Practice MCQs for JEE Physics System Of Particles and Rotational Motion
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Multiple Choice Questions
(a) \( 0.25\times10^{-3} \)
(b) \( 1.5\times10^{-3} \)
(c) \( 1.25\times10^{-3} \)
(d) \( 0.5\times10^{-3} \)
Show Answer & Explanation
Answer: (d) \( 0.5\times10^{-3} \)
Explanation:
1. Conservation of angular momentum: I₁(2) + (10⁻³)(5) = (I₁ + 10⁻³)(4).
2. 2I₁ + 5×10⁻³ = 4I₁ + 4×10⁻³, so I₁ = 0.5×10⁻³ kg m².
Teacher's Note:
1. Conserve Iω for the combined discs.
(a) t = 1 s
(b) t = 0.5 s
(c) t = 0.25 s
(d) t = 2 s
Show Answer & Explanation
Answer: (a) t = 1 s
Explanation:
1. ω = 6t² − 12t and α = 12t − 12.
2. τ = Iα = 0 at t = 1 s.
Teacher's Note:
1. Torque is proportional to angular acceleration.
(a) \( \frac12MR^2 \)
(b) \( \frac32MR^2 \)
(c) \( \frac14MR^2 \)
(d) \( \frac34MR^2 \)
Show Answer & Explanation
Answer: (d) \( \frac34MR^2 \)
Explanation:
1. Centre, perpendicular: ½MR². Edge, perpendicular: (3/2)MR². Diameter: ¼MR².
2. ¾MR² does not correspond to any of the standard axes (tangent in plane gives 5/4 MR²).
Teacher's Note:
1. List the standard axes first.
(a) \( \left(\frac43,1\right) \)
(b) \( \left(\frac12,\frac23\right) \)
(c) \( \left(\frac12,\frac12\right) \)
(d) \( \left(1,\frac43\right) \)
Show Answer & Explanation
Answer: (d) \( \left(1,\frac43\right) \)
Explanation:
1. x = (0 + 3 + 0)/3 = 1 and y = (0 + 0 + 4)/3 = 4/3.
2. So the CM is (1, 4/3).
Teacher's Note:
1. Average the coordinates for equal masses.
(a) 2
(b) zero
(c) 1
(d) −1
Show Answer & Explanation
Answer: (d) −1
Explanation:
1. Angular momentum is conserved if the torque r⃗ × F⃗ = 0, so F⃗ ∥ r⃗.
2. 2/α = −6/3 = −12/6 = −2, so α = −1.
Teacher's Note:
1. Parallel vectors have a zero cross product.
(a) \( \frac14mv_0^2 \)
(b) \( 2mv_0^2 \)
(c) \( \frac12mv_0^2 \)
(d) \( mv_0^2 \)
Show Answer & Explanation
Answer: (b) \( 2mv_0^2 \)
Explanation:
1. The tension is central, so angular momentum is conserved: mv₀R₀ = mv(R₀/2), giving v = 2v₀.
2. K = ½m(2v₀)² = 2mv₀².
Teacher's Note:
1. Conserve angular momentum.
| Column I | Column II |
|---|---|
| (A) Angular momentum | (1) r⃗ × (m v⃗) |
| (B) Impulse | (2) ½ Iω² |
| (C) Torque | (3) r⃗ × F⃗ |
| (D) Rotational energy | (4) m Δv⃗ |
(a) A-1, B-3, C-4, D-2
(b) A-1, B-4, C-3, D-2
(c) A-2, B-4, C-3, D-1
(d) A-4, B-1, C-3, D-2
Show Answer & Explanation
Answer: (b) A-1, B-4, C-3, D-2
Explanation:
1. Angular momentum is r × mv (A-1).
2. Impulse is the change in momentum mΔv (B-4).
3. Torque is r × F (C-3).
4. Rotational energy is ½Iω² (D-2).
Teacher's Note:
1. Each quantity has one defining expression.
(a) \( \frac L2 \)
(b) L
(c) \( \frac L3 \)
(d) \( \frac L4 \)
Show Answer & Explanation
Answer: (a) \( \frac L2 \)
Explanation:
1. L = mr²ω = mr²(2πf).
2. Halving f halves L.
Teacher's Note:
1. L is proportional to frequency at fixed radius.
(a) \( I_{xy}=I_{x'y'} \)
(b) \( I_{xy}=\frac12I_{x'y'} \)
(c) \( I_{x'y'}=\frac34I_{xy} \)
(d) \( I_{xy}=\frac34I_{x'y'} \)
Show Answer & Explanation
Answer: (d) \( I_{xy}=\frac34I_{x'y'} \)
Explanation:
1. Tangent in plane: MR²/2 + MR² = 3MR²/2.
2. Tangent perpendicular to plane: MR² + MR² = 2MR².
3. So I_xy / I_x′y′ = 3/4.
Teacher's Note:
1. Apply the parallel axis theorem to both tangents.
| Column I | Column II |
|---|---|
| (A) Moment of inertia of a solid uniform sphere about the diameter | (1) MR² |
| (B) Moment of inertia of a thin uniform spherical shell about the tangent | (2) ½MR² |
| (C) Moment of inertia of a uniform disc through the centre and perpendicular to its plane | (3) (5/3)MR² |
| (D) Moment of inertia of a disc about a tangent in the plane of the disc | (4) (2/5)MR² |
| (5) (5/4)MR² |
(a) A-4, B-2, C-3, D-5
(b) A-4, B-3, C-5, D-2
(c) A-4, B-5, C-2, D-3
(d) A-4, B-3, C-2, D-5
Show Answer & Explanation
Answer: (d) A-4, B-3, C-2, D-5
Explanation:
1. Solid sphere about a diameter: (2/5)MR² (A-4).
2. Shell about the tangent: (2/3)MR² + MR² = (5/3)MR² (B-3).
3. Disc about the central perpendicular axis: ½MR² (C-2).
4. Disc about a tangent in its plane: ¼MR² + MR² = (5/4)MR² (D-5).
Teacher's Note:
1. Use the parallel axis theorem for tangent axes.
(a) \( I_0+\frac{ML^2}{2} \)
(b) \( I_0+\frac{ML^2}{4} \)
(c) \( I_0+2ML^2 \)
(d) \( I_0+ML^2 \)
Show Answer & Explanation
Answer: (b) \( I_0+\frac{ML^2}{4} \)
Explanation:
1. The end is at distance L/2 from the centre.
2. Parallel axis theorem: I = I₀ + M(L/2)² = I₀ + ML²/4.
Teacher's Note:
1. Shift by half the length.
Assertion (A): If the polar ice melts, days will be longer.
Reason (R): Moment of inertia decreases and thus angular velocity increases.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Melting ice spreads water toward the equator, increasing the moment of inertia, so ω decreases and days get longer.
2. The reason says the moment of inertia decreases, which is false.
Teacher's Note:
1. Mass moving outward raises I.
(a) \( \frac23Mr^2 \)
(b) \( \frac25Mr^2 \)
(c) \( \frac13Mr^2 \)
(d) \( Mr^2 \)
Show Answer & Explanation
Answer: (d) \( Mr^2 \)
Explanation:
1. All the mass of a thin hollow cylinder lies at distance r from the axis.
2. So I = Mr².
Teacher's Note:
1. Mass at one radius gives MR².
(a) increases his linear velocity
(b) decreases his angular velocity
(c) increases his moment of inertia
(d) decreases his moment of inertia
Show Answer & Explanation
Answer: (d) decreases his moment of inertia
Explanation:
1. Bending brings his mass closer to the axis.
2. So the moment of inertia decreases and ω increases.
Teacher's Note:
1. Tuck to spin faster.
(a) 2 : 1
(b) 1 : 2
(c) \( \sqrt2:1 \)
(d) \( 1:\sqrt2 \)
Show Answer & Explanation
Answer: (d) \( 1:\sqrt2 \)
Explanation:
1. K = L²/2I, so for equal K, L ∝ √I.
2. L₁ : L₂ = √I : √(2I) = 1 : √2.
Teacher's Note:
1. K = L²/2I.
(a) 1 : √2
(b) 1 : 3
(c) 2 : 1
(d) \( \sqrt5:\sqrt6 \)
Show Answer & Explanation
Answer: (d) \( \sqrt5:\sqrt6 \)
Explanation:
1. Disc: I = ¼MR² + MR² = (5/4)MR², so k² = 5R²/4.
2. Ring: I = ½MR² + MR² = (3/2)MR², so k² = 3R²/2.
3. k₁ : k₂ = √(5/4) : √(3/2) = √5 : √6.
Teacher's Note:
1. Compute each tangent I with the parallel axis theorem.
(a) \( \frac{(I_1+I_2)\omega}{I_1} \)
(b) \( \frac{I_2\omega}{I_1+I_2} \)
(c) \( \omega \)
(d) \( \frac{I_1\omega}{I_1+I_2} \)
Show Answer & Explanation
Answer: (d) \( \frac{I_1\omega}{I_1+I_2} \)
Explanation:
1. Angular momentum is conserved: I₁ω = (I₁ + I₂)ω′.
2. ω′ = I₁ω/(I₁ + I₂).
Teacher's Note:
1. Conserve angular momentum.
Assertion (A): A particle is moving in a straight line with uniform velocity. Its angular momentum is always zero.
Reason (R): The momentum is not zero when a particle moves with uniform velocity.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Angular momentum about a point not on the line is non-zero (L = mvd), so the assertion is false.
2. The reason is true.
Teacher's Note:
1. L depends on the chosen point.
(a) weight
(b) angular momentum
(c) moment of inertia
(d) torque
Show Answer & Explanation
Answer: (d) torque
Explanation:
1. Force causes linear acceleration and torque causes angular acceleration.
2. Torque is the rotational analogue of force.
Teacher's Note:
1. Torque is to rotation what force is to translation.
Assertion (A): If no external force acts on a system of particles, then the centre of mass will not move in any direction.
Reason (R): If the net external force is zero, then the linear momentum of the system remains constant.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. With zero external force the CM moves with constant velocity, which may be non-zero, so the assertion is false.
2. The reason is true.
Teacher's Note:
1. Constant velocity, not necessarily at rest.
(a) a ring about its axis, perpendicular to the plane of the ring
(b) a disc about its axis, perpendicular to the plane of the disc
(c) a solid sphere about one of its diameters
(d) a spherical shell about one of its diameters
Show Answer & Explanation
Answer: (a) a ring about its axis, perpendicular to the plane of the ring
Explanation:
1. Ring: MR². Disc: ½MR². Solid sphere: (2/5)MR². Spherical shell: (2/3)MR².
2. The ring has the highest value.
Teacher's Note:
1. Mass farther from the axis gives larger I.
(a) \( x=\frac{m_1}{m_2}L \)
(b) \( x=\frac{m_2}{m_1}L \)
(c) \( x=\frac{m_2L}{m_1+m_2} \)
(d) \( x=\frac{m_1L}{m_1+m_2} \)
Show Answer & Explanation
Answer: (c) \( x=\frac{m_2L}{m_1+m_2} \)
Explanation:
1. Work = ½Iω₀², minimum when I is minimum, i.e. the axis passes through the centre of mass.
2. The CM is at x = m₂L/(m₁ + m₂) from m₁.
Teacher's Note:
1. I is least about the centre of mass.
Assertion (A): It is harder to open and shut a door if we apply the force near the hinge.
Reason (R): Torque is maximum at the hinge of the door.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Torque = F × distance from hinge, so near the hinge it is smaller and the door is harder to move; the assertion is true.
2. The torque is zero at the hinge, so the reason is false.
Teacher's Note:
1. Larger lever arm, larger torque.
(a) \( \frac{4MR^2}{9\sqrt3\pi} \)
(b) \( \frac{4MR^2}{3\sqrt3\pi} \)
(c) \( \frac{MR^2}{32\sqrt2\pi} \)
(d) \( \frac{MR^2}{16\sqrt2\pi} \)
Show Answer & Explanation
Answer: (a) \( \frac{4MR^2}{9\sqrt3\pi} \)
Explanation:
1. The cube's diagonal equals the diameter: a√3 = 2R, so a = 2R/√3.
2. Mass of the cube m = M a³/(4πR³/3) = 2M/(√3π).
3. I = ma²/6 = (2M/√3π)(4R²/3)/6 = 4MR²/(9√3π).
Teacher's Note:
1. Find the cube's side from the sphere's diameter.
(a) 22.7 J
(b) 19.6 J
(c) 10.2 J
(d) 9.8 J
Show Answer & Explanation
Answer: (b) 19.6 J
Explanation:
1. Total kinetic energy = mgh = 2 × 9.8 × 3 = 58.8 J.
2. For a solid cylinder, rotational : translational = ½ : 1 (¼mv² : ½mv²), so the rotational share is 1/3.
3. K_rot = 58.8/3 = 19.6 J.
Teacher's Note:
1. For a cylinder, one third of the energy is rotational.
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Practice MCQs for JEE Physics System Of Particles and Rotational Motion
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