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Question

# In unloading grain from the hold of a ship, an elevator lifts the grain through a height of 12 m. Grain is discharged at the top of the elevator at a rate of 2 kg/s, and the discharge speed of each grain particle is 3m/s. Then,the minimum power of the motor that can elevate grain in this way is

A
149 W
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B
249 W
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C
100 W
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D
549 W
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Solution

## The correct option is B 249 WThe work done by motor = Change in Mechanical energy The power output of the motor is Power=ΔK+ΔUt=12m(v2f−v2i+gh)t=mt[12v2f+gh] (∵vi=0)=2[12(3)2+(10)(12)]=249W

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