Efficiency
A motor takes 5 kJ electrical energy to lift a 100 kg mass by 4 m (g=10). Efficiency?
Select the correct option:
Solution
80%
- Definition of Efficiency ((\eta)): Ratio of useful work output to total energy input.
- Energy Input (Ein): 5 kJ =5000 J.
- Useful Work Output (Wout): Lifting a weight requires work against gravity.
- Wout=mgh=100×10×4=4000 J.
- Calculation:
- η=50004000=0.8.
- Result: Efficiency is 80%.
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About This Question
- Subject
- physics
- Chapter
- work, energy and power
- Topic
- efficiency
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
80%
- Definition of Efficiency ((\eta)): Ratio of useful work output to total energy input.
- Energy Input (Ein): 5 kJ =5000 J.
- Useful Work Output (Wout): Lifting a weight requires work against gravity.
- Wout=mgh=100×10×4=4000 J.
- Calculation:
- η=50004000=0.8.
- Result: Efficiency is 80%.
This medium difficulty physics question is from the chapter work, energy and power, covering the topic of efficiency. It appeared in the 2025 exam.
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