Torque
A mechanic applies a force of 50 N perpendicular to a spanner of length 0.30 m to loosen a tight bolt at the far end. What magnitude of torque does the mechanic exert about the bolt?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- torque
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
15 N·m
According to NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), torque is the rotational analogue of force and equals the product of the force, the lever-arm distance, and the sine of the angle between them, τ = rF sinθ. Torque determines how effectively a force produces angular acceleration about a chosen axis. Here the force is applied perpendicular to the spanner, so θ = 90° and sinθ = 1, giving the maximum possible turning effect. Computing: τ = r × F = 0.30 × 50 = 15 N·m. The option 150 N·m is wrong because it ignores the decimal in the 0.30 m length. The option 1.5 N·m results from an extra factor-of-ten error in the length. The option 16.7 N·m comes from incorrectly dividing rather than multiplying. A magnitude check confirms 15 N·m is reasonable for hand-loosening a bolt, and the units N·m correctly represent torque.
This easy difficulty physics question is from the chapter rotational motion, covering the topic of torque. It appeared in the 2025 exam.
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