Torque Study Materials for Competitive Exam Preparation

Torque

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

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

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

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

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

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?

A. 5 Nm
B. 10 Nm
C. 20 Nm
D. 15 Nm

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 point?

A. 5 Nm
B. 10 Nm
C. 20 Nm
D. 15 Nm

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

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

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

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

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

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

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

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

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

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

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 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

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 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

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 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

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

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

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

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

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

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

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

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

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

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

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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Torque

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