JEE Physics System Of Particles and Rotational Motion MCQs Set 29

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Multiple Choice Questions

Question 1: A hollow sphere of mass M and radius R rolls on a horizontal surface without slipping. The ratio of its rotational kinetic energy to the total kinetic energy is
(a) 1 : 2
(b) 3 : 5
(c) 2 : 5
(d) 1 : 5
Show Answer & Explanation

Answer: (c) 2 : 5

Explanation:
1. For a hollow sphere I = (2/3)MR², so K_rot = ½(2/3)MR²(v/R)² = (1/3)Mv².
2. K_trans = ½Mv², so K_total = (5/6)Mv².
3. K_rot/K_total = (1/3)/(5/6) = 2/5.

Teacher's Note:
1. Use ω = v/R.
2. I = (2/3)MR².

Question 2: A disc is rolling on an inclined plane without slipping. What fraction of its total energy is in the form of rotational kinetic energy?
(a) \( \dfrac{1}{3} \)
(b) \( \dfrac{1}{2} \)
(c) \( \dfrac{2}{7} \)
(d) \( \dfrac{2}{5} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{1}{3} \)

Explanation:
1. K_rot = ½(½MR²)(v/R)² = ¼Mv² and K_trans = ½Mv².
2. Fraction = (¼)/(¼ + ½) = 1/3.

Teacher's Note:
1. Use I = ½MR² and ω = v/R.
2. Compare with the total ¾Mv².

Question 3: If the rotational kinetic energy of a rolling body is 50% of its total kinetic energy, the body is a
(a) ring
(b) disc
(c) hollow sphere
(d) solid sphere
Show Answer & Explanation

Answer: (a) ring

Explanation:
1. K_rot = K_trans means ½Iω² = ½mv², so I = mR² (using v = ωR).
2. I = mR² is the moment of inertia of a ring.

Teacher's Note:
1. Set rotational and translational energies equal.
2. Which body has I = mR²?

Question 4: A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K and the radius of the ball is R. The fraction of the total energy associated with its rotational energy is
(a) \( \dfrac{K^2 + R^2}{R^2} \)
(b) \( \dfrac{K^2}{R^2} \)
(c) \( \dfrac{K^2}{K^2 + R^2} \)
(d) \( \dfrac{R^2}{K^2 + R^2} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{K^2}{K^2 + R^2} \)

Explanation:
1. I = MK² and ω = v/R, so K_rot = ½MK²v²/R² and K_trans = ½Mv².
2. Fraction = (K²/R²)/(1 + K²/R²) = K²/(K² + R²).

Teacher's Note:
1. Write I = MK².
2. Divide rotational by total.

Question 5: If the angular velocity of a body rotating about an axis is halved and its moment of inertia is doubled, the rotational kinetic energy changes by a factor of
(a) 4
(b) 2
(c) 1
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. K = ½Iω².
2. K′ = ½(2I)(ω/2)² = ¼Iω² = K/2.

Teacher's Note:
1. Substitute the new values in K = ½Iω².
2. Square the change in ω.

Question 6: When a body starts to roll down an inclined plane, its potential energy is converted into
(a) translational kinetic energy only
(b) translational and rotational kinetic energy
(c) rotational energy only
(d) none of these
Show Answer & Explanation

Answer: (b) translational and rotational kinetic energy

Explanation:
1. A rolling body has both the motion of its centre of mass and rotation about it.
2. So mgh = ½mv² + ½Iω².

Teacher's Note:
1. Rolling means translation plus rotation.
2. Write the energy equation.

Question 7: Which of the following is true about the angular momentum of a cylinder rolling down a slope without slipping?
(a) its magnitude changes but the direction remains the same
(b) both magnitude and direction change
(c) only the direction changes
(d) neither changes
Show Answer & Explanation

Answer: (a) its magnitude changes but the direction remains the same

Explanation:
1. The speed increases as it rolls, so L = Iω (and the angular momentum about a fixed point on the slope) increases in magnitude.
2. The axis of rotation stays horizontal and parallel to itself, so the direction of L stays the same.

Teacher's Note:
1. Speed increases down the slope.
2. The axis of rotation does not turn.

Question 8: A spherical shell rolls down a plane of length L inclined at an angle θ, without slipping. Its velocity on reaching the bottom will be
(a) \( \sqrt{\dfrac{6gL\sin\theta}{5}} \)
(b) \( \sqrt{\dfrac{5gL\sin\theta}{6}} \)
(c) \( \sqrt{\dfrac{10gL\sin\theta}{7}} \)
(d) \( \sqrt{4gL\sin\theta} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\dfrac{6gL\sin\theta}{5}} \)

Explanation:
1. Height h = L sinθ and energy conservation: mgh = ½mv² + ½(2/3 mR²)(v/R)² = (5/6)mv².
2. v² = 6gh/5 = 6gL sinθ/5.

Teacher's Note:
1. Use I = (2/3)mR² for a shell.
2. h = L sinθ.

Question 9: A solid sphere rolls down an inclined plane and its velocity at the bottom is v₁. The same sphere slides down the plane (without friction) and its velocity at the bottom is v₂. Which of the following relations is correct?
(a) v₁ = v₂
(b) \( v_1 = \sqrt{\dfrac{5}{7}}\,v_2 \)
(c) \( v_1 = \sqrt{\dfrac{7}{5}}\,v_2 \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( v_1 = \sqrt{\dfrac{5}{7}}\,v_2 \)

Explanation:
1. Rolling: v₁² = 2gh/(1 + 2/5) = 10gh/7.
2. Sliding: v₂² = 2gh.
3. v₁²/v₂² = 5/7, so v₁ = √(5/7) v₂.

Teacher's Note:
1. Apply energy conservation to both cases.
2. Rolling shares energy with rotation.

Question 10: Two discs having masses in the ratio 1 : 2 and radii in the ratio 1 : 8 roll down without slipping, one by one, from an inclined plane of height h. The ratio of their linear velocities on reaching the ground is
(a) 1 : 16
(b) 1 : 128
(c) \( 1 : 8\sqrt{2} \)
(d) 1 : 1
Show Answer & Explanation

Answer: (d) 1 : 1

Explanation:
1. For a disc, mgh = ¾mv², so v = √(4gh/3).
2. The result is independent of mass and radius, so both discs have the same speed: ratio 1 : 1.

Teacher's Note:
1. Mass and radius cancel in the energy equation.
2. Both have I = ½mR².

Question 11: A spherical shell and a solid cylinder of the same radius roll down an inclined plane. The ratio of their accelerations will be
(a) 15 : 14
(b) 9 : 10
(c) 2 : 3
(d) 3 : 5
Show Answer & Explanation

Answer: (b) 9 : 10

Explanation:
1. a = g sinθ/(1 + I/mR²).
2. Shell: a₁ = g sinθ/(1 + 2/3) = (3/5)g sinθ. Cylinder: a₂ = g sinθ/(1 + 1/2) = (2/3)g sinθ.
3. a₁ : a₂ = (3/5) : (2/3) = 9 : 10.

Teacher's Note:
1. Use a = g sinθ/(1 + k²/R²).
2. Shell: k² = 2R²/3; cylinder: k² = R²/2.

Question 12: A sphere and a disc of the same radius and mass roll on an inclined plane without slipping. Let a_s and a_d be their accelerations and g the acceleration due to gravity. Which statement is correct?
(a) a_s > a_d > g
(b) g > a_s > a_d
(c) a_s > g > a_d
(d) a_d > a_s > g
Show Answer & Explanation

Answer: (b) g > a_s > a_d

Explanation:
1. a_s = g sinθ/(1 + 2/5) = (5/7)g sinθ, a_d = g sinθ/(1 + 1/2) = (2/3)g sinθ.
2. Since 5/7 > 2/3 and sinθ < 1, g > a_s > a_d.

Teacher's Note:
1. Find each acceleration with the rolling formula.
2. Both are less than g sinθ, hence less than g.

Question 13: Calculate the ratio of the times taken by a uniform spherical shell and a solid cylinder of the same mass and the same diameter to roll down through the same distance from rest on an inclined plane.
(a) 10 : 9
(b) 1 : 1
(c) \( \sqrt{10} : \sqrt{9} \)
(d) \( \sqrt{9} : \sqrt{10} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{10} : \sqrt{9} \)

Explanation:
1. a_shell = (3/5)g sinθ and a_cylinder = (2/3)g sinθ.
2. For the same distance, t ∝ 1/√a.
3. t_shell : t_cyl = √(a_cyl/a_shell) = √((2/3)/(3/5)) = √(10/9) = √10 : √9.

Teacher's Note:
1. s = ½at² gives t ∝ a^(-1/2).
2. Use the accelerations for the shell and the cylinder.

Question 14: A disc takes times t₁ and t₂ for sliding down (without friction) and rolling down an inclined plane of length L, respectively, to reach the bottom. The ratio t₁ : t₂ is
(a) \( \sqrt{2} : 1 \)
(b) \( \sqrt{2} : \sqrt{3} \)
(c) 2 : 3
(d) \( \sqrt{3} : 1 \)
Show Answer & Explanation

Answer: (b) \( \sqrt{2} : \sqrt{3} \)

Explanation:
1. Sliding: a₁ = g sinθ. Rolling: a₂ = g sinθ/(1 + ½) = (2/3)g sinθ.
2. L = ½at², so t ∝ 1/√a.
3. t₁ : t₂ = √(a₂/a₁) = √(2/3) = √2 : √3.

Teacher's Note:
1. Use L = ½at² for both cases.
2. Rolling acceleration for a disc is (2/3)g sinθ.

Question 15: A solid sphere is rolling down an inclined plane from rest, and a rectangular block of the same mass is slipping down simultaneously from rest on a similar smooth inclined plane. Then
(a) both will reach the bottom simultaneously
(b) the sphere will reach the bottom first
(c) the rectangular block will reach the bottom first
(d) it depends on the density of the material
Show Answer & Explanation

Answer: (c) the rectangular block will reach the bottom first

Explanation:
1. Block: a = g sinθ. Sphere: a = (5/7)g sinθ.
2. The larger acceleration reaches the bottom first, so the block wins.

Teacher's Note:
1. Compare accelerations.
2. The sphere shares energy with rotation.

Question 16: A solid sphere of mass M and radius R rolls without slipping down an inclined plane of length L and height h. The speed of its centre of mass when it reaches the bottom is
(a) \( \sqrt{10gh} \)
(b) \( \sqrt{7gh} \)
(c) \( \sqrt{\dfrac{7}{10}gh} \)
(d) \( \sqrt{\dfrac{10}{7}gh} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{\dfrac{10}{7}gh} \)

Explanation:
1. Energy conservation: Mgh = ½Mv² + ½(2/5 MR²)(v/R)² = (7/10)Mv².
2. v² = 10gh/7.

Teacher's Note:
1. Add the translational and rotational energies.
2. I = (2/5)MR².

Question 17: When a hollow sphere of moment of inertia I rolls down an inclined plane, the percentage of its total energy which is rotational is approximately
(a) 40 %
(b) 60 %
(c) 100 %
(d) none of these
Show Answer & Explanation

Answer: (a) 40 %

Explanation:
1. For a hollow sphere I = (2/3)mR², so K_rot = (1/3)mv² and K_trans = ½mv².
2. Fraction = (1/3)/(5/6) = 2/5 = 40 %.

Teacher's Note:
1. Use I = (2/3)mR².
2. Fraction = K_rot/(K_rot + K_trans).

Question 18: A uniform rod AB of length ℓ and mass m is free to rotate about the end A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is mℓ²/3, the initial angular acceleration of the rod will be
(a) \( \dfrac{3g}{2\ell} \)
(b) \( \dfrac{2g}{3\ell} \)
(c) \( mg\dfrac{\ell}{2} \)
(d) \( \dfrac{3}{2}g\ell \)
Show Answer & Explanation

Answer: (a) \( \dfrac{3g}{2\ell} \)

Explanation:
1. At release the torque about A is τ = mg × ℓ/2.
2. α = τ/I = (mgℓ/2)/(mℓ²/3) = 3g/(2ℓ).

Teacher's Note:
1. Weight acts at the centre of the rod, ℓ/2 from A.
2. α = τ/I.

Question 19: Four identical thin rods, each of mass M and length ℓ, form a square frame. The moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is
(a) \( \dfrac{1}{3}M\ell^2 \)
(b) \( \dfrac{4}{3}M\ell^2 \)
(c) \( \dfrac{2}{3}M\ell^2 \)
(d) \( \dfrac{13}{3}M\ell^2 \)
Show Answer & Explanation

Answer: (b) \( \dfrac{4}{3}M\ell^2 \)

Explanation:
1. For one rod about the given axis, by the parallel axes theorem: I = Mℓ²/12 + M(ℓ/2)² = Mℓ²/3.
2. For the four rods: 4 × Mℓ²/3 = 4Mℓ²/3.

Teacher's Note:
1. The axis is at distance ℓ/2 from the centre of each rod.
2. Add the four equal contributions.

Question 20: If F is the force acting on a particle having position vector r and τ is the torque of this force about the origin, then
(a) \( \vec r\cdot\vec\tau = 0\ \text{and}\ \vec F\cdot\vec\tau \ne 0 \)
(b) \( \vec r\cdot\vec\tau \ne 0\ \text{and}\ \vec F\cdot\vec\tau = 0 \)
(c) \( \vec r\cdot\vec\tau > 0\ \text{and}\ \vec F\cdot\vec\tau < 0 \)
(d) \( \vec r\cdot\vec\tau = 0\ \text{and}\ \vec F\cdot\vec\tau = 0 \)
Show Answer & Explanation

Answer: (d) \( \vec r\cdot\vec\tau = 0\ \text{and}\ \vec F\cdot\vec\tau = 0 \)

Explanation:
1. τ = r × F is perpendicular to both r and F.
2. So the dot products r·τ and F·τ are both zero.

Teacher's Note:
1. A cross product is perpendicular to both factors.
2. Dot product of perpendicular vectors is zero.

Question 21: The instantaneous angular position of a point on a rotating wheel is given by θ(t) = 2t³ − 6t² (θ in radian, t in second). The torque on the wheel becomes zero at
(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 = dθ/dt = 6t² − 12t and α = dω/dt = 12t − 12.
2. τ = Iα = 0 when 12t − 12 = 0, so t = 1 s.

Teacher's Note:
1. Torque is proportional to the angular acceleration.
2. Differentiate twice.

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