magnitude of a × b is
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Answer: (a) 10 units

|a × b| = ab sin θ = 4 × 5 × sin 30° = 10 units. 17.3 uses cos 30°, which belongs to the scalar product; 20 drops the sine.

Question 10
For the unit vectors of a right-handed system, k̂ × î equals
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Answer: (d) ĵ

In cyclic order î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ. −ĵ is î × k̂, the reversed order; the null vector is a × a.

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