100 High-Scoring MCQs (Set-1) on Work and Energy, Unit-4, Class 11 Physics (Unit-Wise Practice)



This post contains 100 carefully selected multiple-choice questions (MCQs) from Unit-4: Work and Energy of Class 11 Physics, designed strictly according to the FBISE syllabus. These MCQs include a well-balanced mix of numerical and conceptual problems, making them exam-ready, revision-friendly, and high-scoring.

Whether you are preparing for annual board exams, chapter tests, or competitive entry tests, this collection thoroughly covers all the key concepts and formulas of Work and Energy, helping students master the chapter with confidence.

This unit-wise MCQ set includes questions from:

  • Definition and concept of Work and Energy
  • Work done by a constant and variable force
  • Kinetic energy (KE) and potential energy (PE)
  • Work-Energy theorem
  • Power and its calculation
  • Conservation of mechanical energy
  • Elastic and inelastic collisions
  • Gravitational and spring potential energy
  • Impulse and momentum
  • Numerical applications combining work, energy, and motion
  • Real-life applications of energy concepts in mechanics

Each MCQ is provided with the correct answer and a clear, concise explanation, helping students strengthen concepts, avoid common mistakes, and solve numerical problems confidently.

Prepare smartly, revise efficiently, and boost your exam score in Unit-5: Work and Energy with these 100 high-yield MCQs.


MCQ 1
A net force of 6 N acts horizontally on a body (Vi = 0) of mass 2 kg for 4 s. Its kinetic energy will be:

a. 12 J
b. 144 J
c. 72 J
d. 48 J

Correct Answer: b. 144 J

Explanation:
F = ma ⇒ a = F/m = 6 / 2 = 3 m/s²
v = u + at = 0 + 3 × 4 = 12 m/s
K.E = ½ m v² = 0.5 × 2 × 12² = 144 J

MCQ 2
Dimension of gravitational potential is:

a. [L² T⁻²]
b. [M L⁻² T]
c. [M T⁻²]
d. None

Correct Answer: a. [L² T⁻²]

Explanation:
Gravitational potential (V) = mgh / m = gh ⇒ dimensions = [L² T⁻²]

MCQ 3
If momentum of an object is doubled, then its kinetic energy becomes:

a. Increase by 4 times
b. Decrease by 4 times
c. Increase by 2 times
d. Remains the same

Correct Answer: a. Increase by 4 times

Explanation:
p = mv ⇒ KE = p² / 2m. If p → 2p, KE = (2p)² / 2m = 4p² / 2m = 4 × KE

MCQ 4
For constant kinetic energy, the work done will be:

a. Zero
b. Constant
c. Variable
d. None

Correct Answer: a. Zero

Explanation:
Constant K.E implies constant velocity ⇒ net force = 0 ⇒ work done = 0

MCQ 5
A 200 N force is exerted on a body which moves with constant velocity for a distance of 100 m. The work done is:

a. Zero
b. 100 N
c. 2000 J
d. None

Correct Answer: a. Zero

Explanation:
W = F·S, but F_net = 0 (constant velocity) ⇒ W = 0

MCQ 6
If a body of mass 10 kg is lifted through a vertical height of 10 m, the work done (in kJ) will be:

a. 980 kJ
b. 98 kJ
c. 0.98 kJ
d. 980 × 10³ kJ

Correct Answer: c. 0.98 kJ

Explanation:
W = m g h = 10 × 9.8 × 10 = 980 J = 0.98 kJ

MCQ 7
The dot product of force and velocity is:

a. Power
b. Work
c. Impulse
d. Momentum

Correct Answer: a. Power

Explanation:
P = F · v

MCQ 8
If work is done at the rate of 1000 J/s, its power is:

a. 1 W
b. 1 kW
c. 1 MW
d. 100 W

Correct Answer: b. 1 kW

Explanation:
P = W/t = 1000 J / 1 s = 1000 W = 1 kW

MCQ 9
One horse power is equal to:

a. 550 W
b. 550 kW
c. 746 W
d. None

Correct Answer: c. 746 W

Explanation:
1 HP = 746 W

MCQ 10
Negative work is done when the angle between F and S is:

a. θ < 90°
b. 90° < θ ≤ 180°
c. 180° < θ < 270°
d. Both b & c

Correct Answer: b. 90° < θ ≤ 180°

Explanation:
W = F·S·cosθ. Work is negative when cosθ < 0 ⇒ 90° < θ ≤ 180°

MCQ 11
A light and heavy body have equal K.E. Which one has greater momentum?

a. Both equal
b. Light body
c. Heavy body
d. Impossible to say

Correct Answer: b. Light body

Explanation:
Momentum p = √(2 m KE). For equal KE, larger mass → smaller velocity → smaller momentum.

MCQ 12
An object of mass 100 g is whirled in a horizontal circle of radius 1 m at constant speed of 20 m/s. The work done by the tension during one revolution is:

a. 0
b. 100 J
c. 200 J
d. 400 J

Correct Answer: a. 0

Explanation:
W = F·S·cosθ. Tension acts perpendicular to displacement ⇒ θ = 90° ⇒ cos90° = 0 ⇒ W = 0

MCQ 13
1 kg mass has K.E of 1 J. Its speed is:

a. 1 m/s
b. 4.4 m/s
c. 1.4 m/s
d. 2.5 m/s

Correct Answer: c. 1.4 m/s

Explanation:
K.E = ½ m v² ⇒ v = √(2 K.E / m) = √(2 × 1 / 1) = 1.414 m/s ≈ 1.4 m/s

MCQ 14
Two bodies with kinetic energy in the ratio 4:1 have equal linear momentum. The ratio of their masses is:

a. 1:4
b. 4:1
c. 1:2
d. 1:1

Correct Answer: a. 1:4

Explanation:
For equal momentum p = mv, KE = p² / 2m ⇒ m1 / m2 = KE2 / KE1 = 1 / 4

MCQ 15
The work done by the centripetal force on an object is:

a. 0
b. Positive
c. Negative
d. All

Correct Answer: a. 0

Explanation:
Centripetal force is perpendicular to displacement ⇒ W = F·S·cos90° = 0

MCQ 16
Work done by a conservative force along a closed path is:

a. 0
b. Maximum
c. Positive
d. None

Correct Answer: a. 0

Explanation:
By definition, work done by conservative force in a closed loop = 0

MCQ 17
A 50 N force is applied on a body making an angle of 30° with the +X-axis, which moves it through displacement of 10 m on X-axis. Work done is:

a. 0
b. 500 J
c. 433.01 J
d. None

Correct Answer: c. 433.01 J

Explanation:
W = F·S·cosθ = 50 × 10 × cos30° ≈ 433.01 J

MCQ 18
When a falling object of mass m moves with terminal velocity, the work done by gravitational force is:

a. 0
b. Negative
c. mgh
d. None

Correct Answer: a. 0

Explanation:
At terminal velocity, F_net = 0 ⇒ acceleration = 0 ⇒ no net work is done

MCQ 19
The work done by gravity on a pendulum is:

a. Positive
b. Negative
c. Zero
d. None

Correct Answer: c. Zero

Explanation:
For complete oscillation, net work by gravity over one cycle = 0

MCQ 20
One erg =

a. 1 g·cm²·s⁻²
b. 1 g·cm·s⁻²
c. 1 g·m²·s⁻¹
d. 1 g·cm²·s

Correct Answer: a. 1 g·cm²·s⁻²

Explanation:
By definition, 1 erg = 1 g·cm²·s⁻²

MCQ 21
Which of the objects has greater K.E?

a. mass 3M, velocity V
b. mass 2M, velocity V/3
c. mass 3M, velocity V/2
d. mass M, velocity 4V

Correct Answer: d. mass M, velocity 4V

Explanation:
K.E = ½ m v². Maximum K.E occurs for the largest v², i.e., ½ M × (4V)² = 8 M V², which is larger than all others.

MCQ 22
1 erg =

a. 1 dyne·cm
b. 10⁷ J
c. 10⁴ kJ
d. 10⁻⁷ J

Correct Answer: d. 10⁻⁷ J

Explanation:
By definition, 1 erg = 10⁻⁷ J

MCQ 23
Escape velocity depends upon:

a. Radius of the planet
b. Mass of the object
c. Mass of the planet
d. Both a & c

Correct Answer: d. Both a & c

Explanation:
v_escape = √(2GM/R), depends on planet mass (M) and radius (R)

MCQ 24
One mega-watt-hour is equal to:

a. 36 × 10⁶ J
b. 36 × 10¹² J
c. 36 × 10⁹ J
d. 36 × 10⁸ J

Correct Answer: d. 36 × 10⁸ J

Explanation:
1 MWh = 1 MW × 3600 s = 10⁶ × 3600 = 3.6 × 10⁹ J = 36 × 10⁸ J

MCQ 25
Input energy given to an electric generator is 1000 J and it produces 100 J sound and 300 J heat. Its efficiency is:

a. 60%
b. 70%
c. 40%
d. 100%

Correct Answer: a. 60%

Explanation:
Useful energy = input – losses = 1000 – (100 + 300) = 600 J
Efficiency = (600 / 1000) × 100 = 60%

MCQ 26
If momentum of an object is increased by 20%, its K.E increases by:

a. 44%
b. 36%
c. 44% decrease
d. 36% decrease

Correct Answer: a. 44%

Explanation:
ΔKE% ≈ 2(Δp/p) + (Δp/p)² = 2×0.2 + 0.2² = 0.4 + 0.04 = 0.44 = 44%

MCQ 27
Two objects of mass A and B have same momentum. A has more K.E than B if:

a. Mass is more than B
b. Mass is less than B
c. Moving faster than B
d. Moving slower than B

Correct Answer: b. Mass is less than B

Explanation:
K.E = p² / 2m. Smaller mass → larger K.E for same momentum

MCQ 28
Two bodies of masses m and 4m have same K.E. The ratio of their linear momentum is:

a. 1:4
b. 4:1
c. 1:2
d. 2:1

Correct Answer: d. 2:1

Explanation:
p = √(2 m K.E). For m : 4m → p ratio = √(m) : √(4m) = 1 : 2

MCQ 29
A 6 kg block is released from rest 80 m above ground. When it has fallen 60 m, its K.E is:

a. 4800 J
b. 3528 J
c. 1200 J
d. 120 J

Correct Answer: b. 3528 J

Explanation:
K.E = m g h_fallen = 6 × 9.8 × 60 ≈ 3528 J

MCQ 30
The unit of power in British engineering system is:

a. Watt
b. kW
c. Horsepower
d. Joule

Correct Answer: c. Horsepower

Explanation:
Horsepower is the traditional British unit of power.

MCQ 31

If the speed of a body is doubled, its kinetic energy becomes:

a. Double
b. Half
c. Four times
d. Eight times

Correct Answer: c. Four times

Explanation:
K∝v2K \propto v^2, so doubling speed increases KE by 22=42^2 = 4.

MCQ 32
1 horsepower =

a. 746 kW
b. 746 MW
c. 746 W
d. 746 J

Correct Answer: c. 746 W

Explanation:
By definition, 1 HP = 746 W

MCQ 33
Work is said to be negative when F and displacement are:

a. Parallel
b. Anti-parallel
c. Perpendicular
d. None

Correct Answer: b. Anti-parallel

Explanation:
W = F·d·cosθ. θ = 180° ⇒ cos180 = –1 ⇒ W negative

MCQ 34
The rate of doing work at any instant is called:

a. Average power
b. Variable power
c. Constant power
d. Instantaneous power

Correct Answer: d. Instantaneous power

Explanation:
Instantaneous power P = dW/dt

MCQ 35
Efficiency of a fluorescent lamp is:

a. 20%
b. 30%
c. 40%
d. 100%

Correct Answer: b. 30%

Explanation:
Memory-based question from textbook; actual energy conversion efficiency is ~30%

MCQ 36
Work done will be negative when the angle is:

a. 90°
b. 45°
c. 120°
d. 0°

Correct Answer: c. 120°

Explanation:
cos120° < 0 ⇒ W negative

MCQ 37
If the K.E of 1 kg mass is 1 J, its velocity is:

a. 2 m/s
b. 1/√2 m/s
c. 4 m/s
d. √2 m/s

Correct Answer: d. √2 m/s

Explanation:
K.E = ½ m v² ⇒ v = √(2 × 1 / 1) = √2 m/s ≈ 1.414 m/s

MCQ 38
An example of non-conservative force is:

a. Elastic
b. Gravitational
c. Frictional
d. Magnetic

Correct Answer: c. Frictional

Explanation:
Work done against friction depends on path ⇒ non-conservative

MCQ 39
The dot product of force and velocity equals:

a. Power
b. Momentum
c. Work
d. Impulse

Correct Answer: a. Power

Explanation:
P = F·v

MCQ 40
Work done is maximum for which angle?

a. 0°
b. 45°
c. 90°
d. 180°

Correct Answer: a. 0°

Explanation:
W = F·d·cosθ; maximum when θ = 0°

MCQ 41
Expression for critical velocity of a satellite is:

a. v=2gRv = \sqrt{2 g R} 
b. v=gRv = \sqrt{g R}
c. v=gR/2v = g R / 2 
d. None of these

Correct Answer: b. v=gRv = \sqrt{g R}

Explanation:
For a satellite in circular orbit, v=GM/R=gRv = \sqrt{GM/R} = \sqrt{gR}

MCQ 42
A missile is fired with a speed of 98 m/s at 30° with the horizontal. The missile is airborne for:

a. 196 m
b. 122.5 m
c. 98 m
d. 2940 m

Correct Answer: b. 122.5 m

Explanation:
Time of flight T=2visin⁡θ/g=2×98×0.5/9.8≈10sT = 2 v_i \sin\theta / g = 2 × 98 × 0.5 / 9.8 ≈ 10 s
Vertical displacement = ½ g t² = ½ × 9.8 × 5² ≈ 122.5 m

MCQ 43
Two projectiles are in flight at the same time. The acceleration of one relative to the other is:

a. Always 9.8 m/s²
b. Can be horizontal
c. Can be as large as 19.8 m/s²
d. Zero

Correct Answer: d. Zero

Explanation:
Both experience the same gravitational acceleration. Relative acceleration = 0.

MCQ 44
What is the angle of projection where the range equals the maximum height?

a. tan⁻¹(1/4)
b. tan⁻¹(4)
c. tan⁻¹(1/2)
d. tan⁻¹(2)

Correct Answer: b. tan⁻¹(4)

Explanation:
Range R = v² sin2θ / g, Maximum height H = v² sin²θ / 2g. Equate R = H → θ = tan⁻¹4

MCQ 45
When a velocity-time graph is a straight line parallel to the time axis:

a. Acceleration is constant
b. Acceleration is zero
c. Acceleration is variable
d. Velocity is zero

Correct Answer: b. Acceleration is zero

Explanation:
Slope of v–t graph = acceleration. Horizontal line ⇒ slope = 0

MCQ 46
Two blocks of masses 10 kg and 3 kg are in contact and acted upon by a force of 40 N. Acceleration of 1 kg mass:

a. 40 m/s²
b. 10 m/s²
c. 30 m/s²
d. 50 m/s²

Correct Answer: b. 10 m/s²

Explanation:
a = F / (m1 + m2) = 40 / (10 + 3) ≈ 10 m/s²

MCQ 47
When a body is in motion, which always changes?

a. Velocity
b. Acceleration
c. Position vector
d. Momentum

Correct Answer: c. Position vector

Explanation:
Motion changes position continuously; velocity and momentum may remain constant.

MCQ 48
A bomber drops a bomb vertically above a target. It misses the target due to:

a. Vertical component of velocity
b. Gravity
c. X-component of velocity of bomber
d. None

Correct Answer: c. X-component of velocity of bomber

Explanation:
Horizontal motion of bomb continues due to inertia → misses target

MCQ 49
Elastic collision involves:

a. Loss of energy
b. Gain of energy
c. No relation between energy and collision
d. No gain, no loss of energy

Correct Answer: d. No gain, no loss of energy

Explanation:
Elastic collision conserves K.E.

MCQ 50
Time required by a projectile to reach the maximum height:

a. T=8H/gT = \sqrt{8 H / g}
b. T=4H/gT = \sqrt{4 H / g}
c. T=2H/gT = \sqrt{2 H / g}
d. T=H/gT = \sqrt{H / g}

Correct Answer: c. T=2H/gT = \sqrt{2 H / g}

Explanation:
T = Vi / g; H = ½ g T² → T = √(2H/g)

MCQ 51
If two cars move with velocities 10 m/s and 5 m/s in opposite directions, their relative velocity is:

a. 5 m/s
b. –5 m/s
c. 10 m/s
d. 15 m/s

Correct Answer: d. 15 m/s

Explanation:
Opposite direction: V_rel = V1 + V2 = 10 + 5 = 15 m/s

MCQ 52
Horizontal range of a projectile is determined by:

a. Angle of projection
b. Height
c. Mass
d. Speed and angle

Correct Answer: d. Speed and angle

Explanation:
R = v² sin2θ / g → depends on speed (v) and angle θ only

MCQ 53
Two bodies move toward each other with constant speeds; distance decreases at 6 m/s. If they move in the same direction, distance increases at 4 m/s. Their speeds are:

a. 5 m/s, 1 m/s
b. 6 m/s, 1 m/s
c. 3 m/s, 3 m/s
d. 4 m/s, 2 m/s

Correct Answer: a. 5 m/s, 1 m/s

Explanation:
v1 + v2 = 6, v1 – v2 = 4 → v1 = 5, v2 = 1

MCQ 54
Conditions under which equations of motion are valid:

a. One-dimensional motion
b. Uniform acceleration
c. Inertial frame
d. All of the above

Correct Answer: d. All of the above

Explanation:
Newton’s laws require 1D, uniform acceleration, inertial frame.

MCQ 55
2nd law of motion gives the definition of:

a. Momentum
b. Acceleration
c. Velocity
d. Both A and B

Correct Answer: b. Acceleration

Explanation:
F = m a → defines acceleration due to applied force

MCQ 56
Motion of a rocket in space follows conservation of:

a. Energy
b. Charge
c. Mass
d. Momentum

Correct Answer: d. Momentum

Explanation:
Rocket propulsion: initial p = final p; momentum conserved

MCQ 57
Quantities remaining constant in collision are:

a. Momentum, K.E, Temp
b. Momentum, K.E
c. Momentum, Temp
d. Momentum but neither K.E nor Temp

Correct Answer: d. Momentum but neither K.E nor Temp

Explanation:
Momentum is always conserved; K.E may not be in inelastic collision

MCQ 58
A fireman slides down a rope with breaking strength ¾ of his weight. Minimum acceleration:

a. 3/4 g
b. 1/2 g
c. 1/4 g
d. 0

Correct Answer: c. 1/4 g

Explanation:
T = ¾ mg, F = mg – T = ma → a = mg (1 – ¾)/m = 1/4 g

MCQ 59
A 2 N force acts on a mass. If momentum changes by 120 kg·m/s, time interval:

a. 8 s
b. 30 s
c. 60 s
d. 120 s

Correct Answer: c. 60 s

Explanation:
F = Δp / Δt → Δt = Δp / F = 120 / 2 = 60 s

MCQ 60
Property by which a moving object exerts force on a stopping object:

a. Inertia
b. Quantity of motion
c. Acceleration
d. All

Correct Answer: a. Inertia

Explanation:
Inertia = property resisting change in motion

MCQ 61
The force exerted by a wall on water striking normally at 10 m/s and volume 0.0001 m³/s is:

a. 1 N
b. 10 N
c. 100 N
d. None

Correct Answer: a. 1 N

Explanation:
F = Δp / Δt = ρ V Δv = 1000 × 0.0001 × 10 = 1 N

MCQ 62
Distance covered by a body starting from rest in time t:

a. at²/2
b. V t
c. at
d. at/3

Correct Answer: a. at²/2

Explanation:
s = ut + ½ at², u = 0 → s = ½ at²

MCQ 63
Angle between centripetal force and momentum in circular motion:

a. 90°
b. 0°
c. 180°
d. 30°

Correct Answer: a. 90°

Explanation:
Centripetal force acts toward center, momentum tangent → 90°

MCQ 64
Range of projectile is same for which pair of angles?

a. 30° and 60°
b. 20° and 80°
c. 0° and 45°
d. 10° and 90°

Correct Answer: a. 30° and 60°

Explanation:
Complementary angles θ and 90–θ have same range

MCQ 65
Rock dropped from height, after 3 s, its speed:

a. 30 m/s
b. 29.4 m/s
c. 2940 m/s
d. 50 m/s

Correct Answer: b. 29.4 m/s

Explanation:
vf = vi + gt = 0 + 9.8 × 3 ≈ 29.4 m/s

MCQ 66
Relation between time of flight T and height H:

a. H = g T² / 8
b. H = 8 T² / g
c. H = 8 g / T²
d. H = 8 / g T²

Correct Answer: a. H = g T² / 8

Explanation:
Time of flight T = 2 vi sinθ / g; max height H = ½ g (T/2)² = g T² / 8

MCQ 67
Cricket ball hit at 45° with KE = E. KE at highest point:

a. 0
b. E/2
c. E/√2
d. E

Correct Answer: b. E/2

Explanation:
Vertical KE converts to PE → KE_horizontal = ½ E = E/2

MCQ 68
Velocity of projectile at maximum height:

a. Zero
b. Minimum
c. Maximum
d. Between max and min

Correct Answer: b. Minimum

Explanation:
Vertical velocity = 0, horizontal velocity remains → min speed

MCQ 69
Slope of displacement–time graph:

a. Velocity
b. Acceleration
c. Position
d. Momentum

Correct Answer: a. Velocity

Explanation:
Slope of s–t graph = instantaneous velocity

MCQ 70
Shell of mass m moving with velocity V breaks into 2 pieces. Velocity of larger piece (3m/4) if smaller is m/4 at rest:

a. V
b. 2V
c. 4V/3
d. 3V/4

Correct Answer: c. 4V/3

Explanation:
Conservation of momentum: mv = 0 + (3m/4) V2 → V2 = 4V/3

MCQ 71
Rubber ball dropped from height 5 m, rebounds to 1.8 m. Fraction of velocity lost:

a. 16/25
b. 2/5
c. 3/5
d. 9/25

Correct Answer: c. 3/5

Explanation:
v2/v1 = √(h2/h1) → √(1.8/5) ≈ √(9/25) → lost = 3/5

MCQ 72
Distance of x = 20t + 2At². Acceleration:

a. A/4 m/s²
b. 4/A m/s²
c. 4 m/s²
d. 4A m/s²

Correct Answer: d. 4A m/s²

Explanation:
Compare x = ut + ½ at² → ½ a = 2A → a = 4A

MCQ 73
“g” represents acceleration due to free fall. Correct statement:

a. g is gravity
b. g = weight/mass
c. g = weight of object
d. Reduced by air resistance

Correct Answer: b. g = weight/mass

Explanation:
w = mg → g = w/m

MCQ 74
Car travels distance in 2 h and returns in 3 h. Average velocity:

a. 2/5
b. 2.5/5
c. (2/5 + 3/5)
d. 0

Correct Answer: d. 0

Explanation:
Total displacement = 0 → average velocity = 0

MCQ 75
Racing car speeds: 20 m/s for 2 s, 40 m/s for 2 s, 60 m/s for 6 s. Average speed:

a. 12 m/s
b. 13.3 m/s
c. 40 m/s
d. 48 m/s

Correct Answer: d. 48 m/s

Explanation:
Average speed = (Σ vi ti)/(Σ ti) = (20×2 + 40×2 + 60×6)/10 = 48 m/s

MCQ 76
Monkey slides down a branch (breaking strength 75% of weight). Minimum acceleration:

a. 3/4 g
b. 1/2 g
c. 0
d. 1/4 g

Correct Answer: d. 1/4 g

Explanation:
F = w – T → ma = mg – 0.75 mg → a = 0.25 g

MCQ 77
Shape of velocity–time graph for constant acceleration:

a. Straight line
b. Parabola
c. Hyperbola
d. Ellipse

Correct Answer: a. Straight line

Explanation:
v = u + at → linear

MCQ 78
Area under velocity–time graph represents:

a. Force
b. Momentum
c. Distance
d. None

Correct Answer: c. Distance

Explanation:
Area = v × t = displacement

MCQ 79
Mass accelerated uniformly; resultant force:

a. Zero
b. Constant but not zero
c. Increases
d. Both a and b

Correct Answer: b. Constant but not zero

Explanation:
F = ma → constant acceleration → constant F ≠ 0

MCQ 80
3rd law explains:

a. Effect of force
b. Existence of force
c. Loss of force
d. Existence of pair of forces

Correct Answer: d. Existence of pair of forces

Explanation:
Action = reaction, equal and opposite

MCQ 81
Newton 2nd law unit:

a. Work
b. Angular
c. Power
d. Rate of change of momentum

Correct Answer: d. Rate of change of momentum

Explanation:
F = dp/dt

MCQ 82
Helmet increases collision time → decreases:

a. Chance
b. Force
c. Velocity
d. Impulse

Correct Answer: b. Force

Explanation:
I = F Δt → larger Δt → smaller F

MCQ 83
Constant force F acts on body for Δt:

a. -m(Vf² – Vi²)
b. m(Vf – Vi)/Δt
c. Δt/ma
d. (ma)/t

Correct Answer: b. m(Vf – Vi)/Δt

Explanation:
F = Δp / Δt

MCQ 84
Bullet of mass m embeds in block M. Velocity of system:

a. m/(M+m) V
b. (M+m)/m V
c. m/(M+m) × V
d. M/(M–m) V

Correct Answer: c. m/(M+m) × V

Explanation:
Conservation of momentum: mv = (M+m) V

MCQ 85
Conservation of linear momentum equivalent to:

a. Newton 1st law
b. Newton 2nd law
c. Newton 3rd law
d. None

Correct Answer: a. Newton 1st law

Explanation:
Momentum conserved when net force = 0 → Newton 1st law

MCQ 86
Particle of mass m makes head-on elastic collision with equal mass at rest. Velocity after collision:

a. 2V
b. –V
c. +V
d. 0

Correct Answer: d. 0

Explanation:
Elastic collision formula: v1 = (m1–m2)/(m1+m2) V1 + … → v1 = 0

MCQ 87
Horizontal range of projectile:

a. 2 vi² / g
b. 2 vi²/g sin θ
c. vi² / g
d. 2 vi²/g

Correct Answer: b. 2 vi²/g sin θ

Explanation:
R = (vi² sin 2θ)/g formula

MCQ 88
Bomb dropped from plane moving horizontally at 200 mph, reaches ground in 5 s. Altitude:

a. 4 miles
b. 122.5 m
c. 40 m
d. 10 m

Correct Answer: b. 122.5 m

Explanation:
H = ½ g t² = ½ × 9.8 × 5² ≈ 122.5 m

MCQ 89
Study of projectile motion is called:

a. Metaphysics
b. Particle physics
c. Projectile physics
d. Ballistics

Correct Answer: d. Ballistics

Explanation:
Science of motion of projectiles = Ballistics

MCQ 90
To change momentum of an object there must be:

a. Force applied
b. Change in time
c. Change in distance
d. Change in temperature

Correct Answer: a. Force applied

Explanation:
Δp/Δt = F → force required

MCQ 91
Basketball thrown along parabolic path. Acceleration at highest point:

a. Zero
b. ½ horizontal
c. g downward

Correct Answer: c. g downward

Explanation:
Acceleration always vertical downward due to gravity

MCQ 92
Projectile travels maximum range 1000 m. Max height:

a. 500 m
b. 250 m
c. None of these

Correct Answer: c. None of these

Explanation:
H_max = R tan²θ/4 → depends on angle; not exactly 250 or 500

MCQ 93
Quantity not changing when force applied:

a. Mass
b. Velocity
c. Position
d. Acceleration

Correct Answer: a. Mass

Explanation:
Force changes velocity and acceleration, not mass

MCQ 94
Impulse has same unit as:

a. Force
b. Energy
c. Momentum
d. None

Correct Answer: c. Momentum

Explanation:
Impulse = F Δt = Δp → same unit as momentum

MCQ 95
Horizontal component of velocity in projectile motion:

a. Increases
b. Decreases
c. Remains constant
d. None

Correct Answer: c. Remains constant

Explanation:
No horizontal force → horizontal velocity constant

MCQ 96
Projectile travels 1000 m. Maximum height:

a. 400 m
b. 800 m
c. 500 m
d. 250 m

Correct Answer: d. 250 m

Explanation:
H = R tan²θ /4, assume θ = 45° → H = R/4 = 1000/4 = 250 m

MCQ 97
Slope of displacement-time graph:

a. Distance
b. Velocity
c. Acceleration
d. None

Correct Answer: b. Velocity

Explanation:
Slope = instantaneous velocity

MCQ 98
Rate of change in linear momentum for freely falling body =

a. Weight
b. Power
c. Inertia
d. Impulse

Correct Answer: a. Weight

Explanation:
dp/dt = F = mg

MCQ 99
Two bodies with KE ratio 4:1 and equal momentum. Mass ratio:

a. 1:4
b. 4:1
c. 1:2
d. 1:1

Correct Answer: a. 1:4

Explanation:
p = constant, KE = ½ mv² → smaller mass has larger velocity

MCQ 100
1 horsepower =

a. 746 KW
b. 746 MW
c. 746 W
d. 746 J

Correct Answer: c. 746 W

Explanation:
Definition of 1 HP = 746 W