100 Advanced & Numerical MCQs (Level -2) on Motion and Force, Physics (Unit-Wise MCQs Practice):
Whether you are preparing for board examinations, school tests, college assessments, university exams, entry tests, or competitive examinations, this comprehensive Motion and Force MCQs collection is designed to strengthen your conceptual understanding and problem-solving skills.
This chapter covers all the fundamental and advanced topics of motion and force, including scalar and vector quantities, displacement, distance, speed, velocity, acceleration, equations of uniformly accelerated motion, projectile motion, Newton's Laws of Motion, mass, weight, momentum, impulse, collisions, conservation of momentum, graph interpretation, and real-life applications of mechanics.
The MCQs are carefully organized into three progressive difficulty levels to help learners build confidence step by step:
- 100 Basic Conceptual MCQs – Develop a strong foundation by covering definitions, principles, formulas, and essential concepts.
- 100 Advanced & Numerical MCQs – Improve analytical and mathematical problem-solving skills through calculations and application-based questions.
- 50 Higher-Order Thinking Skills (HOTS) MCQs – Challenge your understanding with conceptual reasoning, graph analysis, multi-step problems, and competitive examination-style questions.
- 50 Challenging MCQs Quiz with Answers (1–50) – Carefully selected conceptual, numerical, and HOTS questions to strengthen problem-solving skills and prepare students for board and competitive examinations.
Every MCQ includes:
- ✔ Four carefully designed answer options
- ✔ Instant correct answer
- ✔ Detailed concept-based explanation
- ✔ Exam-focused learning approach
This collection is ideal for students preparing for Physics examinations at school, college, and university levels, as well as candidates appearing in engineering, medical, and other competitive entrance examinations. Teachers can also use these questions for classroom assessments, quizzes, assignments, and revision sessions.
By practicing these 250 carefully selected Motion and Force MCQs, you will improve your conceptual clarity, numerical problem-solving ability, graph interpretation skills, logical reasoning, and confidence in tackling objective-type Physics questions. Whether you are revising before an examination or strengthening your fundamentals, this comprehensive MCQ collection provides an excellent resource for complete chapter preparation and long-term success.
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Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)
MCQ No. 101
A car starts from rest and accelerates uniformly at 3 m/s² for 5 s. Its final velocity is:
a. 8 m/s
b. 10 m/s
c. 15 m/s
d. 20 m/s
Correct Answer: c. 15 m/s
Explanation:
Using the first equation of motion, v = u + at. Since u = 0, a = 3 m/s², and t = 5 s, we get v = 0 + (3 × 5) = 15 m/s.
MCQ No. 102
A body moves with an initial velocity of 10 m/s and accelerates uniformly at 2 m/s² for 6 s. The final velocity is:
a. 18 m/s
b. 20 m/s
c. 22 m/s
d. 24 m/s
Correct Answer: c. 22 m/s
Explanation:
Applying v = u + at, we obtain v = 10 + (2 × 6) = 22 m/s. The velocity increases because the acceleration is positive.
MCQ No. 103
A train accelerates uniformly from rest at 2 m/s² for 8 s. The distance travelled is:
a. 32 m
b. 48 m
c. 64 m
d. 80 m
Correct Answer: c. 64 m
Explanation:
Using s = ut + ½at², where u = 0, we get s = ½ × 2 × 8² = 64 m.
MCQ No. 104
A car moving at 25 m/s comes to rest in 5 s. Its acceleration is:
a. +5 m/s²
b. –5 m/s²
c. –4 m/s²
d. +4 m/s²
Correct Answer: b. –5 m/s²
Explanation:
Using a = (v – u)/t, we get a = (0 – 25)/5 = –5 m/s². The negative sign indicates deceleration.
MCQ No. 105
A body accelerates from 6 m/s to 18 m/s in 4 s. Its acceleration is:
a. 2 m/s²
b. 3 m/s²
c. 4 m/s²
d. 5 m/s²
Correct Answer: b. 3 m/s²
Explanation:
Acceleration is calculated using a = (18 – 6)/4 = 3 m/s².
MCQ No. 106
A body starts from rest and moves with a uniform acceleration of 4 m/s². Its velocity after 10 s is:
a. 20 m/s
b. 30 m/s
c. 40 m/s
d. 50 m/s
Correct Answer: c. 40 m/s
Explanation:
Applying v = u + at, we get v = 0 + (4 × 10) = 40 m/s.
MCQ No. 107
A body moving at 12 m/s accelerates uniformly at 3 m/s². The distance covered in 4 s is:
a. 48 m
b. 60 m
c. 72 m
d. 96 m
Correct Answer: c. 72 m
Explanation:
Using s = ut + ½at², we get s = (12 × 4) + ½ × 3 × 16 = 48 + 24 = 72 m.
MCQ No. 108
The slope of a straight-line velocity–time graph is 5 m/s². This represents the object's:
a. Velocity
b. Displacement
c. Acceleration
d. Momentum
Correct Answer: c. Acceleration
Explanation:
The slope of a velocity–time graph equals acceleration. A constant slope indicates uniform acceleration.
MCQ No. 109
The area under a velocity–time graph between 0 s and 6 s is 90 m. The displacement of the object is:
a. 15 m
b. 45 m
c. 90 m
d. 180 m
Correct Answer: c. 90 m
Explanation:
The area under the velocity–time graph directly represents displacement.
MCQ No. 110
A projectile is launched vertically upward with a speed of 20 m/s. Its maximum height is approximately:
a. 10.2 m
b. 20.4 m
c. 30.6 m
d. 40.8 m
Correct Answer: b. 20.4 m
Explanation:
Using h = u²/2g, we get h = 20²/(2 × 9.8) ≈ 20.4 m.
MCQ No. 111
A projectile is projected vertically upward with a speed of 19.6 m/s. The time taken to reach maximum height is:
a. 1 s
b. 2 s
c. 3 s
d. 4 s
Correct Answer: b. 2 s
Explanation:
Using t = u/g, we get t = 19.6/9.8 = 2 s.
MCQ No. 112
The total time of flight for the projectile in MCQ 111 is:
a. 2 s
b. 3 s
c. 4 s
d. 5 s
Correct Answer: c. 4 s
Explanation:
The time of ascent equals the time of descent. Therefore, total time = 2 × 2 = 4 s.
MCQ No. 113
A projectile is launched at 45° with an initial speed of 20 m/s. Neglecting air resistance, the horizontal range is approximately:
a. 20.4 m
b. 30.6 m
c. 40.8 m
d. 50.2 m
Correct Answer: c. 40.8 m
Explanation:
Using R = u²/g, we get R = 400/9.8 ≈ 40.8 m for a launch angle of 45°.
MCQ No. 114
A 5 kg body moving at 8 m/s has a momentum of:
a. 20 kg·m/s
b. 30 kg·m/s
c. 40 kg·m/s
d. 45 kg·m/s
Correct Answer: c. 40 kg·m/s
Explanation:
Momentum is calculated using p = mv = 5 × 8 = 40 kg·m/s.
MCQ No. 115
A force of 50 N acts for 0.2 s. The impulse is:
a. 5 N·s
b. 10 N·s
c. 15 N·s
d. 20 N·s
Correct Answer: b. 10 N·s
Explanation:
Impulse = Force × Time = 50 × 0.2 = 10 N·s.
MCQ No. 116
A 2 kg object accelerates at 6 m/s². The applied force is:
a. 6 N
b. 8 N
c. 10 N
d. 12 N
Correct Answer: d. 12 N
Explanation:
Using F = ma, we get F = 2 × 6 = 12 N.
MCQ No. 117
A 60 kg person weighs approximately:
a. 588 N
b. 600 N
c. 620 N
d. 680 N
Correct Answer: a. 588 N
Explanation:
Weight is W = mg = 60 × 9.8 = 588 N.
MCQ No. 118
A body moving at 15 m/s is brought to rest by a constant force. Which quantity must decrease continuously?
a. Mass
b. Momentum
c. Weight
d. Density
Correct Answer: b. Momentum
Explanation:
Since momentum depends on velocity, it decreases continuously as the object slows down.
MCQ No. 119
Two equal masses collide head-on with equal speeds and stick together. Their final velocity is:
a. Equal to the initial speed
b. Half the initial speed
c. Double the initial speed
d. Zero
Correct Answer: d. Zero
Explanation:
The two momenta are equal in magnitude but opposite in direction, giving a total momentum of zero. Therefore, the combined body remains at rest.
MCQ No. 120
A body moving with constant velocity experiences:
a. Constant acceleration
b. Zero net force
c. Increasing momentum
d. Variable mass
Correct Answer: b. Zero net force
Explanation:
According to Newton's First Law, constant velocity means the resultant external force acting on the body is zero.
MCQ No. 121
If the mass of a body is doubled while the applied force remains constant, its acceleration becomes:
a. Double
b. Half
c. Four times
d. Unchanged
Correct Answer: b. Half
Explanation:
From F = ma, acceleration is inversely proportional to mass for a constant force.
MCQ No. 122
The momentum of a body becomes zero when its:
a. Mass becomes zero
b. Velocity becomes zero
c. Weight becomes zero
d. Acceleration becomes zero
Correct Answer: b. Velocity becomes zero
Explanation:
Momentum is the product of mass and velocity. If velocity is zero, the momentum is also zero.
MCQ No. 123
A rocket moves forward because:
a. Gravity pulls it upward.
b. Air pushes it upward.
c. Exhaust gases are expelled backward.
d. Its mass continuously increases.
Correct Answer: c. Exhaust gases are expelled backward.
Explanation:
Rocket propulsion is an application of Newton's Third Law. The backward expulsion of gases produces an equal and opposite forward thrust.
MCQ No. 124
The quantity represented by the product of mass and velocity is:
a. Force
b. Impulse
c. Momentum
d. Power
Correct Answer: c. Momentum
Explanation:
Momentum is a vector quantity defined as the product of mass and velocity. It indicates the quantity of motion possessed by a body.
MCQ No. 125
Which of the following equations is most suitable for finding the final velocity when displacement, acceleration, and initial velocity are known?
a. v = u + at
b. s = ut + ½at²
c. v² = u² + 2as
d. a = (v − u)/t
Correct Answer: c. v² = u² + 2as
Explanation:
The equation v² = u² + 2as directly relates final velocity, initial velocity, displacement, and acceleration without involving time. It is particularly useful when the time of motion is unknown.
MCQ No. 126
A car moving at 15 m/s accelerates uniformly at 3 m/s² for 4 s. The distance covered during this time is:
a. 72 m
b. 84 m
c. 96 m
d. 108 m
Correct Answer: b. 84 m
Explanation:
Using s = ut + ½at², where u = 15 m/s, a = 3 m/s², and t = 4 s:
s = (15 × 4) + ½ × 3 × 16 = 60 + 24 = 84 m.
MCQ No. 127
A body moving with an initial velocity of 5 m/s reaches 25 m/s after travelling 150 m. Its acceleration is:
a. 1.5 m/s²
b. 2.0 m/s²
c. 2.5 m/s²
d. 3.0 m/s²
Correct Answer: b. 2.0 m/s²
Explanation:
Using v² = u² + 2as:
25² = 5² + 2a(150)
625 = 25 + 300a
a = 2 m/s².
MCQ No. 128
A train moving at 30 m/s comes to rest with a uniform retardation of 2.5 m/s². The time taken to stop is:
a. 10 s
b. 12 s
c. 15 s
d. 18 s
Correct Answer: b. 12 s
Explanation:
Using v = u + at:
0 = 30 − 2.5t
Therefore, t = 12 s.
MCQ No. 129
A body starts from rest and travels 98 m in 7 s with uniform acceleration. The acceleration is:
a. 2 m/s²
b. 3 m/s²
c. 4 m/s²
d. 5 m/s²
Correct Answer: c. 4 m/s²
Explanation:
Using s = ut + ½at², where u = 0:
98 = ½a(49)
98 = 24.5a
a = 4 m/s².
MCQ No. 130
A projectile is launched vertically upward with an initial speed of 29.4 m/s. Its maximum height is approximately:
a. 22.05 m
b. 33.15 m
c. 44.1 m
d. 55.2 m
Correct Answer: c. 44.1 m
Explanation:
Using h = u²/2g:
h = (29.4)²/(2 × 9.8) = 44.1 m approximately.
MCQ No. 131
A projectile remains in the air for 8 s. The time taken to reach its maximum height is:
a. 2 s
b. 4 s
c. 6 s
d. 8 s
Correct Answer: b. 4 s
Explanation:
For a projectile landing at the same level, the ascent time equals the descent time. Therefore, the time to reach maximum height is 8/2 = 4 s.
MCQ No. 132
A projectile is launched with a speed of 40 m/s at an angle of 45°. The approximate horizontal range is:
a. 81.6 m
b. 122.4 m
c. 163.2 m
d. 200 m
Correct Answer: c. 163.2 m
Explanation:
Using R = u²/g for 45°:
R = 40²/9.8 = 1600/9.8 ≈ 163.2 m.
MCQ No. 133
A force of 100 N acts on a body for 0.15 s. The impulse produced is:
a. 10 N·s
b. 12 N·s
c. 15 N·s
d. 18 N·s
Correct Answer: c. 15 N·s
Explanation:
Impulse = Force × Time
= 100 × 0.15 = 15 N·s.
MCQ No. 134
A 4 kg body moving at 12 m/s collides with a wall and comes to rest. The change in momentum is:
a. 24 kg·m/s
b. 36 kg·m/s
c. 48 kg·m/s
d. 60 kg·m/s
Correct Answer: c. 48 kg·m/s
Explanation:
Initial momentum = 4 × 12 = 48 kg·m/s.
Final momentum = 0.
Therefore, the change in momentum is 48 kg·m/s.
MCQ No. 135
A 3 kg object accelerates at 5 m/s². The net force acting on it is:
a. 10 N
b. 12 N
c. 15 N
d. 18 N
Correct Answer: c. 15 N
Explanation:
According to Newton's Second Law:
F = ma = 3 × 5 = 15 N.
MCQ No. 136
The momentum of a 1200 kg car moving at 20 m/s is:
a. 12,000 kg·m/s
b. 18,000 kg·m/s
c. 24,000 kg·m/s
d. 30,000 kg·m/s
Correct Answer: c. 24,000 kg·m/s
Explanation:
Momentum = mv = 1200 × 20 = 24,000 kg·m/s.
MCQ No. 137
A 50 kg person standing on Earth has a weight of approximately:
a. 390 N
b. 450 N
c. 490 N
d. 540 N
Correct Answer: c. 490 N
Explanation:
Weight is calculated using W = mg.
W = 50 × 9.8 = 490 N.
MCQ No. 138
Two bodies of equal mass move with equal speeds in opposite directions. The total momentum of the system is:
a. Equal to the momentum of one body
b. Twice the momentum of one body
c. Zero
d. Infinite
Correct Answer: c. Zero
Explanation:
Since the two momenta are equal in magnitude but opposite in direction, they cancel each other. Thus, the total momentum of the system is zero.
MCQ No. 139
A body of mass 8 kg experiences a force of 40 N. Its acceleration is:
a. 3 m/s²
b. 4 m/s²
c. 5 m/s²
d. 6 m/s²
Correct Answer: c. 5 m/s²
Explanation:
Using F = ma:
a = 40/8 = 5 m/s².
MCQ No. 140
The graph whose slope represents acceleration is the:
a. Distance-time graph
b. Displacement-time graph
c. Velocity-time graph
d. Force-time graph
Correct Answer: c. Velocity-time graph
Explanation:
Acceleration is defined as the rate of change of velocity. Therefore, the slope of a velocity-time graph represents acceleration.
MCQ No. 141
A 0.5 kg football moving at 16 m/s has a momentum of:
a. 4 kg·m/s
b. 6 kg·m/s
c. 8 kg·m/s
d. 10 kg·m/s
Correct Answer: c. 8 kg·m/s
Explanation:
Momentum = mv = 0.5 × 16 = 8 kg·m/s.
MCQ No. 142
Which physical quantity is represented by the area under a force-time graph?
a. Velocity
b. Acceleration
c. Impulse
d. Momentum
Correct Answer: c. Impulse
Explanation:
The area under a force-time graph equals the impulse delivered to the object. This impulse is equal to the change in momentum.
MCQ No. 143
A 10 kg object moving at 6 m/s has a kinetic energy of:
a. 120 J
b. 180 J
c. 240 J
d. 360 J
Correct Answer: b. 180 J
Explanation:
Using KE = ½mv²:
KE = ½ × 10 × 6² = 5 × 36 = 180 J.
MCQ No. 144
If the momentum of a body is doubled while its mass remains constant, its velocity becomes:
a. Half
b. Double
c. Four times
d. Unchanged
Correct Answer: b. Double
Explanation:
Since p = mv, if the mass remains constant, doubling the momentum doubles the velocity.
MCQ No. 145
A 2 kg body moving at 10 m/s collides elastically with a stationary body. Which quantity is certainly conserved?
a. Potential energy only
b. Momentum only
c. Both momentum and kinetic energy
d. Force only
Correct Answer: c. Both momentum and kinetic energy
Explanation:
In a perfectly elastic collision, both momentum and kinetic energy remain conserved throughout the interaction.
MCQ No. 146
The recoil velocity of a gun is mainly a consequence of:
a. Newton's First Law
b. Conservation of Energy
c. Conservation of Momentum
d. Law of Gravitation
Correct Answer: c. Conservation of Momentum
Explanation:
The forward momentum of the bullet is balanced by the backward momentum of the gun, ensuring the total momentum remains conserved.
MCQ No. 147
A satellite orbiting the Earth is continuously accelerating because:
a. Its speed is increasing.
b. Its direction of motion is continuously changing.
c. Its mass is changing.
d. Gravity is absent.
Correct Answer: b. Its direction of motion is continuously changing.
Explanation:
Even if the speed remains constant, a continuous change in direction means the velocity changes. Since acceleration is the rate of change of velocity, the satellite is continuously accelerating.
MCQ No. 148
The launch angles 30° and 60° produce:
a. Different times of flight but the same range
b. The same maximum height
c. The same range for the same initial speed
d. The same horizontal velocity
Correct Answer: c. The same range for the same initial speed
Explanation:
Complementary angles (θ and 90° − θ) produce the same horizontal range for identical launch speeds when air resistance is neglected.
MCQ No. 149
A body moving with constant speed in a circular path possesses:
a. Zero acceleration
b. Constant velocity
c. Centripetal acceleration
d. Constant displacement
Correct Answer: c. Centripetal acceleration
Explanation:
Although the speed remains constant, the direction of velocity changes continuously. Therefore, the body has centripetal acceleration directed toward the center of the circle.
MCQ No. 150
A 1500 kg car moving at 20 m/s is brought to rest uniformly in 5 s. The average braking force is:
a. 3000 N
b. 4500 N
c. 6000 N
d. 7500 N
Correct Answer: c. 6000 N
Explanation:
First, calculate the acceleration:
a = (0 − 20)/5 = −4 m/s².
Using F = ma:
F = 1500 × (−4) = −6000 N.
The negative sign indicates that the force acts opposite to the direction of motion. The magnitude of the braking force is 6000 N.
MCQ No. 151
A body starts from rest and accelerates uniformly at 5 m/s². The distance travelled in the first 6 s is:
a. 60 m
b. 75 m
c. 90 m
d. 105 m
Correct Answer: c. 90 m
Explanation:
Using the equation s = ut + ½at², where u = 0, a = 5 m/s², and t = 6 s:
s = 0 + ½ × 5 × 6² = 2.5 × 36 = 90 m.
MCQ No. 152
A car moving at 18 m/s accelerates uniformly at 2 m/s² for 8 s. The final velocity is:
a. 30 m/s
b. 32 m/s
c. 34 m/s
d. 36 m/s
Correct Answer: c. 34 m/s
Explanation:
Using v = u + at:
v = 18 + (2 × 8) = 34 m/s. The velocity increases because the acceleration acts in the direction of motion.
MCQ No. 153
A body moving with an initial velocity of 8 m/s accelerates uniformly at 4 m/s². How far will it travel in 5 s?
a. 70 m
b. 80 m
c. 90 m
d. 100 m
Correct Answer: c. 90 m
Explanation:
Using s = ut + ½at²:
s = (8 × 5) + ½ × 4 × 25 = 40 + 50 = 90 m.
MCQ No. 154
A vehicle slows from 25 m/s to 5 m/s in 10 s. Its acceleration is:
a. –1 m/s²
b. –2 m/s²
c. –3 m/s²
d. –4 m/s²
Correct Answer: b. –2 m/s²
Explanation:
Using a = (v − u)/t:
a = (5 − 25)/10 = −20/10 = −2 m/s². The negative sign indicates deceleration.
MCQ No. 155
A projectile is projected vertically upward with an initial speed of 39.2 m/s. The maximum height reached is approximately:
a. 58.8 m
b. 68.6 m
c. 78.4 m
d. 88.2 m
Correct Answer: c. 78.4 m
Explanation:
Using h = u²/2g:
h = (39.2)²/(2 × 9.8) = 78.4 m. At this height, the vertical velocity becomes zero.
MCQ No. 156
A projectile is launched vertically upward with a speed of 24.5 m/s. The total time of flight is:
a. 2 s
b. 3 s
c. 5 s
d. 6 s
Correct Answer: c. 5 s
Explanation:
Time to reach maximum height is t = u/g = 24.5/9.8 = 2.5 s. Therefore, the total time of flight is 2 × 2.5 = 5 s.
MCQ No. 157
A projectile is launched with an initial speed of 28 m/s at an angle of 45°. The approximate horizontal range is:
a. 60 m
b. 70 m
c. 80 m
d. 90 m
Correct Answer: c. 80 m
Explanation:
For a launch angle of 45°, R = u²/g.
R = 28²/9.8 = 784/9.8 = 80 m approximately.
MCQ No. 158
A force of 250 N acts on a body for 0.08 s. The impulse delivered is:
a. 10 N·s
b. 15 N·s
c. 20 N·s
d. 25 N·s
Correct Answer: c. 20 N·s
Explanation:
Impulse is calculated using Impulse = Force × Time.
Therefore, Impulse = 250 × 0.08 = 20 N·s.
MCQ No. 159
A 6 kg object moving at 10 m/s is brought to rest. The magnitude of the change in momentum is:
a. 30 kg·m/s
b. 40 kg·m/s
c. 50 kg·m/s
d. 60 kg·m/s
Correct Answer: d. 60 kg·m/s
Explanation:
Initial momentum = 6 × 10 = 60 kg·m/s. Since the final momentum is zero, the change in momentum is 60 kg·m/s.
MCQ No. 160
A 4 kg object experiences a net force of 28 N. Its acceleration is:
a. 5 m/s²
b. 6 m/s²
c. 7 m/s²
d. 8 m/s²
Correct Answer: c. 7 m/s²
Explanation:
Using F = ma:
a = F/m = 28/4 = 7 m/s².
MCQ No. 161
The momentum of a 1500 kg truck moving at 12 m/s is:
a. 12,000 kg·m/s
b. 15,000 kg·m/s
c. 18,000 kg·m/s
d. 20,000 kg·m/s
Correct Answer: c. 18,000 kg·m/s
Explanation:
Momentum is given by p = mv.
p = 1500 × 12 = 18,000 kg·m/s.
MCQ No. 162
A 70 kg person has a weight of approximately:
a. 588 N
b. 686 N
c. 700 N
d. 784 N
Correct Answer: b. 686 N
Explanation:
Weight is calculated using W = mg.
W = 70 × 9.8 = 686 N.
MCQ No. 163
The area under a force–time graph represents:
a. Work done
b. Momentum
c. Impulse
d. Power
Correct Answer: c. Impulse
Explanation:
The area under a force–time graph gives the impulse applied to an object. Impulse is equal to the change in momentum.
MCQ No. 164
A 3 kg object moving at 4 m/s has kinetic energy equal to:
a. 12 J
b. 18 J
c. 24 J
d. 36 J
Correct Answer: c. 24 J
Explanation:
Using KE = ½mv²:
KE = ½ × 3 × 4² = 1.5 × 16 = 24 J.
MCQ No. 165
If the mass of an object is tripled while its velocity remains unchanged, its momentum becomes:
a. One-third
b. Double
c. Triple
d. Four times
Correct Answer: c. Triple
Explanation:
Momentum is directly proportional to mass when velocity is constant. Therefore, tripling the mass triples the momentum.
MCQ No. 166
A 1500 kg car accelerates at 2 m/s². The net force acting on it is:
a. 1500 N
b. 2000 N
c. 2500 N
d. 3000 N
Correct Answer: d. 3000 N
Explanation:
Applying Newton's Second Law:
F = ma = 1500 × 2 = 3000 N.
MCQ No. 167
A body moving with constant velocity has:
a. Constant net force
b. Zero resultant force
c. Increasing momentum
d. Constant acceleration
Correct Answer: b. Zero resultant force
Explanation:
If the velocity remains constant, there is no acceleration. According to Newton's First Law, the net external force must therefore be zero.
MCQ No. 168
During an explosion in space, the total momentum of the system:
a. Increases
b. Decreases
c. Remains conserved
d. Becomes zero
Correct Answer: c. Remains conserved
Explanation:
In the absence of external forces, momentum is always conserved. Although the fragments move in different directions, the total momentum remains unchanged.
MCQ No. 169
Two balls collide perfectly elastically. Which quantity is conserved in addition to momentum?
a. Temperature
b. Kinetic energy
c. Force
d. Pressure
Correct Answer: b. Kinetic energy
Explanation:
A perfectly elastic collision conserves both momentum and kinetic energy. This distinguishes it from inelastic collisions.
MCQ No. 170
A ball rebounds after striking a wall. Compared with simply stopping, the impulse experienced by the ball is:
a. Smaller
b. The same
c. Greater
d. Zero
Correct Answer: c. Greater
Explanation:
When a ball rebounds, its momentum changes direction. This produces a larger change in momentum than merely coming to rest, resulting in a greater impulse.
MCQ No. 171
A projectile launched at 60° and another at 30° with the same speed will have:
a. The same maximum height
b. The same time of flight
c. The same horizontal range
d. The same vertical velocity throughout
Correct Answer: c. The same horizontal range
Explanation:
Complementary launch angles produce the same horizontal range for identical initial speeds, provided the projectile lands at the same level and air resistance is neglected.
MCQ No. 172
The direction of acceleration of a freely falling object is always:
a. Upward
b. Along its velocity
c. Vertically downward
d. Horizontal
Correct Answer: c. Vertically downward
Explanation:
Gravity always acts toward the center of the Earth. Therefore, the acceleration due to gravity is vertically downward regardless of whether the object is moving upward or downward.
MCQ No. 173
The slope of a displacement–time graph at a particular instant gives the:
a. Average speed
b. Instantaneous velocity
c. Average acceleration
d. Instantaneous acceleration
Correct Answer: b. Instantaneous velocity
Explanation:
The slope of the tangent drawn to a displacement–time graph at a given point represents the instantaneous velocity at that instant.
MCQ No. 174
Which of the following situations best demonstrates Newton's Third Law?
a. A book resting on a table
b. A satellite orbiting the Earth
c. A swimmer pushing water backward to move forward
d. An apple falling from a tree
Correct Answer: c. A swimmer pushing water backward to move forward
Explanation:
The swimmer pushes the water backward, and the water exerts an equal and opposite force that propels the swimmer forward. This is a direct application of Newton's Third Law.
MCQ No. 175
A force acts on a body without changing its speed but changes only its direction of motion. This means the force is:
a. Zero
b. Producing acceleration
c. Conserving momentum
d. Conserving velocity
Correct Answer: b. Producing acceleration
Explanation:
Acceleration is the rate of change of velocity, and velocity depends on both magnitude and direction. Even if the speed remains constant, a change in direction means the velocity changes, so the body is accelerating.
MCQ No. 176
A body starts from rest and accelerates uniformly at 6 m/s² for 5 s. The distance travelled is:
a. 50 m
b. 60 m
c. 75 m
d. 90 m
Correct Answer: c. 75 m
Explanation:
Using the equation s = ut + ½at², where u = 0, a = 6 m/s², and t = 5 s:
s = 0 + ½ × 6 × 25 = 75 m. Since the body starts from rest, the entire displacement is due to its uniform acceleration.
MCQ No. 177
A car increases its speed from 12 m/s to 36 m/s in 8 s. Its acceleration is:
a. 2 m/s²
b. 3 m/s²
c. 4 m/s²
d. 5 m/s²
Correct Answer: b. 3 m/s²
Explanation:
Using a = (v − u)/t, we get:
a = (36 − 12)/8 = 24/8 = 3 m/s². This indicates the car gains 3 m/s of velocity every second.
MCQ No. 178
A body moving at 20 m/s is brought to rest with a uniform deceleration of 4 m/s². The stopping distance is:
a. 40 m
b. 50 m
c. 60 m
d. 70 m
Correct Answer: b. 50 m
Explanation:
Using v² = u² + 2as:
0 = 20² + 2(−4)s
0 = 400 − 8s
s = 50 m. The negative acceleration represents a retarding force.
MCQ No. 179
A projectile is launched vertically upward with an initial speed of 49 m/s. The time taken to reach the highest point is:
a. 2 s
b. 3 s
c. 4 s
d. 5 s
Correct Answer: d. 5 s
Explanation:
The time to reach maximum height is given by t = u/g.
t = 49/9.8 = 5 s. At the highest point, the vertical velocity becomes zero.
MCQ No. 180
A projectile is fired at 45° with a speed of 49 m/s. Neglecting air resistance, its horizontal range is approximately:
a. 196 m
b. 220 m
c. 245 m
d. 260 m
Correct Answer: c. 245 m
Explanation:
For a launch angle of 45°, the range is maximum and is calculated using R = u²/g.
R = 49²/9.8 = 2401/9.8 = 245 m.
MCQ No. 181
A 5 kg object moving at 8 m/s collides with a stationary object and comes to rest. The magnitude of the change in momentum is:
a. 20 kg·m/s
b. 30 kg·m/s
c. 40 kg·m/s
d. 50 kg·m/s
Correct Answer: c. 40 kg·m/s
Explanation:
Initial momentum = 5 × 8 = 40 kg·m/s.
Since the final momentum of the object is zero, the change in momentum is 40 kg·m/s.
MCQ No. 182
A force of 150 N acts on an object for 0.4 s. The impulse delivered is:
a. 40 N·s
b. 50 N·s
c. 60 N·s
d. 70 N·s
Correct Answer: c. 60 N·s
Explanation:
Impulse is calculated using Impulse = Force × Time.
Therefore,
Impulse = 150 × 0.4 = 60 N·s. This impulse is equal to the change in momentum.
MCQ No. 183
A 10 kg body experiences a force of 50 N. Its acceleration is:
a. 2 m/s²
b. 3 m/s²
c. 4 m/s²
d. 5 m/s²
Correct Answer: d. 5 m/s²
Explanation:
Using Newton's Second Law,
F = ma
a = 50/10 = 5 m/s². The acceleration is directly proportional to the applied force.
MCQ No. 184
The momentum of a 2000 kg truck moving at 15 m/s is:
a. 20,000 kg·m/s
b. 25,000 kg·m/s
c. 30,000 kg·m/s
d. 35,000 kg·m/s
Correct Answer: c. 30,000 kg·m/s
Explanation:
Momentum is given by
p = mv = 2000 × 15 = 30,000 kg·m/s. Heavy vehicles possess large momentum because of their large mass.
MCQ No. 185
A 75 kg person has a weight of approximately:
a. 686 N
b. 706 N
c. 735 N
d. 784 N
Correct Answer: c. 735 N
Explanation:
Weight is calculated using W = mg.
W = 75 × 9.8 = 735 N. Weight depends on the gravitational field, whereas mass remains constant.
MCQ No. 186
The slope of a velocity–time graph gives:
a. Displacement
b. Velocity
c. Acceleration
d. Momentum
Correct Answer: c. Acceleration
Explanation:
Acceleration is the rate of change of velocity with time. Therefore, the slope of a velocity–time graph represents acceleration. A straight-line graph indicates uniform acceleration.
MCQ No. 187
The area under a velocity–time graph represents:
a. Force
b. Impulse
c. Displacement
d. Acceleration
Correct Answer: c. Displacement
Explanation:
The area enclosed by a velocity–time graph gives the displacement of the object. This relationship is widely used in graph-based numerical problems.
MCQ No. 188
A 4 kg object moving at 5 m/s has a kinetic energy of:
a. 25 J
b. 40 J
c. 50 J
d. 75 J
Correct Answer: c. 50 J
Explanation:
Using KE = ½mv²:
KE = ½ × 4 × 25 = 2 × 25 = 50 J. Kinetic energy depends on both the mass and the square of the velocity.
MCQ No. 189
If the velocity of a body is doubled while its mass remains constant, its momentum becomes:
a. Half
b. Double
c. Four times
d. Eight times
Correct Answer: b. Double
Explanation:
Momentum is given by p = mv. Since mass remains unchanged, doubling the velocity doubles the momentum.
MCQ No. 190
Two bodies collide perfectly inelastically. Which quantity is always conserved?
a. Kinetic energy
b. Momentum
c. Mechanical energy
d. Speed
Correct Answer: b. Momentum
Explanation:
During a perfectly inelastic collision, the colliding bodies stick together and move with a common velocity. Although kinetic energy decreases, the total momentum remains conserved if no external force acts.
MCQ No. 191
A rocket accelerates upward because:
a. Gravity pushes it upward.
b. Air pressure lifts it.
c. Hot gases are expelled downward.
d. Its mass becomes zero.
Correct Answer: c. Hot gases are expelled downward.
Explanation:
Rocket propulsion is based on Newton's Third Law of Motion. As gases are expelled downward at high speed, an equal and opposite reaction force pushes the rocket upward.
MCQ No. 192
An object moving in a circular path at constant speed has:
a. Zero acceleration
b. Constant velocity
c. Centripetal acceleration directed toward the center
d. No net force
Correct Answer: c. Centripetal acceleration directed toward the center
Explanation:
Although the speed remains constant, the direction of velocity changes continuously. This change in velocity requires a centripetal acceleration directed toward the center of the circular path.
MCQ No. 193
Which launch angle produces the maximum horizontal range for a projectile on level ground?
a. 30°
b. 45°
c. 60°
d. 90°
Correct Answer: b. 45°
Explanation:
For a projectile launched and landing at the same height without air resistance, the maximum horizontal range is achieved at a launch angle of 45°.
MCQ No. 194
A body moving with constant velocity experiences:
a. Constant acceleration
b. Zero resultant force
c. Increasing momentum
d. Variable mass
Correct Answer: b. Zero resultant force
Explanation:
According to Newton's First Law, a body continues moving with constant velocity if no unbalanced external force acts on it. Therefore, the resultant force is zero.
MCQ No. 195
A 2 kg object moving at 15 m/s collides elastically with another object. Which statement is always true?
a. Momentum is conserved but kinetic energy is lost.
b. Only kinetic energy is conserved.
c. Both momentum and kinetic energy are conserved.
d. Both momentum and kinetic energy are lost.
Correct Answer: c. Both momentum and kinetic energy are conserved.
Explanation:
A perfectly elastic collision conserves both momentum and kinetic energy. This property distinguishes elastic collisions from inelastic ones.
MCQ No. 196
A passenger standing in a moving bus falls backward when the bus suddenly starts because of:
a. Gravity
b. Friction
c. Inertia of rest
d. Air resistance
Correct Answer: c. Inertia of rest
Explanation:
When the bus suddenly starts, the passenger's feet move with the bus, but the upper body tends to remain at rest due to inertia. As a result, the passenger appears to fall backward.
MCQ No. 197
The momentum of an isolated system before and after a collision is:
a. Different
b. Zero
c. Equal
d. Infinite
Correct Answer: c. Equal
Explanation:
According to the law of conservation of momentum, the total momentum of an isolated system remains the same before and after any interaction or collision.
MCQ No. 198
A body of mass 3 kg accelerates at 8 m/s². The force acting on it is:
a. 18 N
b. 21 N
c. 24 N
d. 30 N
Correct Answer: c. 24 N
Explanation:
Using F = ma:
F = 3 × 8 = 24 N. A greater acceleration requires a greater force if the mass remains constant.
MCQ No. 199
A stone is thrown vertically upward. At its highest point, which statement is correct?
a. Velocity and acceleration are both zero.
b. Velocity is zero but acceleration is 9.8 m/s² downward.
c. Both velocity and acceleration are upward.
d. Velocity is maximum.
Correct Answer: b. Velocity is zero but acceleration is 9.8 m/s² downward.
Explanation:
At the highest point, the stone momentarily stops before descending, so its velocity becomes zero. However, gravity continues to act downward with an acceleration of 9.8 m/s² throughout the motion.
MCQ No. 200
Which statement best summarizes the Law of Conservation of Momentum?
a. Momentum is conserved only in elastic collisions.
b. Momentum is conserved only when kinetic energy is conserved.
c. The total momentum of an isolated system remains constant regardless of the type of interaction.
d. Momentum is conserved only if the masses of the objects are equal.
Correct Answer: c. The total momentum of an isolated system remains constant regardless of the type of interaction.
Explanation:
The law of conservation of momentum applies to all isolated systems, including elastic collisions, inelastic collisions, and explosions. As long as no external force acts on the system, the total momentum before and after the interaction remains constant.
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