Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 0.8 m, what is the torque about the hinges?
A.
8 Nm
B.
10 Nm
C.
16 Nm
D.
20 Nm
Show solution
Solution
Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.
Correct Answer:
C
— 16 Nm
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Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 1 m, what is the torque about the hinges?
A.
10 Nm
B.
20 Nm
C.
30 Nm
D.
40 Nm
Show solution
Solution
Torque (τ) = F × d = 20 N × 1 m = 20 Nm.
Correct Answer:
B
— 20 Nm
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Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1 m, what is the torque about the hinges?
A.
25 Nm
B.
50 Nm
C.
75 Nm
D.
100 Nm
Show solution
Solution
Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.
Correct Answer:
B
— 50 Nm
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Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1.2 m, what is the torque about the hinges?
A.
60 Nm
B.
50 Nm
C.
70 Nm
D.
40 Nm
Show solution
Solution
Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.
Correct Answer:
A
— 60 Nm
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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
A.
2.0 Nm
B.
5.0 Nm
C.
10.0 Nm
D.
20.0 Nm
Show solution
Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer:
A
— 2.0 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot point?
A.
5 Nm
B.
10 Nm
C.
20 Nm
D.
15 Nm
Show solution
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer:
C
— 20 Nm
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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
20 Nm
D.
15 Nm
Show solution
Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer:
C
— 20 Nm
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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
Show solution
Solution
Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.
Correct Answer:
B
— 17.32 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
A.
7.5 Nm
B.
12.5 Nm
C.
15 Nm
D.
25 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer:
B
— 12.5 Nm
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
Show solution
Solution
Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer:
B
— 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
30 Nm
D.
40 Nm
Show solution
Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer:
B
— 20 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
8.66 Nm
D.
17.32 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer:
C
— 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
17.32 Nm
D.
34.64 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 17.32 Nm
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer:
D
— 52 Nm
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
Show solution
Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer:
C
— 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer:
C
— 34.64 Nm
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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
86.6 Nm
Show solution
Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer:
C
— 86.6 N·m
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 Nm
B.
50 Nm
C.
43.3 Nm
D.
100 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer:
C
— 43.3 Nm
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer:
B
— 43.3 Nm
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Q. A torque of 10 Nm is applied to a wheel. If the radius of the wheel is 0.2 m, what is the force applied tangentially?
A.
50 N
B.
20 N
C.
10 N
D.
5 N
Show solution
Solution
Torque (τ) = F × r; therefore, F = τ / r = 10 Nm / 0.2 m = 50 N.
Correct Answer:
A
— 50 N
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Q. A torque of 12 Nm is applied to a lever arm of 0.4 m. What is the force applied?
A.
30 N
B.
25 N
C.
20 N
D.
15 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer:
C
— 20 N
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Q. A torque of 12 Nm is applied to a wheel of radius 0.4 m. What is the force applied at the edge of the wheel?
A.
30 N
B.
20 N
C.
15 N
D.
10 N
Show solution
Solution
Torque (τ) = r × F, thus F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer:
B
— 20 N
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Q. A torque of 12 Nm is applied to a wheel with a radius of 0.4 m. What is the force applied tangentially to the wheel?
A.
15 N
B.
30 N
C.
40 N
D.
50 N
Show solution
Solution
Using τ = F × r, we have F = τ / r = 12 Nm / 0.4 m = 30 N.
Correct Answer:
B
— 30 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 0.4 m from the pivot. What is the magnitude of the force?
A.
20 N
B.
30 N
C.
40 N
D.
50 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 0.4 m = 30 N.
Correct Answer:
A
— 20 N
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Q. A torque of 12 Nm is produced by a force acting at a distance of 4 m from the pivot. What is the magnitude of the force?
A.
2 N
B.
3 N
C.
4 N
D.
5 N
Show solution
Solution
Force = Torque / Distance = 12 Nm / 4 m = 3 N.
Correct Answer:
B
— 3 N
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Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?
A.
25 N
B.
50 N
C.
45 N
D.
30 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.
Correct Answer:
B
— 50 N
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Q. A torque of 25 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
50 N
B.
25 N
C.
75 N
D.
100 N
Show solution
Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer:
A
— 50 N
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Q. A torque of 25 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
A.
10 N
B.
25 N
C.
50 N
D.
5 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 25 Nm / 0.5 m = 50 N.
Correct Answer:
A
— 10 N
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Showing 1 to 30 of 75 (3 Pages)
Torque MCQ & Objective Questions
Understanding torque is crucial for students preparing for school exams and competitive tests. Torque is a fundamental concept in physics that often appears in various forms in objective questions. Practicing MCQs related to torque not only enhances your grasp of the topic but also boosts your confidence in tackling important questions during exams.
What You Will Practise Here
Definition of torque and its significance in rotational motion.
Key formulas related to torque, including torque = force x distance.
Understanding the concept of the moment of inertia and its relation to torque.
Applications of torque in real-life scenarios and mechanical systems.
Diagrams illustrating the direction and magnitude of torque.
Common units of torque and conversions between them.
Problem-solving techniques for torque-related MCQs.
Exam Relevance
Torque is a significant topic in various examinations, including CBSE, State Boards, NEET, and JEE. It frequently appears in the form of conceptual questions, numerical problems, and application-based scenarios. Students can expect to encounter questions that require them to calculate torque, analyze rotational systems, or interpret diagrams. Familiarity with torque MCQ questions can greatly enhance your performance in these exams.
Common Mistakes Students Make
Confusing torque with linear force, leading to incorrect application of formulas.
Overlooking the direction of torque, which is essential for solving problems accurately.
Neglecting to consider the point of application of force when calculating torque.
Misunderstanding the relationship between torque and angular acceleration.
FAQs
Question: What is the formula for calculating torque?Answer: The formula for torque is τ = r × F, where τ is torque, r is the distance from the pivot point to the point of force application, and F is the applied force.
Question: How does torque relate to rotational motion?Answer: Torque is the rotational equivalent of linear force and is responsible for causing an object to rotate about an axis.
Get ready to enhance your understanding of torque! Start solving practice MCQs today to test your knowledge and prepare effectively for your exams. Remember, mastering torque will not only help you in physics but also in achieving your academic goals.
