A pump is used to lift $500 k g$ of water from a depth of $80 m$ in $10 s$ . Calculate the power rating of the pump if its efficiency is $40 %$ :

100 k W

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

40 k W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16 k W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $100 k W$
$F o r c e = m \times g$
$F = 500 \times 10 = 5000 N$
$W o r k = F \times s$
$W = 5000 \times 80 = 4 \times 10^{5} j o u l e$ .
$P o w e r = w o r k / t i m e$
$P = 4 \times 10^{5} / 10 = 4 \times 10^{4} = 40 k W$ .
$P o w e r R a t i n g = U s e f u l p o w e r / e f f i c i e n c y$
Power Rating = $40 k / 0.4 = 100 k W$

Suggest Corrections

Similar questions

A pump is used to lift 500 kg of water from a depth of 80 m in 10 s.Calculate:

(a) the work done by the pump,

(b) the power at which the pump works, and

500 k g

80 m

10 s

g = 10 m s^{- 2}

500 k g

80 m

10 s

g = 10 m s^{- 2}

500

80

10

40 %

g = 10

s^{- 2}

Join BYJU'S Learning Program