JEE Physics System Of Particles and Rotational Motion MCQs Set 28

Answer: (a) Graph (a)

Explanation:
1. J = Iω with constant I for a given body, so J is directly proportional to ω.
2. The graph is a straight line through the origin with positive slope I: graph (a).

Teacher's Note:
1. Write J in terms of ω.
2. A direct proportion is a straight line through the origin.

Answer: (a) the centre of the circle

Explanation:
1. About the centre, r is constant and perpendicular to v, so L = mvr is constant in magnitude and direction.
2. About any other point, the distance and direction of the position vector keep changing, so L changes in magnitude or direction.

Teacher's Note:
1. Check what happens to r × p during the motion.
2. Only the centre gives a constant r ⟂ v.

Answer: (a) increase

Explanation:
1. The water spreads away from the pole towards the equator, away from the axis of rotation.
2. Mass moves to larger r, so I = Σmr² increases.

Teacher's Note:
1. Consider where the melted water goes.
2. Mass farther from the axis means larger I.

Answer: (b) decrease

Explanation:
1. The atmosphere also rotates with the earth and contributes to the total I; without it I decreases.
2. No external torque, so L = Iω is constant and ω increases.
3. T = 2π/ω decreases.

Teacher's Note:
1. Use conservation of angular momentum.
2. Smaller I gives larger ω.

Answer: (a) increase

Explanation:
1. Extra mass at the distance R from the axis adds mR² to the moment of inertia.

Teacher's Note:
1. I is the sum of mr² of all parts.
2. Added mass adds to I.

Answer: (b) decrease

Explanation:
1. No external torque, so L = Iω is constant.
2. I increases, therefore ω decreases.

Teacher's Note:
1. Angular momentum is conserved.
2. I goes up, so ω goes down.

Answer: (b) the applied net torque on the body is zero

Explanation:
1. τ = dL/dt, so L is constant only when the net external torque is zero.
2. Zero net force does not guarantee zero torque (for example a couple).

Teacher's Note:
1. Use τ = dL/dt.
2. A constant torque still changes L.

Answer: (c) increase

Explanation:
1. No external torque, so the total angular momentum is conserved.
2. Moving inwards decreases the moment of inertia of the system, so ω increases.

Teacher's Note:
1. Conserve angular momentum.
2. Smaller r gives smaller I.

Answer: (c) its angular momentum is constant but linear momentum is variable

Explanation:
1. The direction of velocity keeps changing, so linear momentum p = mv is not constant.
2. The tension acts along the string through the centre, so its torque about the centre is zero; L = mvr is constant.

Teacher's Note:
1. Linear momentum is a vector and its direction changes.
2. Central force gives no torque about the centre.

Answer: (d) 12ω

Explanation:
1. Angular momentum is conserved: mr²(3ω) = m(r/2)²ω′.
2. ω′ = 3ω × 4 = 12ω.

Teacher's Note:
1. The pull acts through the centre, so L is conserved.
2. I = mr².

Answer: (c) 6 hours

Explanation:
1. I ∝ R², so I becomes one-fourth.
2. Iω is constant, so ω becomes 4 times and T = 24 h/4 = 6 h.

Teacher's Note:
1. Use conservation of angular momentum.
2. I is proportional to R².

Answer: (b) \( \dfrac{M\omega}{M + 8m} \)

Explanation:
1. Initial I = ½Mr². After placing the masses at the rim: I′ = ½Mr² + 4mr².
2. ½Mr²ω = (½Mr² + 4mr²)ω′, so ω′ = Mω/(M + 8m).

Teacher's Note:
1. Add 4mr² to the moment of inertia.
2. Conserve angular momentum.

Answer: (c) \( \dfrac{4}{5} \)ω

Explanation:
1. I₁ = ½MR² and the added disc has I₂ = ½(M/4)R² = MR²/8.
2. ½MR²ω = (½MR² + ⅛MR²)ω′ = (5/8)MR²ω′.
3. ω′ = (1/2)/(5/8) ω = 4ω/5.

Teacher's Note:
1. Conserve angular momentum.
2. Moments of inertia add for coaxial discs.

Answer: (a) \( \dfrac{\pi}{15} \)

Explanation:
1. Initial ω = 2π/20 = π/10 rad/s.
2. New I = 200 + 25 × 2² = 300 kg m².
3. 200 × π/10 = 300 ω′, so ω′ = π/15 rad/s.

Teacher's Note:
1. Add mr² for the person.
2. Conserve angular momentum.

Answer: (a) 1000 rpm

Explanation:
1. The chain has the same linear speed on both wheels: v = ωR.
2. ω₁R₁ = ω₂R₂, so ω₂ = 200 × 0.5/0.1 = 1000 rpm.

Teacher's Note:
1. Same chain speed on both rims.
2. Rotation rate is inversely proportional to radius.

Answer: (d) K increases but L remains constant

Explanation:
1. No external torque, so L remains constant.
2. I decreases and ω increases as Iω = constant.
3. K = L²/(2I) increases when I decreases (the person does work in pulling the arms in).

Teacher's Note:
1. L is conserved.
2. Use K = L²/2I.

Answer: (d) All of the above

Explanation:
1. Pure rolling: ω = v/R = 4/2 = 2 rad/s, and the contact point A is at rest, v_A = 0.
2. B and D: the speed is √(v² + (ωR)²) = 4√2 m/s for both.
3. C: v_C = v + ωR = 8 m/s = 2v₀ (v₀ = speed of the centre).
4. All the three statements are correct.

Teacher's Note:
1. Velocity of any point = velocity of centre plus ω × r.
2. Use ωR = v.

Answer: (a) \( 2v\sin\dfrac{\theta}{2} \)

Explanation:
1. Rolling without slipping is, at every instant, pure rotation about the contact point with ω = v/R.
2. Distance of P from the contact point = 2R sin(θ/2).
3. Speed = ω × 2R sin(θ/2) = 2v sin(θ/2).

Teacher's Note:
1. Pure rolling is rotation about the contact point.
2. Chord length = 2R sin(θ/2).

Answer: (c) \( 3\sqrt{2}\ \text{m/s} \)

Explanation:
1. For a point at the end of the horizontal diameter: velocity of centre v = 3 m/s (horizontal) and rotational velocity ωR = v = 3 m/s (vertical).
2. Resultant = √(3² + 3²) = 3√2 m/s.

Teacher's Note:
1. Add translational and rotational velocities as vectors.
2. They are perpendicular at those points.

Answer: (c) 3 × 10⁻³ J

Explanation:
1. K = ½mv² + ½Iω² = ½mv² + ¼mv² = ¾mv².
2. K = ¾ × 0.4 × (0.1)² = 3 × 10⁻³ J.

Teacher's Note:
1. For a rolling disc K = (3/4)mv².
2. Convert 10 cm/s to 0.1 m/s.

Answer: (d) 3E/2

Explanation:
1. Translational energy = ½mv² = E.
2. Rotational energy = ½(½mR²)(v/R)² = ¼mv² = E/2.
3. Total = E + E/2 = 3E/2.

Teacher's Note:
1. Find rotational energy in terms of translational energy.
2. For a disc K_rot = ½ K_trans.