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Question
40000 litre of oil of density 0.9 g/cc is pumped from an oil tanker ship into a storage tank at 10 m higher level than the ship in half an hour. What should be the power of the pump?
Numerical
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Solution
Given:
h = 10 m, ρ = 0.9 g/cc = 900 kg/m3, g = 10 m/s2,
V = 40000 litre = 40000 × 103 × 10-6 m3 = 40 m3
T = 30 min = 1800 s
To find: Power(P)
Formula: P = `"W"/"t" = ("h"rho"gV")/"t"`
Calculation: From formula,
P = `(10 xx 900 xx 10 xx 40)/1800`
∴ P = 2000 W
∴ P = 2 kW
The power of the pump is 2 kW.
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Chapter 4: Laws of Motion - Exercises [Page 76]
