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An elevator weighing 500 kg is to be lifted up at a constant velociyt of 0.20 m/s. What would be the minimum horsepower of the motor to be used?A. `10.30 hp`B. `5.15 hp`C. 2.62 hpD. 1.31 hp |
Answer» Correct Answer - D (d) Given, mass of elevator=500 kg velocity=0.20 `ms^(-1)` Weight of elevator`=500xx9.8=F` Now,power, `P=Fv=500xx9.8xx0.20=980W` Therefore, hp-rating of motor`=(1)/(746)xx980=1.31hp` |
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