CBSE Class 10 Physics Dynamics MCQs Set 29

Practice MCQs for Class 10 Physics Dynamics

Explore reliable objective questions for Dynamics tailored for Class 10 learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: An object of mass 1 kg moves with an initial velocity of 20 m s-1. What force (in N) is required to change its velocity to 30 m s-1 in 1 ms?
(a) 103
(b) 104
(c) 102
(d) 105
Show Answer & Explanation

Answer: (b) 104

Explanation:
1. F = m(v - u)/t = 1 × (30 - 20)/(10-3) = 104 N.

Teacher's Note:
1. Force equals the rate of change of momentum.
2. 1 ms = 10-3 s.

Question 2: A force of 10 N is applied on a body of mass 1 kg. If the friction acting on it is 2 N, what is the acceleration of the body?
(a) 8 m s-2
(b) 4 m s-2
(c) 16 m s-2
(d) 12 m s-2
Show Answer & Explanation

Answer: (a) 8 m s-2

Explanation:
1. Net force = 10 - 2 = 8 N.
2. a = F/m = 8/1 = 8 m s-2.

Teacher's Note:
1. Friction opposes the applied force.
2. Use Fnet = ma.

Question 3: A car of mass 2000 kg moves uniformly at 20 m s-1. If the engine gives it an extra momentum of 1000 kg m s-1, what is its new velocity?
(a) 25 m s-1
(b) 20.5 m s-1
(c) 14.5 m s-1
(d) 17.5 m s-1
Show Answer & Explanation

Answer: (b) 20.5 m s-1

Explanation:
1. Initial momentum = 2000 × 20 = 40000 kg m s-1.
2. New momentum = 41000, so v = 41000/2000 = 20.5 m s-1.

Teacher's Note:
1. Add the extra momentum to the initial momentum.
2. Divide by the mass for the velocity.

Question 4: A body of mass 2 kg is dropped from a height of 10 m. What is its kinetic energy (in J) on reaching the ground? (g = 10 m s-2)
(a) 400
(b) 300
(c) 200
(d) 100
Show Answer & Explanation

Answer: (c) 200

Explanation:
1. By conservation of energy, KE = mgh = 2 × 10 × 10 = 200 J.

Teacher's Note:
1. Loss of potential energy equals gain in kinetic energy.
2. Air resistance is ignored.

Question 5: The momentum of a body of mass 3 kg is 25 kg m s-1. What is its kinetic energy (in J)?
(a) 180.2
(b) 104.16
(c) 275.8
(d) 325.6
Show Answer & Explanation

Answer: (b) 104.16

Explanation:
1. KE = p2/(2m) = 625/6 = 104.16 J.

Teacher's Note:
1. KE = p2/2m links momentum and kinetic energy.
2. Check: v = 25/3 m s-1, ½mv2 = 104.16 J.

Question 6: A force of 100 N acts on a body at an angle of 30° to the direction of displacement. If the displacement is 2 m, what is the work done?
(a) \( 100\sqrt{3} \) J
(b) 50 J
(c) 150 J
(d) \( 150\sqrt{2} \) J
Show Answer & Explanation

Answer: (a) \( 100\sqrt{3} \) J

Explanation:
1. W = Fs cosθ = 100 × 2 × cos30° = 200 × (\( \sqrt{3} \)/2) = \( 100\sqrt{3} \) J.

Teacher's Note:
1. Only the component of force along the displacement does work.
2. cos30° = √3/2.

Question 7: What power is developed by an engine in lifting 2000 L of water from a well 20 m deep in 20 s? (1 L of water = 1 kg, g = 10 m s-2)
(a) 6 kW
(b) 20 kW
(c) 4 kW
(d) 2 kW
Show Answer & Explanation

Answer: (b) 20 kW

Explanation:
1. Mass = 2000 kg, work = mgh = 2000 × 10 × 20 = 4 × 105 J.
2. Power = 4 × 105/20 = 20 kW.

Teacher's Note:
1. Power is work divided by time.
2. 1 L of water has a mass of 1 kg.

Question 8: What is the work done by gravity when an object of mass 50 kg is lifted upwards through 20 m? (g = 10 m s-2)
(a) 104 J
(b) -104 J
(c) 102 J
(d) -102 J
Show Answer & Explanation

Answer: (b) -104 J

Explanation:
1. Gravity acts downwards while the displacement is upwards, so θ = 180°.
2. W = mgh cos180° = -50 × 10 × 20 = -104 J.

Teacher's Note:
1. Work by gravity is negative when the body goes up.
2. The lifting force does +104 J.

Question 9: Match Column A with Column B.
(A) Acceleration due to gravity at the centre of the Earth (B) Escape velocity on the Earth (C) Orbital velocity near the Earth (D) Acceleration due to gravity on the Moon
(P) 1.63 m s-2 (Q) 11.2 km s-1 (R) 7.8 km s-1 (S) zero
(a) A → S; B → Q; C → R; D → P
(b) A → S; B → R; C → Q; D → P
(c) A → R; B → Q; C → S; D → P
(d) A → P; B → S; C → Q; D → R
Show Answer & Explanation

Answer: (a) A → S; B → Q; C → R; D → P

Explanation:
1. g is zero at the centre of the Earth (S).
2. Escape velocity is 11.2 km s-1 (Q), orbital velocity near the surface is about 7.8 km s-1 (R).
3. g on the Moon is 1.63 m s-2 (P).

Teacher's Note:
1. Escape velocity = √2 × orbital velocity.
2. Moon's g is about one-sixth of Earth's.

Question 10: A particle moves in a circular path of radius 20 cm with an angular velocity of 2π rad s-1. What is its speed?
(a) 0.4π m s-1
(b) 4π m s-1
(c) 2 m s-1
(d) 4 m s-1
Show Answer & Explanation

Answer: (a) 0.4π m s-1

Explanation:
1. v = rω = 0.2 × 2π = 0.4π m s-1.

Teacher's Note:
1. Convert 20 cm to 0.2 m.
2. v = rω.

Question 11: An object of mass m hangs from the ceiling of a cabin that is moving down with an acceleration a (a < g). What is the tension in the string?
(a) m(g - a)
(b) m(g + a)
(c) mg
(d) ma
Show Answer & Explanation

Answer: (a) m(g - a)

Explanation:
1. Taking downward as positive: mg - T = ma.
2. T = m(g - a).

Teacher's Note:
1. Downward acceleration reduces the tension.
2. Upward acceleration would give m(g + a).

Question 12: A bullet of mass 100 g moving at 100 m s-1 is stopped by a plank after penetrating 10 cm. What is the average force exerted by the plank?
(a) 100 N
(b) 1000 N
(c) 2500 N
(d) 5000 N
Show Answer & Explanation

Answer: (d) 5000 N

Explanation:
1. By the work-energy theorem, F × 0.1 = ½ × 0.1 × 1002 = 500 J.
2. F = 500/0.1 = 5000 N.

Teacher's Note:
1. Work done by the retarding force equals the loss of kinetic energy.
2. Use the mass in kg.

Question 13: A car of mass 1500 kg takes a turn on a level curved road of radius 20 m at 36 km h-1. What is the static friction acting on it?
(a) 3200 N
(b) 4840 N
(c) 7500 N
(d) 8500 N
Show Answer & Explanation

Answer: (c) 7500 N

Explanation:
1. 36 km h-1 = 10 m s-1.
2. Friction supplies the centripetal force: f = mv2/r = 1500 × 100/20 = 7500 N.

Teacher's Note:
1. On an unbanked road, friction provides the centripetal force.
2. Convert km h-1 to m s-1.

Question 14: A body slides from rest on a smooth hemisphere of radius R starting at the topmost point A. At point B the radius to the body makes an angle θ with the vertical. What is the speed of the body at B?
(a) R - R cosθ
(b) \( \sqrt{gR(\cos\theta - 1)} \)
(c) \( \sqrt{2gR(1 - \cos\theta)} \)
(d) \( \sqrt{gR\cos\theta} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{2gR(1 - \cos\theta)} \)

Explanation:
1. The body falls through a vertical height R - R cosθ.
2. mg(R - R cosθ) = ½mv2, so v = \( \sqrt{2gR(1 - \cos\theta)} \).

Teacher's Note:
1. Loss of potential energy gives kinetic energy.
2. Height below the top = R(1 - cosθ).

Question 15: A body of mass m revolves around a body of mass M in a circle of radius R under gravitational attraction. Arrange the steps to find its speed.
(P) Determine the speed of the body.
(Q) Determine the gravitational force on the body.
(R) Determine the centripetal force.
(S) Equate the gravitational force with the centripetal force.
(a) P R Q S
(b) P R S Q
(c) P S Q R
(d) Q R S P
Show Answer & Explanation

Answer: (d) Q R S P

Explanation:
1. First find the gravitational force GMm/R2 (Q) and the centripetal force mv2/R (R).
2. Equate them (S) and solve for v = \( \sqrt{GM/R} \) (P).
3. The order is Q, R, S, P.

Teacher's Note:
1. Gravity supplies the centripetal force.
2. v = √(GM/R) does not depend on the mass of the orbiting body.

Question 16: A car of mass 2000 kg is moving at 18 km h-1. What force is required to change its speed to 36 km h-1 in 2 s?
(a) 2800 N
(b) 1500 N
(c) 2500 N
(d) 5000 N
Show Answer & Explanation

Answer: (d) 5000 N

Explanation:
1. u = 5 m s-1, v = 10 m s-1.
2. F = m(v - u)/t = 2000 × 5/2 = 5000 N.

Teacher's Note:
1. Convert km h-1 to m s-1 by dividing by 3.6.

Question 17: Forces of 20 N and 5 N act simultaneously on a body of mass 2 kg in opposite directions. What is its acceleration?
(a) 12.5 m s-2
(b) 2 m s-2
(c) 7.5 m s-2
(d) 5 m s-2
Show Answer & Explanation

Answer: (c) 7.5 m s-2

Explanation:
1. Net force = 20 - 5 = 15 N.
2. a = 15/2 = 7.5 m s-2.

Teacher's Note:
1. Opposite forces subtract.

Question 18: An object of mass 20 kg moves with a velocity of 200 m s-1. A force is applied so that its momentum decreases by 100 kg m s-1. What is the new velocity?
(a) 75 m s-1
(b) 205 m s-1
(c) 100 m s-1
(d) 195 m s-1
Show Answer & Explanation

Answer: (d) 195 m s-1

Explanation:
1. Initial momentum = 20 × 200 = 4000 kg m s-1.
2. New momentum = 3900, so v = 3900/20 = 195 m s-1.

Teacher's Note:
1. Momentum = mass × velocity.

Question 19: A body of mass 2 kg is thrown up with a velocity of 10 m s-1. What is the maximum potential energy gained?
(a) 100 J
(b) 200 J
(c) 300 J
(d) 400 J
Show Answer & Explanation

Answer: (a) 100 J

Explanation:
1. At the highest point all the kinetic energy is converted to potential energy.
2. PEmax = ½mv2 = ½ × 2 × 100 = 100 J.

Teacher's Note:
1. Energy is conserved without air resistance.

Question 20: The kinetic energy of a body of mass 5 kg is 200 J. What is its momentum (in kg m s-1)?
(a) \( 20\sqrt{5} \)
(b) \( \sqrt{125} \)
(c) \( \sqrt{175} \)
(d) \( \sqrt{200} \)
Show Answer & Explanation

Answer: (a) \( 20\sqrt{5} \)

Explanation:
1. p = \( \sqrt{2mK} \) = \( \sqrt{2 \times 5 \times 200} \) = \( \sqrt{2000} \) = \( 20\sqrt{5} \).

Teacher's Note:
1. From K = p2/2m, p = √(2mK).

Question 21: The work done by a force of 20 N when the displacement is 4 m is 40 J. What is the angle between the force and the displacement?
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Show Answer & Explanation

Answer: (c) 60°

Explanation:
1. W = Fs cosθ gives 40 = 20 × 4 × cosθ.
2. cosθ = 1/2, so θ = 60°.

Teacher's Note:
1. Use W = Fs cosθ.

Question 22: What power is developed by an engine if it imparts a velocity of 10 m s-1 to a body of mass 2 kg in 0.1 s?
(a) 1 kW
(b) 2 kW
(c) 3 kW
(d) 4 kW
Show Answer & Explanation

Answer: (a) 1 kW

Explanation:
1. Work done = gain in kinetic energy = ½ × 2 × 100 = 100 J.
2. P = 100/0.1 = 1000 W = 1 kW.

Teacher's Note:
1. Power = work / time.

Question 23: What is the work done by gravity on a 20 kg object when it is moved horizontally by applying a force of 20 N to produce a displacement of 2 m?
(a) 10 J
(b) 20 J
(c) 40 J
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. Gravity acts vertically downwards and the displacement is horizontal, so the angle is 90°.
2. W = Fs cos90° = 0.

Teacher's Note:
1. A force perpendicular to the displacement does no work.

Question 24: Match Column A with Column B (R is the radius of the orbit or body).
(A) Orbital velocity (B) Escape velocity (C) Centripetal force (D) Impulse
(P) \( \sqrt{2GM/R} \) (Q) \( \sqrt{GM/R} \) (R) force × time (S) Mv2/R
(a) A → P; B → R; C → Q; D → S
(b) A → S; B → Q; C → R; D → P
(c) A → Q; B → P; C → S; D → R
(d) A → Q; B → S; C → R; D → P
Show Answer & Explanation

Answer: (c) A → Q; B → P; C → S; D → R

Explanation:
1. Orbital velocity is \( \sqrt{GM/R} \) (Q).
2. Escape velocity is \( \sqrt{2GM/R} \) (P).
3. Centripetal force is Mv2/R (S) and impulse is force × time (R).

Teacher's Note:
1. Escape velocity is √2 times orbital velocity.
2. Impulse equals change in momentum.

Question 25: A particle moves at 10 m s-1 along a circular path of radius 30 cm. What is its angular velocity (in rad s-1)?
(a) 121.2
(b) 11.1
(c) 33.3
(d) 22.2
Show Answer & Explanation

Answer: (c) 33.3

Explanation:
1. ω = v/r = 10/0.3 = 33.3 rad s-1.

Teacher's Note:
1. Convert cm to m.
2. ω = v/r.

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Practice MCQs for Class 10 Physics Dynamics

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