Question

1. A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. Calculate
(I) Work done by the pump.
(II) The power at which the pump works.
(III) The power rating of the pump if the efficiency is 40%.
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Answer 1

Explanation:

(a) W =mg×h = mgh

w =500×10×80=4×10⁵ j

(b) power at which pump works =work done/time taken =MGT/t

=50×10×80j/10s=4×10⁵/10=40 kw

(c)Efficiency =useful power/power input

efficiency=40%=0.4

0.4=40kw/power input

power input=40/0.4 100kw

Answer 2

a) W=FS

W=mg×S

500×10×80

4,00,000J

b) P=W/t

=400000/10=40000W=40kW

c) efficiency=useful power/power input

40%=40/Power input

Power input = 40 × 100/40

Power input = 40kW