ICSE Class 10 Answered
An electric motor of power 125 w is switched on for 1 minute 40 second if 75% of energy of motor is useful then find useful work done b.load lifted by it through the height of 4 m
Asked by Samrat | 30 Jun, 2018, 09:28: PM
Expert Answer
workdone = Power × time × efficiency = 125×100×.75 = 9375 J
to get the load to be lifted, m×g×h = 9375 ; hence m = 9375/(9.8×4) = 239 kg
Answered by Thiyagarajan K | 01 Jul, 2018, 08:24: PM
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