JEE Physics System Of Particles and Rotational Motion MCQs Set 65

Multiple Choice Questions (MCQs) for JEE Physics: System Of Particles and Rotational Motion

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Multiple Choice Questions

Question 1:
Assertion (A): The moment of inertia of a particle is the same whatever the axis of rotation.
Reason (R): Moment of inertia depends on the mass and distance of the particles.
(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. I = mr² depends on the axis through r, so the assertion is false.
2. The reason is true.

Teacher's Note:
1. I depends on the axis.

Question 2: Choose the wrong statement from the following.
(a) The centre of mass of a uniform circular ring is at its geometric centre
(b) Moment of inertia is a scalar quantity
(c) Radius of gyration is a vector quantity
(d) For the same mass and radius, the moment of inertia of a ring is twice that of a uniform disc
Show Answer & Explanation

Answer: (c) Radius of gyration is a vector quantity

Explanation:
1. The radius of gyration is a length and is a scalar, so (c) is wrong.
2. Ring MR² is twice disc ½MR², so (d) is right.

Teacher's Note:
1. k = √(I/M) is a scalar.

Question 3: Which of the following is incorrect?
(a) \( \vec v=\vec\omega\times\vec r \)
(b) \( \vec\tau=\vec F\times\vec r \)
(c) \( \vec L=\vec r\times\vec p \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \vec\tau=\vec F\times\vec r \)

Explanation:
1. The correct relation is τ⃗ = r⃗ × F⃗.
2. F⃗ × r⃗ has the opposite sign.

Teacher's Note:
1. The order of the cross product matters.

Question 4: A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad/s, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is
(a) \( 14.4\ \text{kg m}^2\text{s}^{-1} \)
(b) \( 8.64\ \text{kg m}^2\text{s}^{-1} \)
(c) \( 20.16\ \text{kg m}^2\text{s}^{-1} \)
(d) \( 11.52\ \text{kg m}^2\text{s}^{-1} \)
Show Answer & Explanation

Answer: (a) \( 14.4\ \text{kg m}^2\text{s}^{-1} \)

Explanation:
1. Speed v = ωr = 12 × 0.6 = 7.2 m/s.
2. The position vector from the ground point has magnitude √(0.6² + 0.8²) = 1 m and is perpendicular to v⃗ (the velocity is tangential, the position has vertical and radial parts).
3. L = m v × 1 = 2 × 7.2 × 1 = 14.4 kg m²/s.

Teacher's Note:
1. Take the distance from the chosen point, not the circle's radius.

Question 5:
Assertion (A): When no external torque acts on a body, its angular velocity remains constant as long as the moment of inertia is constant.
Reason (R): Torque τ = dL/dt = 0, so L = Iω is 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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With τ = 0, L = Iω is constant.
2. If I is constant, then ω is constant, so the reason explains the assertion.

Teacher's Note:
1. L = Iω.

Question 6: Four point masses, each of value m, are placed at the corners of a square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
(a) \( 2ml^2 \)
(b) \( \sqrt3\,ml^2 \)
(c) \( 3ml^2 \)
(d) \( ml^2 \)
Show Answer & Explanation

Answer: (c) \( 3ml^2 \)

Explanation:
1. The distance of B and D from the axis is l/√2 each.
2. A is on the axis and C is at distance √2 l.
3. I = m(l²/2) + m(l²/2) + m(2l²) = 3ml².

Teacher's Note:
1. Find each perpendicular distance to the axis.

Question 7: A planet is moving around the sun in an elliptical orbit. Its speed is
(a) the same at all points of the orbit
(b) maximum when it is nearest to the sun
(c) maximum when it is farthest from the sun
(d) cannot say
Show Answer & Explanation

Answer: (b) maximum when it is nearest to the sun

Explanation:
1. Angular momentum mvr is conserved.
2. The speed is greatest where r is least, at the nearest point.

Teacher's Note:
1. Kepler's second law.

Question 8: Consider a system of two particles having masses m₁ and m₂. If the particle of mass m₁ is pushed towards the centre of mass of the particles through a distance d, by what distance would the particle of mass m₂ move so as to keep the centre of mass of the particles at the original position?
(a) \( \frac{m_2}{m_1}d \)
(b) \( \frac{m_1}{m_1+m_2}d \)
(c) \( \frac{m_1}{m_2}d \)
(d) d
Show Answer & Explanation

Answer: (c) \( \frac{m_1}{m_2}d \)

Explanation:
1. The CM stays fixed: m₁d = m₂x.
2. So x = (m₁/m₂)d, moving toward the CM as well.

Teacher's Note:
1. Mass times distance must balance.

Question 9:
Assertion (A): The earth is slowing down and as a result the moon is coming nearer to it.
Reason (R): The angular momentum of the earth-moon system is conserved.
(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. As the earth slows down, angular momentum is transferred to the moon, which moves farther away, so the assertion is false.
2. Conservation of angular momentum of the system is true.

Teacher's Note:
1. The moon is receding.

Question 10: The moment of inertia of a circular disc of mass M and radius R about an axis passing through the centre of mass is I₀. The moment of inertia of another circular disc of the same mass and thickness but half the density about the same axis is
(a) \( \frac{I_0}{8} \)
(b) \( \frac{I_0}{4} \)
(c) \( 8I_0 \)
(d) \( 2I_0 \)
Show Answer & Explanation

Answer: (d) \( 2I_0 \)

Explanation:
1. Same thickness and half density with the same mass needs twice the area, so R′² = 2R².
2. I′ = ½MR′² = ½M(2R²) = 2I₀.

Teacher's Note:
1. Half the density at equal mass doubles the area.

Question 11: Which of the following is an invalid equation? (τ, ω, L and α have their usual meanings)
(a) \( \vec\tau=I\vec\omega \)
(b) \( \vec L=I\vec\omega \)
(c) \( \vec\tau=I\vec\alpha \)
(d) all of these are valid
Show Answer & Explanation

Answer: (a) \( \vec\tau=I\vec\omega \)

Explanation:
1. The correct relations are L = Iω and τ = Iα.
2. τ = Iω is dimensionally wrong.

Teacher's Note:
1. Torque goes with angular acceleration.

Question 12: A wheel having moment of inertia 2 kg m² about its vertical axis rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be
(a) \( \frac{\pi}{18}\ \text{N m} \)
(b) \( \frac{2\pi}{15}\ \text{N m} \)
(c) \( \frac{\pi}{12}\ \text{N m} \)
(d) \( \frac{\pi}{15}\ \text{N m} \)
Show Answer & Explanation

Answer: (d) \( \frac{\pi}{15}\ \text{N m} \)

Explanation:
1. ω = 2π rad/s and it must stop in 60 s, so α = 2π/60 = π/30 rad/s².
2. τ = Iα = 2 × π/30 = π/15 N m.

Teacher's Note:
1. τ = Iα.

Question 13: According to the principle of conservation of angular momentum, if the moment of inertia of a rotating body decreases, then its angular velocity
(a) decreases
(b) increases
(c) remains constant
(d) becomes zero
Show Answer & Explanation

Answer: (b) increases

Explanation:
1. Iω is constant when no external torque acts.
2. If I decreases, ω increases.

Teacher's Note:
1. L = Iω stays fixed.

Question 14:
Assertion (A): The centre of mass of a two-particle system lies on the line joining the two particles, being closer to the heavier particle.
Reason (R): The product of the mass of a particle and its distance from the centre of mass is numerically equal to the product of the mass of the other particle and its distance from the centre of mass.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. m₁r₁ = m₂r₂ gives r₁/r₂ = m₂/m₁.
2. So the heavier particle is closer to the CM, and the reason explains the assertion.

Teacher's Note:
1. m₁r₁ = m₂r₂.

Question 15: The correct relation between moment of inertia I, radius of gyration k and mass M of the body is
(a) \( k=I^2M \)
(b) \( k=IM^2 \)
(c) \( k=\sqrt{M/I} \)
(d) \( k=\sqrt{I/M} \)
Show Answer & Explanation

Answer: (d) \( k=\sqrt{I/M} \)

Explanation:
1. I = Mk².
2. So k = √(I/M).

Teacher's Note:
1. Radius of gyration: I = Mk².

Question 16: A rod PQ of mass M and length L is hinged at end P and held horizontal by a string at Q. When the string is cut, the initial angular acceleration of the rod is
(a) \( \frac gL \)
(b) \( \frac{2g}{L} \)
(c) \( \frac{2g}{3L} \)
(d) \( \frac{3g}{2L} \)
Show Answer & Explanation

Answer: (d) \( \frac{3g}{2L} \)

Explanation:
1. Torque about P = Mg (L/2).
2. I = ML²/3, so α = (MgL/2)/(ML²/3) = 3g/2L.

Teacher's Note:
1. Use the moment of inertia about the hinge.

Question 17: A man standing on a rotating table is holding two masses at arm's length. Without moving his arms, he drops the two masses. His angular speed will
(a) increase
(b) decrease
(c) become zero
(d) remain constant
Show Answer & Explanation

Answer: (d) remain constant

Explanation:
1. The masses leave with the same tangential velocity and carry away their angular momentum in proportion to their share of I.
2. There is no change in the angular speed of the man and table.

Teacher's Note:
1. Dropping the masses exerts no torque on the table.

Question 18: A stick is thrown in the air and lands on the ground at some distance from the thrower. The centre of mass of the stick will move along a parabolic path
(a) in all cases
(b) only if the stick is uniform
(c) only if the stick has linear motion but no rotational motion
(d) only if the stick has a shape such that its centre of mass is located at some point on it and not outside it
Show Answer & Explanation

Answer: (a) in all cases

Explanation:
1. The only external force is gravity, so the CM moves as a projectile.
2. It does so whatever the shape and whatever the rotation.

Teacher's Note:
1. CM follows a parabola for any shape.

Question 19: A ring of mass m and radius r is melted and then moulded into a sphere. The moment of inertia of the sphere will be
(a) more than that of the ring
(b) less than that of the ring
(c) equal to that of the ring
(d) none of these
Show Answer & Explanation

Answer: (b) less than that of the ring

Explanation:
1. The ring's moment of inertia about its axis is mr².
2. The sphere has the same mass but a radius smaller than r, and I = (2/5)mR² is clearly smaller.

Teacher's Note:
1. Compare mr² with (2/5)mR².

Question 20:
Assertion (A): The centre of mass of a system of n particles is the weighted average of the position vectors of the n particles making up the system.
Reason (R): The position of the centre of mass of a system is independent of the coordinate system chosen.
(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: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. R⃗ = Σmᵢr⃗ᵢ/Σmᵢ is a weighted average, so the assertion is true.
2. The CM position relative to the particles is the same in any coordinate system, so the reason is true but does not explain the weighted-average definition.

Teacher's Note:
1. Both true, no explanation link.

Question 21: The sum of moments of all the particles in a system about the centre of mass is always
(a) maximum
(b) minimum
(c) zero
(d) infinite
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. Σ mᵢ r⃗ᵢ = 0 when the position vectors are measured from the centre of mass.
2. So the sum of the mass moments is zero.

Teacher's Note:
1. This is the defining property of the CM.

Question 22: Three particles, each of mass m, are situated at the vertices of an equilateral triangle ABC of side l. The moment of inertia of the system about a line AX through A, perpendicular to AB and lying in the plane of ABC, is
(a) \( \frac32ml^2 \)
(b) \( \frac34ml^2 \)
(c) \( 2ml^2 \)
(d) \( \frac54ml^2 \)
Show Answer & Explanation

Answer: (d) \( \frac54ml^2 \)

Explanation:
1. A is on the axis. B is at distance l from the axis along AB.
2. C's perpendicular distance from AX is its projection along AB, l/2.
3. I = ml² + m(l/2)² = (5/4)ml².

Teacher's Note:
1. Find perpendicular distances from the axis.

Question 23: A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at a distance x from A. The normal reaction on A is
(a) \( \frac{Wd}{x} \)
(b) \( \frac{W(d-x)}{x} \)
(c) \( \frac{W(d-x)}{d} \)
(d) \( \frac{Wx}{d} \)
Show Answer & Explanation

Answer: (c) \( \frac{W(d-x)}{d} \)

Explanation:
1. Take torques about B: N_A × d = W(d − x).
2. So N_A = W(d − x)/d.

Teacher's Note:
1. Choose the pivot at the other knife edge.

Question 24: The moment of inertia of a circular wire of mass M and radius R about its diameter is
(a) \( MR^2/2 \)
(b) \( MR^2 \)
(c) \( 2MR^2 \)
(d) \( MR^2/4 \)
Show Answer & Explanation

Answer: (a) \( MR^2/2 \)

Explanation:
1. For a ring, I about the axis perpendicular to its plane is MR².
2. By the perpendicular axis theorem, the two diameters share it equally, so I = MR²/2.

Teacher's Note:
1. Use the perpendicular axis theorem.

Question 25: A uniform thin rod AB of length L has linear mass density μ(x) = a + bx/L, where x is measured from A. If the centre of mass of the rod lies at a distance of (7/12)L from A, then a and b are related as
(a) \( a=2b \)
(b) \( 2a=b \)
(c) \( a=b \)
(d) \( 3a=2b \)
Show Answer & Explanation

Answer: (b) \( 2a=b \)

Explanation:
1. x_cm = ∫xμ dx / ∫μ dx = L(a/2 + b/3)/(a + b/2) = 7L/12.
2. 6a + 4b = 7a + 3.5b gives b = 2a.

Teacher's Note:
1. Set up the CM integral.

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