CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
249 W
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
100 W
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
549 W
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 249 W
The work done by motor = Change in Mechanical energy
The power output of the motor is

Power=ΔK+ΔUt=12m(v2fv2i+gh)t=mt[12v2f+gh] (vi=0)=2[12(3)2+(10)(12)]=249W

flag
Suggest Corrections
thumbs-up
13
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Power
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon