Free OAR Full-length Practice Test – Officer Aptitude Rating Test Prep 2026
This is Our full-length OAR practice test with Answers and Explanations designed to help students prepare for the OAR exam with confidence. This OAR mock test includes 80 questions covering Math Skills, Reading Comprehension, and Mechanical Comprehension, giving you complete practice in one place.
If you are looking for the best OAR practice test online, this page offers realistic OAR exam preparation, detailed explanations, and a structured way to improve your score.
This OAR full-length practice test is ideal for candidates who want to improve speed, accuracy, and test-taking strategy before the real exam.
By solving these 80 OAR practice questions, you can identify weak areas, revise important concepts, and build stronger performance under exam conditions. Whether you need an OAR mock exam, OAR practice questions, or OAR exam preparation, this test helps you practice smarter and rank better in your preparation journey.
Quiz-summary
0 of 80 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
Information
Test Instructions
No. of Questions: 80
Questions: Multiple choice with 4 options (A, B, C, D)
Review: You can flag questions for review and return to them
Submission: Click “Submit Test” when ready to finish.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 80 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- Answered
- Review
-
Question 1 of 80
1. Question
Solve the inequality: 3x − 7 > 2x + 5.
Correct
Subtract 2x: x − 7 > 5. Add 7: x > 12.
Incorrect
Subtract 2x: x − 7 > 5. Add 7: x > 12.
-
Question 2 of 80
2. Question
An item costs $320 before an 8.5 % sales tax is added. What is the total price?
Correct
Tax = 320 × 0.085 = $27.20. Total Price = $320 + $27.20 = $347.20.
Incorrect
Tax = 320 × 0.085 = $27.20. Total Price = $320 + $27.20 = $347.20.
-
Question 3 of 80
3. Question
A trapezoidal deck section has parallel sides of 8 m and 12 m with a perpendicular height of 5 m. What is its area?
Correct
Area = 1/2(a + b) x h = [(8 + 12) ÷ 2] × 5 = 10 × 5 = 50 m².
Incorrect
Area = 1/2(a + b) x h = [(8 + 12) ÷ 2] × 5 = 10 × 5 = 50 m².
-
Question 4 of 80
4. Question
Three workers complete a job in 8 days. How many workers are needed to finish the same job in 6 days?
Correct
Total work = 3 × 8 = 24 man-days. Workers needed = 24 ÷ 6 = 4.
Incorrect
Total work = 3 × 8 = 24 man-days. Workers needed = 24 ÷ 6 = 4.
-
Question 5 of 80
5. Question
A mixture contains oil and water in a 2:3 ratio. If the total mixture is 50 litres, how many litres of oil are present?
Correct
Oil fraction = 2/5. Oil = 50 × 2/5 = 20 litres.
Incorrect
Oil fraction = 2/5. Oil = 50 × 2/5 = 20 litres.
-
Question 6 of 80
6. Question
A quadratic has factors (x − 3) and (x + 5). What is the sum of its two roots?
Correct
Roots are x = 3 and x = −5. Sum = 3 + (−5) = −2.
Incorrect
Roots are x = 3 and x = −5. Sum = 3 + (−5) = −2.
-
Question 7 of 80
7. Question
A navigator drives 60 miles at 40 mph then 60 miles at 60 mph. What is the average speed for the entire trip?
Correct
Total time = 60/40 + 60/60 = 1.5 + 1 = 2.5 h. Average speed = 120 ÷ 2.5 = 48 mph.
Incorrect
Total time = 60/40 + 60/60 = 1.5 + 1 = 2.5 h. Average speed = 120 ÷ 2.5 = 48 mph.
-
Question 8 of 80
8. Question
Evaluate: √(144 ÷ 25).
Correct
√144 = 12; √25 = 5. Result = 12/5 = 2.4.
Incorrect
√144 = 12; √25 = 5. Result = 12/5 = 2.4.
-
Question 9 of 80
9. Question
Value A is 30 % greater than Value B. By approximately what percentage is B less than A?
Correct
B/A = 1/1.30 ≈ 0.769, so B is about 23.1 % less than A.
Incorrect
B/A = 1/1.30 ≈ 0.769, so B is about 23.1 % less than A.
-
Question 10 of 80
10. Question
Solve: 2 ÷ (x − 1) = 4.
Correct
2 = 4(x − 1) → 4x = 6 → x = 1.5.
Incorrect
2 = 4(x − 1) → 4x = 6 → x = 1.5.
-
Question 11 of 80
11. Question
A rectangular compartment measures 6 m × 4 m × 3 m. What is its volume?
Correct
Volume = 6 × 4 × 3 = 72 m³.
Incorrect
Volume = 6 × 4 × 3 = 72 m³.
-
Question 12 of 80
12. Question
Two standard six-sided dice are rolled. What is the probability the sum equals 7?
Correct
Favorable pairs: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) = 6 of 36. P = 1/6.
Incorrect
Favorable pairs: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) = 6 of 36. P = 1/6.
-
Question 13 of 80
13. Question
An investment of $1 000 earns compound interest at 10 % per year. What is the value after 2 years?
Correct
Year 1: $1 000 × 1.10 = $1 100. Year 2: $1 100 × 1.10 = $1 210.
Incorrect
Year 1: $1 000 × 1.10 = $1 100. Year 2: $1 100 × 1.10 = $1 210.
-
Question 14 of 80
14. Question
A geometric sequence starts 3, 6, 12, 24 … What is the sum of the first 6 terms?
Correct
Sum of ‘n’ terms in G.P with common ration ‘r’ = a(rⁿ – 1)/(r – 1) , where a = 3 , Common ration r = a₂/a₁ = 6/3 = 2 and n = 6. therefore , Sum = 3(2⁶ − 1)/(2 − 1) = 3 × 63 = 189.
Incorrect
Sum of ‘n’ terms in G.P with common ration ‘r’ = a(rⁿ – 1)/(r – 1) , where a = 3 , Common ration r = a₂/a₁ = 6/3 = 2 and n = 6. therefore , Sum = 3(2⁶ − 1)/(2 − 1) = 3 × 63 = 189.
-
Question 15 of 80
15. Question
John is currently twice Mary’s age. In 5 years, John will be 1.5 times Mary’s age. How old is Mary now?
Correct
Let M = Mary’s age, J = 2M. In 5 years: 2M+5 = 1.5(M+5) → 0.5M = 2.5 → M = 5.
Incorrect
Let M = Mary’s age, J = 2M. In 5 years: 2M+5 = 1.5(M+5) → 0.5M = 2.5 → M = 5.
-
Question 16 of 80
16. Question
A stock rises from $80 to $92. What is the percentage increase?
Correct
increase = $12. Percentage = (12 ÷ 80) × 100 = 15 %.
Incorrect
increase = $12. Percentage = (12 ÷ 80) × 100 = 15 %.
-
Question 17 of 80
17. Question
Two-thirds of a group are women, and three-quarters of the women wear glasses. What fraction of the total group wears glasses?
Correct
(2/3) × (3/4) = 1/2 of the total group wears glasses.
Incorrect
(2/3) × (3/4) = 1/2 of the total group wears glasses.
-
Question 18 of 80
18. Question
A line passes through (2, 3) and (6, 11). What is the equation of the line?
Correct
Slope = 8/4 = 2. Using (2,3): 3 = 4 + b → b = −1. Equation: y = 2x − 1.
Incorrect
Slope = 8/4 = 2. Using (2,3): 3 = 4 + b → b = −1. Equation: y = 2x − 1.
-
Question 19 of 80
19. Question
Red and blue balls are in a 3:5 ratio, totaling 48 balls. How many are red?
Correct
Red = 48 × (3/8) = 18.
Incorrect
Red = 48 × (3/8) = 18.
-
Question 20 of 80
20. Question
Ship A leaves port at 30 knots. Ship B leaves the same port in the same direction 2 hours later at 45 knots. How many hours after Ship B departs does it catch Ship A?
Correct
Head start = 30 × 2 = 60 nm. Closing speed = 15 knots. Time = 60 ÷ 15 = 4 hours.
Incorrect
Head start = 30 × 2 = 60 nm. Closing speed = 15 knots. Time = 60 ÷ 15 = 4 hours.
-
Question 21 of 80
21. Question
Given (x + y)² = 49 and (x − y)² = 9, what is the value of xy?
Correct
Expanding and subtracting: 4xy = 49 − 9 = 40 → xy = 10.
Incorrect
Expanding and subtracting: 4xy = 49 − 9 = 40 → xy = 10.
-
Question 22 of 80
22. Question
What is the volume of a sphere with radius 3? (Use π ≈ 3.14)
Correct
V = (4/3)πr³ = (4/3) × 3.14 × 27 ≈ 113.10.
Incorrect
V = (4/3)πr³ = (4/3) × 3.14 × 27 ≈ 113.10.
-
Question 23 of 80
23. Question
15 is what percentage of 60?
Correct
(15 ÷ 60) × 100 = 25 %.
Incorrect
(15 ÷ 60) × 100 = 25 %.
-
Question 24 of 80
24. Question
In an arithmetic sequence, the 5th term is 29 and the 9th term is 53. What is the common difference?
Correct
difference(d) = (53 − 29) ÷ (9 − 5) = 24 ÷ 4 = 6.
Incorrect
difference(d) = (53 − 29) ÷ (9 − 5) = 24 ÷ 4 = 6.
-
Question 25 of 80
25. Question
A supply order includes 3 items at $12.50 each and 5 items at $8.75 each. What is the total cost?
Correct
3 × $12.50 = $37.50. 5 × $8.75 = $43.75. Total = $81.25.
Incorrect
3 × $12.50 = $37.50. 5 × $8.75 = $43.75. Total = $81.25.
-
Question 26 of 80
26. Question
Solve: (3x + 2)/4 = (x + 5)/2.
Correct
Multiply both sides by 4: 3x+2 = 2(x+5) → 3x+2 = 2x+10 → x = 8.
Incorrect
Multiply both sides by 4: 3x+2 = 2(x+5) → 3x+2 = 2x+10 → x = 8.
-
Question 27 of 80
27. Question
A semicircle has a diameter of 10 m. What is its area? (Use π ≈ 3.14)
Correct
Full-circle area = π × 5² = 78.5 m². Semicircle = 78.5 ÷ 2 = 39.25 m².
Incorrect
Full-circle area = π × 5² = 78.5 m². Semicircle = 78.5 ÷ 2 = 39.25 m².
-
Question 28 of 80
28. Question
In how many ways can 4 books be arranged on a shelf?
Correct
4! = 4 × 3 × 2 × 1 = 24.
Incorrect
4! = 4 × 3 × 2 × 1 = 24.
-
Question 29 of 80
29. Question
Solve: |2x − 3| ≤ 7. Which range contains all solutions?
Correct
−7 ≤ 2x−3 ≤ 7 → −4 ≤ 2x ≤ 10 → −2 ≤ x ≤ 5.
Incorrect
−7 ≤ 2x−3 ≤ 7 → −4 ≤ 2x ≤ 10 → −2 ≤ x ≤ 5.
-
Question 30 of 80
30. Question
A store sells Item A (120 units at $5), Item B (80 units at $12), and Item C (40 units at $20). What is the total revenue?
Correct
Total revenue = 120×5 + 80×12 + 40×20 = 600 + 960 + 800 = $2 360.
Incorrect
Total revenue = 120×5 + 80×12 + 40×20 = 600 + 960 + 800 = $2 360.
-
Question 31 of 80
31. Question
A passage explains that effective leadership in high-stress naval environments demands both technical competence and emotional resilience, arguing that officers lacking either quality make measurably poorer decisions under pressure. What is the main idea?
Correct
The passage presents both qualities jointly as necessary—neither alone is sufficient.
Incorrect
The passage presents both qualities jointly as necessary—neither alone is sufficient.
-
Question 32 of 80
32. Question
A passage describes a team that repeatedly postponed structural inspections citing budget constraints, until a minor crack was found during routine cleaning. What can be inferred?
Correct
Repeated postponement followed by a discovered defect implies inspection delays allowed the problem to go unnoticed.
Incorrect
Repeated postponement followed by a discovered defect implies inspection delays allowed the problem to go unnoticed.
-
Question 33 of 80
33. Question
An admiral is described as being ‘circumspect in her assessment, weighing each factor methodically before committing.’ What does ‘circumspect’ most nearly mean?
Correct
Methodical weighing before committing indicates careful, cautious deliberation.
Incorrect
Methodical weighing before committing indicates careful, cautious deliberation.
-
Question 34 of 80
34. Question
A passage details three case studies where poor communication aboard ships led to accidents, using each to illustrate a specific failure mode. What is the author’s primary purpose?
Correct
Structured case studies each targeting a failure mode signals analytical and explanatory intent.
Incorrect
Structured case studies each targeting a failure mode signals analytical and explanatory intent.
-
Question 35 of 80
35. Question
A passage first establishes that a new radar system was expensive, then traces how it reduced navigation errors over two years, and finally projects long-term cost savings from fewer accidents. What pattern is used?
Correct
Expenditure (cause) linked to reduced errors and savings (effects) is a cause-and-effect structure.
Incorrect
Expenditure (cause) linked to reduced errors and savings (effects) is a cause-and-effect structure.
-
Question 36 of 80
36. Question
A passage about a rescue mission uses phrases like ‘remarkable composure,’ ‘seamless coordination,’ and ‘outstanding valor.’ What is the dominant tone?
Correct
Phrases praising composure, coordination, and valor convey admiration and celebration.
Incorrect
Phrases praising composure, coordination, and valor convey admiration and celebration.
-
Question 37 of 80
37. Question
A report states: ‘The new hull design reduced drag by 12 % in controlled tests, making it the single greatest engineering advance in the fleet’s history.’ What best describes this statement?
Correct
Measured test results are empirical; characterising it as the ‘single greatest advance’ is a sweeping evaluative judgment.
Incorrect
Measured test results are empirical; characterising it as the ‘single greatest advance’ is a sweeping evaluative judgment.
-
Question 38 of 80
38. Question
An essay argues that physical exercise improves cognitive performance under stress. Which evidence best supports this?
Correct
Empirical data directly linking exercise to measurable error reduction under simulated stress is the strongest support.
Incorrect
Empirical data directly linking exercise to measurable error reduction under simulated stress is the strongest support.
-
Question 39 of 80
39. Question
Vessels with automated damage-control systems suffered 40 % fewer flooding-related casualties than those using only manual procedures, across 50 incident reports. What is the best conclusion?
Correct
The data shows a strong association without claiming elimination or total failure.
Incorrect
The data shows a strong association without claiming elimination or total failure.
-
Question 40 of 80
40. Question
A passage states the admiral’s decision was ‘pragmatic rather than idealistic, prioritising operational readiness over theoretical perfection.’ What does ‘pragmatic’ most nearly mean?
Correct
Choosing readiness over theoretical perfection reflects a practical, results-focused approach.
Incorrect
Choosing readiness over theoretical perfection reflects a practical, results-focused approach.
-
Question 41 of 80
41. Question
A passage argues that advanced simulation training is cost-effective because it reduces equipment damage, lowers fuel costs, and permits more repetitions per training hour. What is the main idea?
Correct
The passage focuses on several cost-saving dimensions of simulation training taken together.
Incorrect
The passage focuses on several cost-saving dimensions of simulation training taken together.
-
Question 42 of 80
42. Question
A passage notes a crew received updated safety briefings every two weeks, and injury rates fell by half over the following year. What can most reasonably be inferred?
Correct
The timeline linking regular briefings to a halved injury rate supports a reasonable inference of contributory causation.
Incorrect
The timeline linking regular briefings to a halved injury rate supports a reasonable inference of contributory causation.
-
Question 43 of 80
43. Question
A passage opens with outdated navigation equipment, examines how each limitation caused operational problems, then closes with the capabilities of a modern replacement. How is it organised?
Correct
Identifying a problem (old equipment’s limits) and presenting a solution (modern replacement) is the problem-solution pattern.
Incorrect
Identifying a problem (old equipment’s limits) and presenting a solution (modern replacement) is the problem-solution pattern.
-
Question 44 of 80
44. Question
A passage about a training accident uses measured language, listing contributing factors and recommending corrective actions without attributing blame. What is the tone?
Correct
Systematic listing of factors and recommendations without blame indicates an analytical, objective tone.
Incorrect
Systematic listing of factors and recommendations without blame indicates an analytical, objective tone.
-
Question 45 of 80
45. Question
A passage contends that properly maintained equipment extends a vessel’s operational service life significantly. Which detail most directly supports this?
Correct
Longitudinal service data directly linking maintenance to longer operational life is the most specific and relevant support.
Incorrect
Longitudinal service data directly linking maintenance to longer operational life is the most specific and relevant support.
-
Question 46 of 80
46. Question
A passage catalogues navigation tools from sextants to GPS satellites, providing technical descriptions and historical dates for each without commentary on which is superior. What is the author’s purpose?
Correct
Technical descriptions and historical dates without evaluation signal an informative purpose.
Incorrect
Technical descriptions and historical dates without evaluation signal an informative purpose.
-
Question 47 of 80
47. Question
A passage states: ‘Fuel consumption increased by 18 % following the engine retrofit, and this result unambiguously proves the retrofit was a complete failure.’ What best describes this?
Correct
Measured fuel consumption is empirical; characterising the outcome as a ‘complete failure’ based solely on that is a judgment call.
Incorrect
Measured fuel consumption is empirical; characterising the outcome as a ‘complete failure’ based solely on that is a judgment call.
-
Question 48 of 80
48. Question
A study finds junior officers who receive structured feedback within 24 hours of an exercise perform significantly better on the next exercise than those who receive feedback after one week. What is the most valid conclusion?
Correct
The data compares early versus late feedback and finds better outcomes for early feedback.
Incorrect
The data compares early versus late feedback and finds better outcomes for early feedback.
-
Question 49 of 80
49. Question
A passage notes that ship communication protocols were ‘antiquated, causing systemic delays across all command levels.’ What does ‘antiquated’ most nearly mean?
Correct
Causing systemic delays signals the protocols are old and no longer suitable.
Incorrect
Causing systemic delays signals the protocols are old and no longer suitable.
-
Question 50 of 80
50. Question
A passage describes a submarine crew that maintained optimal performance throughout a demanding 90-day deployment; the author notes the crew had undergone an intensive pre-deployment psychological screening and cohesion programme. What can be inferred?
Correct
Juxtaposing the programme and optimal performance over a demanding deployment suggests a likely positive contribution.
Incorrect
Juxtaposing the programme and optimal performance over a demanding deployment suggests a likely positive contribution.
-
Question 51 of 80
51. Question
A class-1 lever has a 800 N load placed 0.5 m from the fulcrum, and the effort arm is 2 m long. What effort force balances it?
Correct
800 × 0.5 = F × 2 → F = 400 ÷ 2 = 200 N.
Incorrect
800 × 0.5 = F × 2 → F = 400 ÷ 2 = 200 N.
-
Question 52 of 80
52. Question
A block-and-tackle system uses 5 rope segments supporting a 750 N load. Ignoring friction, what input force is required?
Correct
Input force = 750 ÷ 5 = 150 N.
Incorrect
Input force = 750 ÷ 5 = 150 N.
-
Question 53 of 80
53. Question
A driving gear has 45 teeth at 80 RPM and meshes with a driven gear having 15 teeth. What is the driven gear’s speed?
Correct
Speed_driven = Speed_driver × (Teeth_driver ÷ Teeth_driven). therefore, Driver (Gear A): 45 teeth at 80 RPM Driven (Gear B): 15 teeth Speed of B = 80 × (45 ÷ 15) = 80 × 3 = 240 RPM
Incorrect
Speed_driven = Speed_driver × (Teeth_driver ÷ Teeth_driven). therefore, Driver (Gear A): 45 teeth at 80 RPM Driven (Gear B): 15 teeth Speed of B = 80 × (45 ÷ 15) = 80 × 3 = 240 RPM
-
Question 54 of 80
54. Question
A hydraulic press has a small cylinder of 10 cm² and a large cylinder of 150 cm². If 60 N is applied to the small piston, what force acts on the large piston?
Correct
F₂ = F₁ × (A₂ ÷ A₁).therefore, F₁ = 60 N (input), A₁ = 10 cm² (small piston), A₂ = 150 cm² (large piston). F₂ = 60 × (150 ÷ 10) = 60 × 15 = 900 N
Incorrect
F₂ = F₁ × (A₂ ÷ A₁).therefore, F₁ = 60 N (input), A₁ = 10 cm² (small piston), A₂ = 150 cm² (large piston). F₂ = 60 × (150 ÷ 10) = 60 × 15 = 900 N
-
Question 55 of 80
55. Question
Two resistors of 8 Ω and 24 Ω are connected in parallel. What is the total resistance?
Correct
for two parallel resistors: 1/R_total = 1/R₁ + 1/R₂ . therefore: 1/R = 1/8 + 1/24 Find a common denominator: 1/8 = 3/24 1/R = 3/24 + 1/24 = 4/24 R = 24 ÷ 4 = 6 Ω
Incorrect
for two parallel resistors: 1/R_total = 1/R₁ + 1/R₂ . therefore: 1/R = 1/8 + 1/24 Find a common denominator: 1/8 = 3/24 1/R = 3/24 + 1/24 = 4/24 R = 24 ÷ 4 = 6 Ω
-
Question 56 of 80
56. Question
A submarine hatch has an area of 0.5 m² and is subjected to water pressure of 300 000 Pa. What total force acts on the hatch?
Correct
force: F = P × A . therefore: Pressure = 300 000 Pa (Pa = N/m²), Area = 0.5 m². Force = 300 000 × 0.5 = 150 000 N
Incorrect
force: F = P × A . therefore: Pressure = 300 000 Pa (Pa = N/m²), Area = 0.5 m². Force = 300 000 × 0.5 = 150 000 N
-
Question 57 of 80
57. Question
A ramp is 12 m long and rises 3 m. A 600 N crate is pushed up the ramp (no friction). What force is required?
Correct
MECHANICAL ADVANTAGE (MA) = Ramp length ÷ Rise height. Required force = Load ÷ MA. therefore: MA = 12 ÷ 3 = 4 -> Force = 600 ÷ 4 = 150 N
Incorrect
MECHANICAL ADVANTAGE (MA) = Ramp length ÷ Rise height. Required force = Load ÷ MA. therefore: MA = 12 ÷ 3 = 4 -> Force = 600 ÷ 4 = 150 N
-
Question 58 of 80
58. Question
A pipe wrench applies 200 N perpendicular to a handle 0.25 m long. What is the resulting torque?
Correct
torque, τ = F × d × sin(θ) — when force is perpendicular, θ = 90°, sin(90°) = 1, so: τ = F × d. therefore: Force (F) = 200 N Distance (d) = 0.25 m Torque = 200 × 0.25 = 50 N.m
Incorrect
torque, τ = F × d × sin(θ) — when force is perpendicular, θ = 90°, sin(90°) = 1, so: τ = F × d. therefore: Force (F) = 200 N Distance (d) = 0.25 m Torque = 200 × 0.25 = 50 N.m
-
Question 59 of 80
59. Question
A 10 kg mass falls from rest through 20 m. Using conservation of energy (g = 10 m/s²), what is its kinetic energy just before impact?
Correct
When an object falls (ignoring air resistance), all gravitational potential energy (GPE) converts to kinetic energy (KE). GPE at top = KE at bottom mgh = ½mv². therefore (using energy): GPE = m × g × h = 10 × 10 × 20 = 2 000 J KE at impact = 2 000 J (all GPE converted to KE)
Incorrect
When an object falls (ignoring air resistance), all gravitational potential energy (GPE) converts to kinetic energy (KE). GPE at top = KE at bottom mgh = ½mv². therefore (using energy): GPE = m × g × h = 10 × 10 × 20 = 2 000 J KE at impact = 2 000 J (all GPE converted to KE)
-
Question 60 of 80
60. Question
A wooden block floats with 60 % of its volume submerged in fresh water (density 1 000 kg/m³). What is the wood’s density?
Correct
ARCHIMEDES’ PRINCIPLE: A floating object displaces its own weight of fluid. For a floating object, the fraction submerged equals the ratio of the object’s density to the fluid’s density. Fraction submerged = ρ_object ÷ ρ_fluid. therefore: Fraction submerged = 60 % = 0.60, ρ_water = 1 000 kg/m³ ρ_wood = 0.60 × 1 000 = 600 kg/m³
Incorrect
ARCHIMEDES’ PRINCIPLE: A floating object displaces its own weight of fluid. For a floating object, the fraction submerged equals the ratio of the object’s density to the fluid’s density. Fraction submerged = ρ_object ÷ ρ_fluid. therefore: Fraction submerged = 60 % = 0.60, ρ_water = 1 000 kg/m³ ρ_wood = 0.60 × 1 000 = 600 kg/m³
-
Question 61 of 80
61. Question
A 200 N force is required to slide a 500 N crate at constant velocity. What is the coefficient of kinetic friction?
Correct
Friction force = μₖ × Normal force → μₖ = F_friction ÷ F_normal. KEY: At CONSTANT VELOCITY, acceleration = 0, so net force = 0. This means the applied force exactly equals the friction force. therefore: F_friction = 200 N (equal to applied force at constant velocity) F_normal = 500 N (weight of crate on a flat surface) μₖ = 200 ÷ 500 = 0.40
Incorrect
Friction force = μₖ × Normal force → μₖ = F_friction ÷ F_normal. KEY: At CONSTANT VELOCITY, acceleration = 0, so net force = 0. This means the applied force exactly equals the friction force. therefore: F_friction = 200 N (equal to applied force at constant velocity) F_normal = 500 N (weight of crate on a flat surface) μₖ = 200 ÷ 500 = 0.40
-
Question 62 of 80
62. Question
Two meshing gears have a speed ratio of 3:1 (driver to driven). If the driver has 12 teeth, how many teeth does the driven gear have?
Correct
Gear speed ratio and tooth ratio are INVERSE of each other: Speed_driver / Speed_driven = Teeth_driven / Teeth_driver If the driver spins 3× faster than the driven, the driven must have 3× MORE teeth. therefore: Speed ratio (driver : driven) = 3 : 1 Tooth ratio (driver : driven) = 1 : 3 (inverse) Driver has 12 teeth → Driven has 12 × 3 = 36 teeth
Incorrect
Gear speed ratio and tooth ratio are INVERSE of each other: Speed_driver / Speed_driven = Teeth_driven / Teeth_driver If the driver spins 3× faster than the driven, the driven must have 3× MORE teeth. therefore: Speed ratio (driver : driven) = 3 : 1 Tooth ratio (driver : driven) = 1 : 3 (inverse) Driver has 12 teeth → Driven has 12 × 3 = 36 teeth
-
Question 63 of 80
63. Question
A spring with k = 250 N/m is compressed 0.20 m. How much elastic potential energy is stored?
Correct
HOOKE’S LAW: The force needed to stretch or compress a spring is proportional to the displacement: F = kx. The ENERGY stored in a compressed or stretched spring is: PE_spring = ½ × k × x². therefore: k = 250 N/m (spring constant) x = 0.20 m (compression distance) x² = 0.20² = 0.04 m² PE = ½ × 250 × 0.04 = 125 × 0.04 = 5.0 J
Incorrect
HOOKE’S LAW: The force needed to stretch or compress a spring is proportional to the displacement: F = kx. The ENERGY stored in a compressed or stretched spring is: PE_spring = ½ × k × x². therefore: k = 250 N/m (spring constant) x = 0.20 m (compression distance) x² = 0.20² = 0.04 m² PE = ½ × 250 × 0.04 = 125 × 0.04 = 5.0 J
-
Question 64 of 80
64. Question
A driving pulley of radius 15 cm rotates at 100 RPM and drives a pulley of radius 5 cm via a belt. What is the driven pulley’s speed?
Correct
RPM_driven = RPM_driver × (r_driver ÷ r_driven). Here, RPM_driver = 100 RPM, r_driver = 15 cm, r_driven = 5 cm. So, RPM_driven = 100 × (15 ÷ 5) = 100 × 3 = 300 RPM
Incorrect
RPM_driven = RPM_driver × (r_driver ÷ r_driven). Here, RPM_driver = 100 RPM, r_driver = 15 cm, r_driven = 5 cm. So, RPM_driven = 100 × (15 ÷ 5) = 100 × 3 = 300 RPM
-
Question 65 of 80
65. Question
A hydraulic jack must lift a 12 000 N car. The large piston is 200 cm² and the small piston is 20 cm². What force must be applied to the small piston?
Correct
Pascal’s Principle in a hydraulic jack: pressure is equal on both pistons. P₁ = P₂ → F₁/A₁ = F₂/A₂ Rearranged: F₁ = F₂ × (A₁ ÷ A₂). Here: F₂ (output) = 12 000 N (car’s weight), A₂ (large piston) = 200 cm², A₁ (small piston) = 20 cm². So, F₁ = 12 000 × (20 ÷ 200) = 12 000 × 0.10 = 1 200 N
Incorrect
Pascal’s Principle in a hydraulic jack: pressure is equal on both pistons. P₁ = P₂ → F₁/A₁ = F₂/A₂ Rearranged: F₁ = F₂ × (A₁ ÷ A₂). Here: F₂ (output) = 12 000 N (car’s weight), A₂ (large piston) = 200 cm², A₁ (small piston) = 20 cm². So, F₁ = 12 000 × (20 ÷ 200) = 12 000 × 0.10 = 1 200 N
-
Question 66 of 80
66. Question
A circuit operates at 12 V with a total resistance of 4 Ω. What current flows through it?
Correct
OHM’S LAW: Voltage = Current × Resistance (V = IR) Rearranged for current: I = V ÷ R. Here: V = 12 V , R = 4 Ω .So, I = 12 ÷ 4 = 3 A UNIT
Incorrect
OHM’S LAW: Voltage = Current × Resistance (V = IR) Rearranged for current: I = V ÷ R. Here: V = 12 V , R = 4 Ω .So, I = 12 ÷ 4 = 3 A UNIT
-
Question 67 of 80
67. Question
A ship’s wheel has a radius of 40 cm; the connected axle has a radius of 8 cm. What is the mechanical advantage?
Correct
Mechanical Advantage (MA) = Radius of wheel ÷ Radius of axle. Here: Wheel radius = 40 cm, Axle radius = 8 cm. So, MA = 40 ÷ 8 = 5
Incorrect
Mechanical Advantage (MA) = Radius of wheel ÷ Radius of axle. Here: Wheel radius = 40 cm, Axle radius = 8 cm. So, MA = 40 ÷ 8 = 5
-
Question 68 of 80
68. Question
A 60 kg person stands 1 m from one end of a uniform 4 m plank weighing 20 kg. How far from that same end must a support be placed to balance the plank?
Correct
he CENTER OF MASS (CM) is the balance point of a system. For a uniform plank, its own CM is at its midpoint. Placing the support at the system’s CM creates balance. SETUP: Let x = 0 be the end where the person stands. Person: 60 kg at x = 1 m Plank CM: 20 kg at x = 2 m (midpoint of 4 m plank) FORMULA: CM = (m₁x₁ + m₂x₂) ÷ (m₁ + m₂). So, CM = (60×1 + 20×2) ÷ (60 + 20) = (60 + 40) ÷ 80 = 100 ÷ 80 = 1.25 m from the person’s end
Incorrect
he CENTER OF MASS (CM) is the balance point of a system. For a uniform plank, its own CM is at its midpoint. Placing the support at the system’s CM creates balance. SETUP: Let x = 0 be the end where the person stands. Person: 60 kg at x = 1 m Plank CM: 20 kg at x = 2 m (midpoint of 4 m plank) FORMULA: CM = (m₁x₁ + m₂x₂) ÷ (m₁ + m₂). So, CM = (60×1 + 20×2) ÷ (60 + 20) = (60 + 40) ÷ 80 = 100 ÷ 80 = 1.25 m from the person’s end
-
Question 69 of 80
69. Question
The gauge pressure from water alone at a certain depth is 200 000 Pa. With water density 1 000 kg/m³ and g = 10 m/s², what is the depth?
Correct
Gauge pressure due to a liquid column: P = ρ × g × h Where ρ = density, g = gravitational acceleration, h = depth. Rearranged to find depth: h = P ÷ (ρ × g) Here: P = 200 000 Pa, ρ = 1 000 kg/m³, g = 10 m/s². So, h = 200 000 ÷ (1 000 × 10) = 200 000 ÷ 10 000 = 20 m.
Incorrect
Gauge pressure due to a liquid column: P = ρ × g × h Where ρ = density, g = gravitational acceleration, h = depth. Rearranged to find depth: h = P ÷ (ρ × g) Here: P = 200 000 Pa, ρ = 1 000 kg/m³, g = 10 m/s². So, h = 200 000 ÷ (1 000 × 10) = 200 000 ÷ 10 000 = 20 m.
-
Question 70 of 80
70. Question
A pump lifts 500 kg of water to a height of 6 m in 30 seconds. What is its power output? (g = 10 m/s²)
Correct
WORK = Force × Distance = mgh (for vertical lifting), POWER = Work ÷ Time (how fast work is done) Unit: Watts (W) = Joules per second. therefore: Work = m × g × h = 500 × 10 × 6 = 30 000 J and Power = Work ÷ Time = 30 000 ÷ 30 = 1 000 W
Incorrect
WORK = Force × Distance = mgh (for vertical lifting), POWER = Work ÷ Time (how fast work is done) Unit: Watts (W) = Joules per second. therefore: Work = m × g × h = 500 × 10 × 6 = 30 000 J and Power = Work ÷ Time = 30 000 ÷ 30 = 1 000 W
-
Question 71 of 80
71. Question
A rotating shaft completes 300 revolutions per minute. What is its angular speed in radians per second?
Correct
ANGULAR SPEED conversion: RPM → rad/s One full revolution = 2π radians One minute = 60 seconds. Formula: ω (rad/s) = RPM × (2π ÷ 60) = RPM × π/30. therefore: 300 RPM ÷ 60 s/min = 5 revolutions per second 5 rev/s × 2π rad/rev = 10π rad/s
Incorrect
ANGULAR SPEED conversion: RPM → rad/s One full revolution = 2π radians One minute = 60 seconds. Formula: ω (rad/s) = RPM × (2π ÷ 60) = RPM × π/30. therefore: 300 RPM ÷ 60 s/min = 5 revolutions per second 5 rev/s × 2π rad/rev = 10π rad/s
-
Question 72 of 80
72. Question
Water flows at 4 m/s through a pipe with cross-section 0.05 m² that narrows to 0.02 m². What is the water speed in the narrower section?
Correct
THE CONTINUITY EQUATION: For an incompressible fluid (like water), the volume flow rate is constant throughout a pipe, regardless of changes in cross-section: A₁ × v₁ = A₂ × v₂ (Area × Velocity = constant). Here: A₁ = 0.05 m², v₁ = 4 m/s A₂ = 0.02 m², v₂ = ? So, 0.05 × 4 = 0.02 × v₂ => 0.20 = 0.02 × v₂ => v₂ = 0.20 ÷ 0.02 = 10 m/s
Incorrect
THE CONTINUITY EQUATION: For an incompressible fluid (like water), the volume flow rate is constant throughout a pipe, regardless of changes in cross-section: A₁ × v₁ = A₂ × v₂ (Area × Velocity = constant). Here: A₁ = 0.05 m², v₁ = 4 m/s A₂ = 0.02 m², v₂ = ? So, 0.05 × 4 = 0.02 × v₂ => 0.20 = 0.02 × v₂ => v₂ = 0.20 ÷ 0.02 = 10 m/s
-
Question 73 of 80
73. Question
A simple machine requires 50 N of input to move a 250 N load. What is its actual mechanical advantage?
Correct
MECHANICAL ADVANTAGE (MA) tells you how much a machine multiplies your input force. FORMULA: MA = Output force ÷ Input force = Load ÷ Effort. Here : Output force (load) = 250 N, Input force (effort) = 50 N. So, MA = 250 ÷ 50 = 5
Incorrect
MECHANICAL ADVANTAGE (MA) tells you how much a machine multiplies your input force. FORMULA: MA = Output force ÷ Input force = Load ÷ Effort. Here : Output force (load) = 250 N, Input force (effort) = 50 N. So, MA = 250 ÷ 50 = 5
-
Question 74 of 80
74. Question
A 60 W light bulb operates on a 120 V circuit. What current does it draw?
Correct
POWER in an electrical circuit: P = V × I (Power = Voltage × Current) Rearranged for current: I = P ÷ V . Here: P = 60 W ,V = 120 V. So, I = 60 ÷ 120 = 0.5 A
Incorrect
POWER in an electrical circuit: P = V × I (Power = Voltage × Current) Rearranged for current: I = P ÷ V . Here: P = 60 W ,V = 120 V. So, I = 60 ÷ 120 = 0.5 A
-
Question 75 of 80
75. Question
A block rests on a surface inclined at 30°. The coefficient of static friction is 0.6. Will the block slide? (tan 30° ≈ 0.577)
Correct
SLIDING CONDITION: tan(θ) > μₛ , NO SLIDING: tan(θ) ≤ μₛ ,WHY tan(θ)? On an incline: Gravity component along slope = mg sin θ Maximum static friction = μₛ × Normal force = μₛ × mg cos θ, Sliding when: mg sin θ > μₛ × mg cos θ Simplify: sin θ / cos θ > μₛ → tan θ > μₛ. Here: tan 30° ≈ 0.577, μₛ = 0.6 . So, Compare: 0.577 < 0.6 → tan θ < μₛ → NO SLIDING
Incorrect
SLIDING CONDITION: tan(θ) > μₛ , NO SLIDING: tan(θ) ≤ μₛ ,WHY tan(θ)? On an incline: Gravity component along slope = mg sin θ Maximum static friction = μₛ × Normal force = μₛ × mg cos θ, Sliding when: mg sin θ > μₛ × mg cos θ Simplify: sin θ / cos θ > μₛ → tan θ > μₛ. Here: tan 30° ≈ 0.577, μₛ = 0.6 . So, Compare: 0.577 < 0.6 → tan θ < μₛ → NO SLIDING
-
Question 76 of 80
76. Question
A worm gear has 1 thread; its mating worm wheel has 40 teeth. If the worm rotates at 400 RPM, how fast does the worm wheel rotate?
Correct
Wheel RPM = Worm RPM × (Number of worm starts ÷ Number of wheel teeth). Here: Worm starts = 1 (single thread), Wheel teeth = 40, Gear ratio = 40 ÷ 1 = 40 : 1 (worm turns 40× faster than wheel). So, Wheel RPM = 400 ÷ 40 = 10 RPM
Incorrect
Wheel RPM = Worm RPM × (Number of worm starts ÷ Number of wheel teeth). Here: Worm starts = 1 (single thread), Wheel teeth = 40, Gear ratio = 40 ÷ 1 = 40 : 1 (worm turns 40× faster than wheel). So, Wheel RPM = 400 ÷ 40 = 10 RPM
-
Question 77 of 80
77. Question
A sealed gas at 2 atm occupies 5 L and is compressed to 2 L at constant temperature. What is the new pressure?
Correct
BOYLE’S LAW: At constant temperature, the pressure and volume of a fixed amount of gas are inversely proportional. FORMULA: P₁ × V₁ = P₂ × V₂ INTUITION: Squeeze the same number of gas molecules into a smaller space → they hit the walls more often → pressure rises. Here: P₁ = 2 atm, V₁ = 5 L, V₂ = 2 L. So, P₁V₁ = P₂V₂ → 2 × 5 = P₂ × 2 10 = 2 P₂ P₂ = 10 ÷ 2 = 5 atm
Incorrect
BOYLE’S LAW: At constant temperature, the pressure and volume of a fixed amount of gas are inversely proportional. FORMULA: P₁ × V₁ = P₂ × V₂ INTUITION: Squeeze the same number of gas molecules into a smaller space → they hit the walls more often → pressure rises. Here: P₁ = 2 atm, V₁ = 5 L, V₂ = 2 L. So, P₁V₁ = P₂V₂ → 2 × 5 = P₂ × 2 10 = 2 P₂ P₂ = 10 ÷ 2 = 5 atm
-
Question 78 of 80
78. Question
A compressed spring stores 180 J and releases all of it to a 2 kg ball. What is the ball’s speed when fully released?
Correct
CONSERVATION OF ENERGY: All elastic potential energy (PE) of the spring converts to kinetic energy (KE) of the ball. PE_spring = KE_ball => ½mv² = PE_spring → v = √(2 × PE ÷ m). Here: PE = 180 J, m = 2 kg. So, ½ × 2 × v² = 180 => v² = 180 → v = √180 => √180 = √(36 × 5) = 6√5 ≈ 6 × 2.236 ≈ 13.4 m/s
Incorrect
CONSERVATION OF ENERGY: All elastic potential energy (PE) of the spring converts to kinetic energy (KE) of the ball. PE_spring = KE_ball => ½mv² = PE_spring → v = √(2 × PE ÷ m). Here: PE = 180 J, m = 2 kg. So, ½ × 2 × v² = 180 => v² = 180 → v = √180 => √180 = √(36 × 5) = 6√5 ≈ 6 × 2.236 ≈ 13.4 m/s
-
Question 79 of 80
79. Question
A single fixed pulley is used to lift a 200 N load. Ignoring friction, what input force is needed?
Correct
A SINGLE FIXED PULLEY changes only the DIRECTION of the force — it provides no mechanical advantage. The rope tension is the same on both sides. Mechanical Advantage of a single fixed pulley = 1 , Input force = Load ÷ MA = 200 ÷ 1 = 200 N
Incorrect
A SINGLE FIXED PULLEY changes only the DIRECTION of the force — it provides no mechanical advantage. The rope tension is the same on both sides. Mechanical Advantage of a single fixed pulley = 1 , Input force = Load ÷ MA = 200 ÷ 1 = 200 N
-
Question 80 of 80
80. Question
A tugboat exerts a constant 5 000 N force on a barge moving at a steady 3 m/s. What is the tugboat’s power output?
Correct
POWER is the rate of doing work. When force and velocity are constant and in the same direction: FORMULA: P = F × v (Power = Force × velocity). Here: F = 5 000 N, v = 3 m/s. So, P = 5 000 × 3 = 15 000 W
Incorrect
POWER is the rate of doing work. When force and velocity are constant and in the same direction: FORMULA: P = F × v (Power = Force × velocity). Here: F = 5 000 N, v = 3 m/s. So, P = 5 000 × 3 = 15 000 W
Next –
OAR Exam Study Guide > >