Question

An engine pumps up `100 kg` water through a height of `10 m` in `5 s`. If efficiency of the engine is `60%`. What is the power of the engine? `Take g = 10 ms^(2)`.

Answer 1

Here, `m=100kg, h=10m, t=5s`
`g=10m//s^(2),eta=60%`
output power, `P_(0)=(mgh)/(l)=(100xx10xx10)/(5)`
Input power `P_(i)=?`
As `eta=(P_(o))/(P_(i)) :.P_(i)=(P_(o))/(eta)=(2)/(60//100)=3.33kW`