Work done by a force F on a body displaced by d is the dot product: W = F · d = Fd cosθ where θ is the angle between force and displacement.
Answer: B
2
A body moves on a smooth horizontal surface under no horizontal force. The work done by gravity is:
Easy
A. mgh
B. mgd
C. Zero
D. Negative
✅ Solution
Since the body moves horizontally and gravity acts vertically downward, the angle between force and displacement is 90°. W = Fd cos90° = 0
Answer: C
3
The work-energy theorem states that the net work done on a body equals:
Easy
A. Change in momentum
B. Change in potential energy
C. Change in kinetic energy
D. Change in total energy
✅ Solution
The Work-Energy Theorem: Wnet = Kf − Ki = ΔKE The net work done on a body equals the change in its kinetic energy.
Answer: C
4
The SI unit of power is:
Easy
A. Joule
B. Newton-metre
C. Watt
D. Erg
✅ Solution
Power = Work / Time. Its SI unit is Watt (W), named after James Watt. 1 W = 1 J/s = 1 kg·m²/s³ Note: 1 hp = 746 W
Answer: C
5
A force of 10 N acts at an angle of 60° to the displacement of 5 m. The work done is:
Moderate
A. 50 J
B. 25 J
C. 43.3 J
D. 25√3 J
✅ Solution
W = Fd cosθ = 10 × 5 × cos60° = 50 × 0.5 = 25 J
Answer: B — 25 J
6
Kinetic energy of a body of mass m moving with velocity v is:
Easy
A. mv
B. mv²
C. ½ mv²
D. 2mv²
✅ Solution
KE = ½mv² Kinetic energy is a scalar quantity. It is always positive (or zero) and depends on mass and the square of velocity.
Answer: C
7
If the velocity of a body is doubled, its kinetic energy becomes:
Moderate
A. Doubled
B. Halved
C. Four times
D. Unchanged
✅ Solution
KE ∝ v². If v → 2v: KE' = ½m(2v)² = 4 × (½mv²) = 4 × KESo the kinetic energy becomes 4 times.
Answer: C — Four times
8
Gravitational potential energy of an object at height h is:
Easy
A. mgh²
B. mgh
C. ½mgh
D. mg/h
✅ Solution
V(h) = mghThis is the work done against gravity to raise the object to height h. It is stored as potential energy.
Answer: B — mgh
9
The spring constant k has SI unit:
Easy
A. N·m
B. N/m
C. J/m
D. kg/s²
✅ Solution
From Hooke's Law, F = −kx, so k = F/x. Unit of k = N/m = kg/s²Both B and D are equivalent, but N/m (Newton per metre) is the standard form.
Answer: B — N/m
10
In a completely inelastic collision:
Moderate
A. Both KE and momentum conserved
B. Only momentum conserved
C. Only KE conserved
D. Neither is conserved
✅ Solution
In all collisions, linear momentum is conserved (Newton's 3rd Law). In a completely inelastic collision, the bodies stick together and kinetic energy is not conserved (lost as heat/sound).
Answer: B — Only momentum conserved
11
When a conservative force does positive work on a body, its potential energy:
Moderate
A. Increases
B. Decreases
C. Remains same
D. Becomes zero
✅ Solution
ΔV = −W (for conservative forces). If W is positive (force does positive work), then ΔV is negative, meaning potential energy decreases. (e.g., object falling — gravity does positive work, PE decreases)
Answer: B — Decreases
12
A dot product of two perpendicular vectors is:
Easy
A. AB
B. −AB
C. Zero
D. AB/2
✅ Solution
For perpendicular vectors, θ = 90°. A · B = AB cos90° = AB × 0 = 0Also: î·ĵ = ĵ·k̂ = k̂·î = 0
Answer: C — Zero
13
The potential energy stored in a spring stretched by x is:
Easy
A. kx
B. kx²
C. ½kx²
D. 2kx²
✅ Solution
V(x) = ½kx²This is obtained by integrating the spring force F = −kx over displacement.
Answer: C — ½kx²
14
1 kWh (kilowatt-hour) of energy in Joules equals:
Moderate
A. 3.6 × 10³ J
B. 3.6 × 10⁵ J
C. 3.6 × 10⁶ J
D. 3.6 × 10⁴ J
✅ Solution
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ JkWh is a unit of energy used in electricity bills.
Answer: C — 3.6 × 10⁶ J
15
Two equal masses undergo an elastic head-on collision. After collision:
Moderate
A. First stops, second moves with same speed
B. Both move together
C. Both stop
D. First bounces back
✅ Solution
For elastic collision with equal masses (m₁ = m₂): v₁f = 0, v₂f = v₁iThe first mass comes to rest and the second moves with the initial speed of the first.
Answer: A
16
Friction is an example of a:
Easy
A. Conservative force
B. Non-conservative force
C. Central force
D. Electrostatic force
✅ Solution
Friction is a non-conservative force because the work done by friction on a closed path is NOT zero. It dissipates energy as heat and depends on the path taken.
Answer: B — Non-conservative force
17
A weightlifter holds a 150 kg mass steadily on his shoulder for 30 seconds. Work done on the load is:
Moderate
A. 150 × 10 × 30 J
B. 150 × 10 J
C. Zero
D. 45000 J
✅ Solution
Since there is no displacement of the load, work done = zero. W = F × d × cosθ = F × 0 = 0Despite muscles exerting force, no physical work is done on the load.
Answer: C — Zero
18
The principle of conservation of mechanical energy holds when:
Moderate
A. All forces are non-conservative
B. Only conservative forces do work
C. Friction is present
D. External forces act
✅ Solution
Mechanical energy (KE + PE) is conserved only when all forces doing work are conservative. K + V = constant (when only conservative forces act)
Answer: B
19
Instantaneous power is defined as:
Easy
A. P = W/t²
B. P = F · v
C. P = W × t
D. P = F/v
✅ Solution
P = dW/dt = F · (dr/dt) = F · vInstantaneous power is the dot product of force and instantaneous velocity.
Answer: B — P = F · v
20
Which of the following has the highest kinetic energy? (Refer to NCERT Table 5.2)
Moderate
A. Running athlete (70 kg, 10 m/s)
B. Bullet (0.05 kg, 200 m/s)
C. Car (2000 kg, 25 m/s)
D. Stone dropped from 10 m (1 kg)
✅ Solution
Car: ½ × 2000 × 25² = 6.25 × 10⁵ J ✓ (Highest)Athlete: ½ × 70 × 100 = 3500 JBullet: ½ × 0.05 × 40000 = 1000 J
Answer: C — Car has highest KE ≈ 6.3 × 10⁵ J
21
Work done by the spring force in a complete cycle (from xᵢ back to xᵢ) is:
Moderate
A. ½kxᵢ²
B. kxᵢ²
C. −½kxᵢ²
D. Zero
✅ Solution
Spring force is conservative. Work done by a conservative force over a closed path is zero. W_s = ½kxᵢ² − ½kxᵢ² = 0
Answer: D — Zero
22
In an elastic collision between a very heavy mass M (at rest) and a light ball m (moving with v), after collision the ball approximately:
Hard
A. Stops completely
B. Moves forward with 2v
C. Reverses with speed ≈ v
D. Both move together
✅ Solution
When m₂ >> m₁ (Case II in NCERT), from elastic collision formula: v₁f ≈ −v₁i v₂f ≈ 0The light ball reverses with approximately the same speed. The heavy mass remains nearly undisturbed.
Answer: C — Reverses with speed ≈ v
23
A body is initially at rest and undergoes constant acceleration. Power delivered to it at time t is proportional to:
Hard
A. t^(1/2)
B. t
C. t^(3/2)
D. t²
✅ Solution
v = at (constant acceleration, starts from rest). P = Fv = (ma)(at) = ma²tSo P ∝ t (linear relationship with time).
Answer: B — t
24
At the maximum compression of a spring in a spring-collision problem (no friction), the kinetic energy of the block is:
Moderate
A. Maximum
B. Equal to initial KE
C. Zero
D. Half of initial KE
✅ Solution
At maximum compression, the block momentarily comes to rest. All kinetic energy has been converted to elastic potential energy of the spring: KE = 0, PE = ½kx²_max = Initial KE
Answer: C — Zero
25
A ball is dropped from height H. Just before hitting the ground, assuming no air resistance, its speed is:
Moderate
A. √(gH)
B. √(2gH)
C. 2gH
D. gH
✅ Solution
By conservation of mechanical energy (no friction): mgH = ½mv² → v = √(2gH)This is derived from PE at top = KE at bottom.
Answer: B — √(2gH)
🔴
Section B — Numerical Problems
15 Questions × 3 Marks
1
A force F = (3î + 4ĵ − 5k̂) N acts on an object which is displaced by d = (5î + 4ĵ + 3k̂) m. Find (i) the work done and (ii) the angle between F and d.
Moderate3 Marks
A raindrop of mass 1.00 g falls from a height of 1.00 km and hits the ground with a speed of 50 m/s. Find: (a) work done by gravity, (b) work done by the resistive force.
Moderate3 Marks
✅ Solution
Given: m = 1g = 10⁻³ kg, h = 10³ m, v = 50 m/sW_gravity = mgh = 10⁻³ × 10 × 10³ = 10.0 JChange in KE: ΔK = ½mv² − 0 = ½ × 10⁻³ × 2500 = 1.25 JW_resistive = ΔK − W_gravity = 1.25 − 10.0 = −8.75 J
(a) W_gravity = 10.0 J; (b) W_resistive = −8.75 J
3
A bullet of mass 50 g is fired with speed 200 m/s into soft wood. It emerges with only 10% of its initial KE. Find the emergent speed of the bullet.
Moderate3 Marks
✅ Solution
Initial KE: ½ × 0.05 × 200² = ½ × 0.05 × 40000 = 1000 JFinal KE: 10% of 1000 = 100 J½mv_f² = 100 → v_f = √(2 × 100 / 0.05) = √4000 = 63.2 m/sNote: Speed reduced by ~68%, not 90%! (KE ∝ v²)
v_f = 63.2 m/s ≈ 63 m/s
4
A body of mass 2 kg initially at rest moves under an applied horizontal force of 7 N on a table with coefficient of kinetic friction μ = 0.1. Find the work done by (a) applied force, (b) friction, (c) net force, in 10 seconds.
Moderate3 Marks
A car of mass 1000 kg moving at 18 km/h on a smooth road collides with a spring of spring constant 5.25 × 10³ N/m. Find the maximum compression of the spring.
Moderate3 Marks
✅ Solution
Convert: 18 km/h = 5 m/sKE = ½mv² = ½ × 1000 × 25 = 12500 J = 1.25 × 10⁴ JAt max compression: KE = PE of spring½kx_m² = 12500 → x_m² = (2 × 12500) / 5250 = 4.76 → x_m = 2.00 m
Maximum compression x_m = 2.00 m
6
An elevator of maximum load 1800 kg moves up at constant speed 2 m/s. Frictional force opposing motion is 4000 N. Find the minimum power delivered by the motor (in watts and horsepower).
Moderate3 Marks
✅ Solution
Total downward force = mg + Friction = 1800×10 + 4000 = 22000 NAt constant speed, motor must supply force = 22000 NP = F × v = 22000 × 2 = 44000 W = 44 kWIn hp: 44000 / 746 ≈ 59 hp
P = 44,000 W = 44 kW ≈ 59 hp
7
A bob of mass m is suspended by a string of length L and given a horizontal velocity v₀ at the lowest point. Find v₀ such that the string becomes slack only at the topmost point. Also find the speed at the top.
Moderate3 Marks
✅ Solution
At top C, tension T = 0 → centripetal force = gravity:mg = mv_C²/L → v_C = √(gL)Total mechanical energy at C = ½mv_C² + mg(2L) = ½mgL + 2mgL = 5mgL/2Equating to KE at bottom:½mv₀² = 5mgL/2 → v₀ = √(5gL)
v₀ = √(5gL); Speed at top v_C = √(gL)
8
A pump on the ground floor fills a tank of volume 30 m³ in 15 min. The tank is 40 m above the ground. Efficiency of pump is 30%. Find the electric power consumed. (ρ_water = 1000 kg/m³)
Moderate3 Marks
✅ Solution
Mass of water = ρV = 1000 × 30 = 30,000 kgW = mgh = 30000 × 10 × 40 = 1.2 × 10⁷ JUseful power = W/t = 1.2×10⁷ / (15×60) = 1.333 × 10⁴ WEfficiency = Useful Power / Input Power → P_input = 1.333×10⁴ / 0.30 = 4.44 × 10⁴ W ≈ 44.4 kW
Electric power consumed ≈ 44.4 kW
9
A body of mass 0.5 kg travels in a straight line with velocity v = 5x^(3/2) m/s. Find the work done by the net force during its displacement from x = 0 to x = 2 m.
Moderate3 Marks
✅ Solution
Use Work-Energy Theorem: W = ΔKE = KE_f − KE_iAt x = 0: v = 0, KE_i = 0At x = 2: v = 5 × (2)^(3/2) = 5 × 2√2 = 10√2 m/sKE_f = ½ × 0.5 × (10√2)² = ½ × 0.5 × 200 = 50 JW = KE_f − KE_i = 50 − 0 = 50 J
Work done = 50 J
10
A bob of a pendulum is released from a horizontal position. Length of pendulum is 1.5 m. It dissipates 5% of its energy against air resistance. Find the speed at the lowest point.
Moderate3 Marks
✅ Solution
Initial PE (taking lowest point as reference) = mgL = mg × 1.5Energy available at bottom = 95% of mgL (5% lost)½mv² = 0.95 × mgL → v² = 2 × 0.95 × 10 × 1.5 = 28.5v = √28.5 ≈ 5.34 m/s
Speed at lowest point ≈ 5.34 m/s
11
A person trying to lose weight lifts a 10 kg mass to height 0.5 m, one thousand times. Assume PE lost while lowering is dissipated. (a) Total work done against gravity. (b) Fat supplies 3.8 × 10⁷ J/kg with 20% efficiency — how much fat is burned?
Moderate3 Marks
✅ Solution
(a) W = mgh × n = 10 × 10 × 0.5 × 1000 = 50,000 J = 5 × 10⁴ J(b) Mechanical energy from fat (20% efficiency):Energy needed = W / 0.20 = 5×10⁴ / 0.20 = 2.5 × 10⁵ JFat burned = 2.5×10⁵ / (3.8×10⁷) ≈ 6.6 × 10⁻³ kg ≈ 6.6 g
(a) 5 × 10⁴ J; (b) ≈ 6.6 grams of fat
12
Two masses m₁ = 3 kg (moving at 4 m/s) and m₂ = 1 kg (at rest) undergo a perfectly inelastic collision. Find (a) final velocity and (b) loss in kinetic energy.
Moderate3 Marks
The blades of a windmill sweep a circle of area A = 30 m². Wind flows at v = 36 km/h perpendicular to the circle. Density of air = 1.2 kg/m³. If 25% of energy is converted to electrical energy, find the electrical power produced.
Moderate3 Marks
✅ Solution
v = 36 km/h = 10 m/sMass of air in time t: m = ρ × A × v × t = 1.2 × 30 × 10 × t = 360t kgKE of air per second = ½ × 360 × (10)² = 18000 W = 18 kWElectrical power = 25% × 18 kW = 4.5 kW
Electrical power produced = 4.5 kW
14
A block of mass m = 1 kg enters a rough patch from x = 0.1 m to x = 2.01 m with initial speed 2 m/s. Retarding force F_r = −k/x where k = 0.5 J. Find the final kinetic energy and speed as the block exits the patch.
Hard3 Marks
✅ Solution
Using Work-Energy theorem with variable force:K_f = K_i + ∫(−k/x)dx from 0.1 to 2.01K_f = ½(1)(2²) − k·ln(2.01/0.1) = 2 − 0.5 × ln(20.1)= 2 − 0.5 × 3.0 = 2 − 1.5 = 0.5 Jv_f = √(2K_f/m) = √(2 × 0.5 / 1) = 1 m/s
K_f = 0.5 J; v_f = 1 m/s
15
A family solar panel system: Solar intensity = 200 W/m², 20% conversion efficiency. Family uses 8 kW of power. (a) What area of solar panels is needed? (b) A typical roof is about 50 m² — is it sufficient?
Moderate3 Marks
✅ Solution
(a) Effective power from panels = 20% × 200 × A = 40A watts40A = 8000 W → A = 8000/40 = 200 m²(b) A typical roof ≈ 50 m². Needed area = 200 m².200 m² >> 50 m². So a typical roof area is NOT sufficient. The family would need ~4 houses worth of roof space!
(a) Area needed = 200 m²; (b) Not sufficient — typical roof (50 m²) is too small