AP Physics 1 Practice Test 17 – 60 Questions with Answers

Test yourself with 60 original AP Physics 1 practice questions covering all 8 units of the course: kinematics, force and translational dynamics, work, energy, and power, linear momentum, torque and rotational dynamics, energy and momentum of rotating systems, oscillations, and fluids. Every answer includes the full working. All questions are algebra-based (no calculus). Take g = 9.8 m/s2 unless stated otherwise.

Unit 1: Kinematics

Q1. A cyclist covers a distance of 350 m in 25 s on a straight road. What is the cyclist’s average speed?

  • (A) 8750 m/s
  • (B) 0.07 m/s
  • (C) 14 m/s
  • (D) 7 m/s
Show Answer

Answer: C – Average speed = distance ÷ time = 350 ÷ 25 = 14 m/s.

Q2. A delivery scooter starts from a velocity of 5 m/s and accelerates uniformly at 3 m/s2 for 5 s. What is its final velocity? (Ignore air resistance.)

  • (A) 8 m/s
  • (B) 15 m/s
  • (C) 35 m/s
  • (D) 20 m/s
Show Answer

Answer: D – v = v0 + at = 5 + (3 × 5) = 5 + 15 = 20 m/s. (Simplification: air resistance is ignored, as stated.)

Q3. A small ball is dropped from rest from a tall tower. Ignoring air resistance, how far does it fall in the first 3 s?

  • (A) 44.1 m
  • (B) 22.05 m
  • (C) 88.2 m
  • (D) 66.15 m
Show Answer

Answer: A – Distance fallen = 0.5gt2 = 0.5 × 9.8 × 32 = 0.5 × 9.8 × 9 = 44.1 m. (Simplification: air resistance is ignored, as stated.)

Q4. A stone is dropped from rest from a bridge 44.1 m above a river. Ignoring air resistance, how long does it take to reach the water?

  • (A) 4 s
  • (B) 3 s
  • (C) 6 s
  • (D) 1.5 s
Show Answer

Answer: B – h = 0.5gt2 gives t = √(2h ÷ g) = √(88.2 ÷ 9.8) = √9 = 3 s. (Simplification: air resistance is ignored, as stated.)

Q5. A train increases its speed by 24 m/s in 5 s with constant acceleration. What is the magnitude of its acceleration?

  • (A) 120 m/s2
  • (B) 2.4 m/s2
  • (C) 4.8 m/s2
  • (D) 24 m/s2
Show Answer

Answer: C – a = Δv ÷ Δt = 24 ÷ 5 = 4.8 m/s2.

Q6. A kart starts from rest and accelerates uniformly at 2.5 m/s2. How far does it travel in 4 s?

  • (A) 10 m
  • (B) 40 m
  • (C) 30 m
  • (D) 20 m
Show Answer

Answer: D – x = v0t + 0.5at2 = 0 + 0.5 × 2.5 × 42 = 0.5 × 2.5 × 16 = 20 m.

Q7. A car travelling at 22 m/s brakes with a constant deceleration of 5 m/s2 on a dry road. What is its stopping distance?

  • (A) 48.4 m
  • (B) 96.8 m
  • (C) 4.4 m
  • (D) 24.2 m
Show Answer

Answer: A – v2 = v02 + 2aΔx with v = 0 gives 0 = 222 – 2 × 5 × Δx, so Δx = 484 ÷ 10 = 48.4 m.

Unit 2: Force and Translational Dynamics

Q8. A net force acts on a 7 kg toolbox and gives it an acceleration of 3 m/s2 across a smooth floor. What is the magnitude of the net force?

  • (A) 10 N
  • (B) 21 N
  • (C) 42 N
  • (D) 2.33 N
Show Answer

Answer: B – Fnet = ma = 7 × 3 = 21 N. (Simplification: the floor is treated as smooth, as stated.)

Q9. What is the weight of a 16 kg suitcase on Earth?

  • (A) 16 N
  • (B) 160 N
  • (C) 156.8 N
  • (D) 78.4 N
Show Answer

Answer: C – Weight = mg = 16 × 9.8 = 156.8 N.

Q10. A crate on a horizontal floor experiences a normal force of 150 N. If the coefficient of kinetic friction is 0.25, what is the magnitude of the kinetic friction force?

  • (A) 0.25 N
  • (B) 150 N
  • (C) 75 N
  • (D) 37.5 N
Show Answer

Answer: D – Friction f = μN = 0.25 × 150 = 37.5 N.

Q11. A 10 kg wagon is pulled to the right with a 70 N force while a 20 N friction force acts to the left. What is the wagon’s acceleration?

  • (A) 5 m/s2 to the right
  • (B) 9 m/s2 to the right
  • (C) 7 m/s2 to the right
  • (D) 5 m/s2 to the left
Show Answer

Answer: A – Fnet = 70 – 20 = 50 N to the right. a = Fnet ÷ m = 50 ÷ 10 = 5 m/s2 to the right.

Q12. A 10 kg parcel sits on a frictionless ramp inclined at 30° to the horizontal. Using sin 30° = 0.5, what is the component of the parcel’s weight down the ramp?

  • (A) 98 N
  • (B) 49 N
  • (C) 73.5 N
  • (D) 24.5 N
Show Answer

Answer: B – Weight = mg = 10 × 9.8 = 98 N. Component down the ramp = mg sin Θ = 98 × 0.5 = 49 N. (Simplification: the ramp is frictionless, as stated.)

Q13. Two masses of 6 kg and 4 kg hang from opposite ends of a light string over a frictionless pulley. What is the magnitude of their acceleration? (Ignore the mass of the pulley.)

  • (A) 9.8 m/s2
  • (B) 5.88 m/s2
  • (C) 1.96 m/s2
  • (D) 3.92 m/s2
Show Answer

Answer: C – Net driving force = (6 – 4) × 9.8 = 19.6 N. Total mass = 10 kg. a = F ÷ m = 19.6 ÷ 10 = 1.96 m/s2. (Simplification: the pulley is massless and frictionless, as stated.)

Q14. A spring with spring constant 250 N/m is stretched 0.12 m from equilibrium. What is the magnitude of the spring force?

  • (A) 2083.33 N
  • (B) 250.12 N
  • (C) 60 N
  • (D) 30 N
Show Answer

Answer: D – F = kx = 250 × 0.12 = 30 N.

Q15. A 65 kg student stands on a bathroom scale in a lift that accelerates upward at 2 m/s2. What does the scale read, in newtons?

  • (A) 767 N
  • (B) 637 N
  • (C) 507 N
  • (D) 130 N
Show Answer

Answer: A – Apparent weight = m(g + a) = 65 × (9.8 + 2) = 65 × 11.8 = 767 N.

Unit 3: Work, Energy, and Power

Q16. A gardener pushes a wheelbarrow with a constant horizontal force of 60 N through a distance of 12 m. How much work is done by the push?

  • (A) 72 J
  • (B) 720 J
  • (C) 5 J
  • (D) 1440 J
Show Answer

Answer: B – W = Fd = 60 × 12 = 720 J.

Q17. What is the kinetic energy of a 3 kg medicine ball moving at 10 m/s?

  • (A) 30 J
  • (B) 300 J
  • (C) 150 J
  • (D) 75 J
Show Answer

Answer: C – KE = 0.5mv2 = 0.5 × 3 × 102 = 0.5 × 3 × 100 = 150 J.

Q18. A 6 kg box is raised 10 m above the ground at constant speed. By how much does its gravitational potential energy increase?

  • (A) 60 J
  • (B) 58.8 J
  • (C) 294 J
  • (D) 588 J
Show Answer

Answer: D – ΔPE = mgh = 6 × 9.8 × 10 = 588 J.

Q19. A crane does 1400 J of work in 20 s while lifting a load. What is its average power output?

  • (A) 70 W
  • (B) 28000 W
  • (C) 700 W
  • (D) 140 W
Show Answer

Answer: A – Power = work ÷ time = 1400 ÷ 20 = 70 W.

Q20. A diver drops from rest from a platform 20 m above the water. Ignoring air resistance, what is the diver’s speed just before hitting the water?

  • (A) 196 m/s
  • (B) 19.8 m/s
  • (C) 14 m/s
  • (D) 9.9 m/s
Show Answer

Answer: B – Energy: mgh = 0.5mv2, so v = √(2gh) = √(2 × 9.8 × 20) = √392 = 19.8 m/s. (Simplification: air resistance is ignored, as stated.)

Q21. A spring with spring constant 250 N/m is compressed 0.2 m from equilibrium. How much elastic potential energy is stored in the spring?

  • (A) 50 J
  • (B) 10 J
  • (C) 5 J
  • (D) 7.5 J
Show Answer

Answer: C – PE = 0.5kx2 = 0.5 × 250 × 0.22 = 0.5 × 250 × 0.04 = 5 J.

Q22. A motor lifts a 25 kg crate 5 m in 10 s at constant speed. What is the average power output of the motor?

  • (A) 1225 W
  • (B) 12250 W
  • (C) 61.25 W
  • (D) 122.5 W
Show Answer

Answer: D – Work = mgh = 25 × 9.8 × 5 = 1225 J. Power = work ÷ time = 1225 ÷ 10 = 122.5 W.

Unit 4: Linear Momentum

Q23. What is the magnitude of the momentum of a 4 kg bowling ball rolling at 12 m/s?

  • (A) 48 kg·m/s
  • (B) 16 kg·m/s
  • (C) 576 kg·m/s
  • (D) 24 kg·m/s
Show Answer

Answer: A – p = mv = 4 × 12 = 48 kg·m/s.

Q24. A constant force of 50 N acts on a hockey puck for 3 s. What is the impulse delivered to the puck?

  • (A) 16.67 N·s
  • (B) 150 N·s
  • (C) 53 N·s
  • (D) 300 N·s
Show Answer

Answer: B – J = FΔt = 50 × 3 = 150 N·s.

Q25. A 5 kg trolley moving at 6 m/s collides with a stationary 2 kg trolley and the two stick together. What is their common velocity after the collision? (Ignore friction.)

  • (A) 6 m/s
  • (B) 15 m/s
  • (C) 4.29 m/s
  • (D) 8.57 m/s
Show Answer

Answer: C – Momentum: m1v1 = (m1 + m2)vf, so vf = (5 × 6) ÷ (5 + 2) = 30 ÷ 7 = 4.29 m/s. (Simplification: friction is ignored, as stated.)

Q26. An astronaut of mass 65 kg, floating at rest in space, throws a 2 kg tool away at 10 m/s. What is the astronaut’s recoil speed? (Ignore external forces.)

  • (A) 10 m/s
  • (B) 325 m/s
  • (C) 0.62 m/s
  • (D) 0.31 m/s
Show Answer

Answer: D – Total momentum stays zero, so mastronautvrecoil = mtoolvtool. vrecoil = (2 × 10) ÷ 65 = 20 ÷ 65 = 0.31 m/s. (Simplification: external forces are ignored, as stated.)

Q27. A 1.5 kg squash ball travelling at 5 m/s hits a wall and rebounds at 5 m/s in the opposite direction. What is the magnitude of the change in the ball’s momentum?

  • (A) 15 kg·m/s
  • (B) 7.5 kg·m/s
  • (C) 0 kg·m/s
  • (D) 22.5 kg·m/s
Show Answer

Answer: A – Take the rebound direction as positive: Δp = m(vf – vi) = 1.5 × (5 – (-5)) = 1.5 × 10 = 15 kg·m/s in magnitude.

Q28. A footballer’s foot changes a ball’s momentum by 150 kg·m/s during a contact lasting 4 s. What is the average force exerted on the ball?

  • (A) 600 N
  • (B) 37.5 N
  • (C) 150 N
  • (D) 18.75 N
Show Answer

Answer: B – Average force = Δp ÷ Δt = 150 ÷ 4 = 37.5 N.

Q29. Two bumper cars move in the same direction and stick together on collision. The first has mass 3 kg and velocity 6 m/s; the second has mass 2 kg and velocity 4 m/s. What is their velocity after the collision? (Ignore friction.)

  • (A) 5 m/s
  • (B) 6 m/s
  • (C) 5.2 m/s
  • (D) 10.4 m/s
Show Answer

Answer: C – Total momentum before = 3 × 6 + 2 × 4 = 18 + 8 = 26 kg·m/s. Total mass = 5 kg, so vf = 26 ÷ 5 = 5.2 m/s. (Simplification: friction is ignored, as stated.)

Q30. Two skaters glide in the same direction. One has mass 3 kg and velocity 3 m/s; the other has mass 4 kg and velocity 2 m/s. What is the total momentum of the pair?

  • (A) 9 kg·m/s
  • (B) 8 kg·m/s
  • (C) 22 kg·m/s
  • (D) 17 kg·m/s
Show Answer

Answer: D – Total p = m1v1 + m2v2 = (3 × 3) + (4 × 2) = 9 + 8 = 17 kg·m/s.


Unit 5: Torque and Rotational Dynamics

Q31. A wrench handle is 0.5 m long. A force of 50 N is applied at right angles to the handle at its end. What is the magnitude of the torque about the nut?

  • (A) 25 N·m
  • (B) 100 N·m
  • (C) 50.5 N·m
  • (D) 50 N·m
Show Answer

Answer: A – τ = rF sin Θ = 0.5 × 50 × sin 90° = 0.5 × 50 × 1 = 25 N·m.

Q32. What is the moment of inertia of a 3 kg point mass at a distance of 0.5 m from an axis of rotation?

  • (A) 1.5 kg·m2
  • (B) 0.75 kg·m2
  • (C) 3 kg·m2
  • (D) 2.25 kg·m2
Show Answer

Answer: B – I = mr2 = 3 × 0.52 = 3 × 0.25 = 0.75 kg·m2. (Simplification: the mass is treated as a point mass, as stated.)

Q33. A net torque of 18 N·m acts on a wheel with moment of inertia 3 kg·m2. What is the wheel’s angular acceleration?

  • (A) 54 rad/s2
  • (B) 0.17 rad/s2
  • (C) 6 rad/s2
  • (D) 3 rad/s2
Show Answer

Answer: C – α = τ ÷ I = 18 ÷ 3 = 6 rad/s2.

Q34. A spinning top has an initial angular velocity of 3 rad/s and a constant angular acceleration of 3 rad/s2. What is its angular velocity after 4 s?

  • (A) 6 rad/s
  • (B) 12 rad/s
  • (C) 30 rad/s
  • (D) 15 rad/s
Show Answer

Answer: D – ω = ω0 + αt = 3 + (3 × 4) = 3 + 12 = 15 rad/s.

Q35. A uniform plank is balanced on a pivot. A weight of 120 N sits 0.5 m to the left of the pivot. What downward force must be applied 2 m to the right of the pivot to keep the plank balanced?

  • (A) 30 N
  • (B) 120 N
  • (C) 480 N
  • (D) 60 N
Show Answer

Answer: A – Balance requires clockwise torque = anticlockwise torque: F × 2 = 120 × 0.5 = 60 N·m, so F = 60 ÷ 2 = 30 N. (Simplification: the plank’s own weight acts at the pivot, as stated.)

Q36. A merry-go-round starts with an angular velocity of 2 rad/s and accelerates uniformly at 2 rad/s2. Through what angle does it rotate in 4 s?

  • (A) 8 rad
  • (B) 24 rad
  • (C) 16 rad
  • (D) 48 rad
Show Answer

Answer: B – Θ = ω0t + 0.5αt2 = (2 × 4) + 0.5 × 2 × 42 = 8 + 16 = 24 rad.

Q37. Two torques act on a door about its hinges: 40 N·m clockwise and 10 N·m anticlockwise. What is the net torque?

  • (A) 50 N·m clockwise
  • (B) 30 N·m anticlockwise
  • (C) 30 N·m clockwise
  • (D) 10 N·m clockwise
Show Answer

Answer: C – Net torque = 40 – 10 = 30 N·m, in the clockwise direction (the larger torque).

Unit 6: Energy and Momentum of Rotating Systems

Q38. A flywheel has moment of inertia 3 kg·m2 and rotates at 4 rad/s. What is its rotational kinetic energy?

  • (A) 12 J
  • (B) 48 J
  • (C) 36 J
  • (D) 24 J
Show Answer

Answer: D – KErot = 0.5Iω2 = 0.5 × 3 × 42 = 0.5 × 3 × 16 = 24 J.

Q39. A spinning disc has moment of inertia 4 kg·m2 and angular velocity 2 rad/s. What is its angular momentum?

  • (A) 8 kg·m2/s
  • (B) 6 kg·m2/s
  • (C) 2 kg·m2/s
  • (D) 16 kg·m2/s
Show Answer

Answer: A – L = Iω = 4 × 2 = 8 kg·m2/s.

Q40. An ice skater spins with moment of inertia 4 kg·m2 at 4 rad/s. She pulls her arms in, reducing her moment of inertia to 2 kg·m2. What is her new angular velocity? (Ignore external torques.)

  • (A) 4 rad/s
  • (B) 8 rad/s
  • (C) 2 rad/s
  • (D) 16 rad/s
Show Answer

Answer: B – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = (4 × 4) ÷ 2 = 16 ÷ 2 = 8 rad/s. (Simplification: external torques are ignored, as stated.)

Q41. A hoop of mass 3 kg rolls without slipping at 3 m/s. Its moment of inertia about its centre is mr2. What is its total kinetic energy?

  • (A) 13.5 J
  • (B) 9 J
  • (C) 27 J
  • (D) 54 J
Show Answer

Answer: C – Translational KE = 0.5mv2 = 0.5 × 3 × 9 = 13.5 J. Rotational KE = 0.5Iω2 = 0.5(mr2)(v/r)2 = 0.5mv2 = 13.5 J. Total = 13.5 + 13.5 = 27 J.

Q42. A constant torque of 15 N·m rotates a shaft through an angle of 4 rad. How much work is done by the torque?

  • (A) 3.75 J
  • (B) 19 J
  • (C) 120 J
  • (D) 60 J
Show Answer

Answer: D – W = τΘ = 15 × 4 = 60 J.

Q43. A solid disc of mass 4 kg and radius 0.5 m rotates at 12 rad/s. Its moment of inertia is 0.5mr2. What is its rotational kinetic energy?

  • (A) 36 J
  • (B) 6 J
  • (C) 72 J
  • (D) 18 J
Show Answer

Answer: A – I = 0.5mr2 = 0.5 × 4 × 0.52 = 0.5 kg·m2. KErot = 0.5Iω2 = 0.5 × 0.5 × 122 = 0.5 × 0.5 × 144 = 36 J.

Q44. A 2 kg particle moves at 6 m/s at a perpendicular distance of 0.4 m from a fixed point. What is its angular momentum about that point?

  • (A) 12 kg·m2/s
  • (B) 4.8 kg·m2/s
  • (C) 0.8 kg·m2/s
  • (D) 9.6 kg·m2/s
Show Answer

Answer: B – L = mvr = 2 × 6 × 0.4 = 4.8 kg·m2/s.

Q45. A turbine wheel with moment of inertia 6 kg·m2 speeds up from 2 rad/s to 7 rad/s. By how much does its rotational kinetic energy increase?

  • (A) 147 J
  • (B) 12 J
  • (C) 135 J
  • (D) 270 J
Show Answer

Answer: C – Initial KE = 0.5 × 6 × 22 = 12 J. Final KE = 0.5 × 6 × 72 = 147 J. Increase = 147 – 12 = 135 J.

Unit 7: Oscillations

Q46. A mass of 9 kg hangs from a spring of spring constant 100 N/m and oscillates vertically. Using π = 3.14, what is the period of oscillation?

  • (A) 20.93 s
  • (B) 0.94 s
  • (C) 3.77 s
  • (D) 1.88 s
Show Answer

Answer: D – T = 2π√(m ÷ k) = 2 × 3.14 × √(9 ÷ 100) = 6.28 × 0.3 = 1.88 s.

Q47. A simple pendulum has length 2.45 m. Using π = 3.14 and g = 9.8 m/s2, what is its period for small oscillations?

  • (A) 3.14 s
  • (B) 9.83 s
  • (C) 1.57 s
  • (D) 6.28 s
Show Answer

Answer: A – T = 2π√(L ÷ g) = 2 × 3.14 × √(2.45 ÷ 9.8) = 6.28 × 0.5 = 3.14 s. (Simplification: small oscillations only, as stated.)

Q48. A vibrating string completes 25 oscillations in 10 s. What is its frequency?

  • (A) 0.4 Hz
  • (B) 2.5 Hz
  • (C) 250 Hz
  • (D) 5 Hz
Show Answer

Answer: B – Frequency = number of oscillations ÷ time = 25 ÷ 10 = 2.5 Hz.

Q49. A mass on a spring oscillates with amplitude 0.25 m and angular frequency 5 rad/s. What is its maximum speed?

  • (A) 20 m/s
  • (B) 0.31 m/s
  • (C) 1.25 m/s
  • (D) 2.5 m/s
Show Answer

Answer: C – vmax = ωA = 5 × 0.25 = 1.25 m/s.

Q50. A spring of spring constant 200 N/m oscillates with amplitude 0.15 m. What is the total mechanical energy of the oscillation?

  • (A) 30 J
  • (B) 4.5 J
  • (C) 3.38 J
  • (D) 2.25 J
Show Answer

Answer: D – E = 0.5kA2 = 0.5 × 200 × 0.152 = 0.5 × 200 × 0.02 = 2.25 J.

Q51. A mass-spring system has a period of 1.5 s. If the mass is quadrupled while the spring constant stays the same, what is the new period?

  • (A) 3 s
  • (B) 1.5 s
  • (C) 6 s
  • (D) 0.75 s
Show Answer

Answer: A – T is proportional to √m, so quadrupling m multiplies T by √4 = 2. New T = 2 × 1.5 = 3 s.

Q52. A particle in simple harmonic motion has angular frequency 4 rad/s and amplitude 0.3 m. What is its maximum acceleration?

  • (A) 1.2 m/s2
  • (B) 4.8 m/s2
  • (C) 16 m/s2
  • (D) 9.6 m/s2
Show Answer

Answer: B – amax = ω2A = 42 × 0.3 = 16 × 0.3 = 4.8 m/s2.

Unit 8: Fluids

Q53. A metal block has mass 50 kg and volume 0.02 m3. What is its density?

  • (A) 1 kg/m3
  • (B) 0 kg/m3
  • (C) 2500 kg/m3
  • (D) 1250 kg/m3
Show Answer

Answer: C – Density = mass ÷ volume = 50 ÷ 0.02 = 2500 kg/m3.

Q54. A force of 250 N acts uniformly over an area of 0.5 m2. What is the pressure?

  • (A) 125 Pa
  • (B) 250.5 Pa
  • (C) 1000 Pa
  • (D) 500 Pa
Show Answer

Answer: D – P = F ÷ A = 250 ÷ 0.5 = 500 Pa.

Q55. What is the gauge pressure due to water at a depth of 3 m below the surface? (Density of water = 1000 kg/m3, g = 9.8 m/s2.)

  • (A) 29400 Pa
  • (B) 3000 Pa
  • (C) 29.4 Pa
  • (D) 14700 Pa
Show Answer

Answer: A – P = ρgh = 1000 × 9.8 × 3 = 29400 Pa.

Q56. A block displaces 0.02 m3 of water when fully submerged. What is the buoyant force on it? (Density of water = 1000 kg/m3.)

  • (A) 20 N
  • (B) 196 N
  • (C) 0.2 N
  • (D) 392 N
Show Answer

Answer: B – Buoyant force = ρVg = 1000 × 0.02 × 9.8 = 196 N.

Q57. Water flows through a pipe of cross-sectional area 0.04 m2 at 2 m/s, then enters a narrower or wider section of area 0.02 m2. What is the water speed in the second section?

  • (A) 2 m/s
  • (B) 6 m/s
  • (C) 4 m/s
  • (D) 7 m/s
Show Answer

Answer: C – Continuity: A1v1 = A2v2, so v2 = (0.04 × 2) ÷ 0.02 = 0.08 ÷ 0.02 = 4 m/s.

Q58. Water flows through a hose of cross-sectional area 0.02 m2 at 4 m/s. What is the volume flow rate?

  • (A) 200 m3/s
  • (B) 4.02 m3/s
  • (C) 0.16 m3/s
  • (D) 0.08 m3/s
Show Answer

Answer: D – Volume flow rate = Av = 0.02 × 4 = 0.08 m3/s.

Q59. In a horizontal pipe, water speeds up from 2 m/s to 5 m/s. By how much does the pressure drop? (Density of water = 1000 kg/m3.)

  • (A) 10500 Pa
  • (B) 3000 Pa
  • (C) 5250 Pa
  • (D) 21000 Pa
Show Answer

Answer: A – Bernoulli (horizontal): ΔP = 0.5ρ(v22 – v12) = 0.5 × 1000 × (52 – 22) = 500 × (25 – 4) = 500 × 21 = 10500 Pa.

Q60. A wooden block of density 300 kg/m3 floats in water of density 1000 kg/m3. What percentage of the block’s volume is submerged?

  • (A) 40%
  • (B) 30%
  • (C) 15%
  • (D) 60%
Show Answer

Answer: B – For a floating object, fraction submerged = density of object ÷ density of fluid = 300 ÷ 1000 = 0.3. As a percentage, that is 30%.

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