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4m^3 of water is to be pumped to a height of 20 m and forced into a reservoir at a pressure of 2*10^5 Nm^2. The work done by the motor is(external pressure =10^5 Nm^2)

Answers (1)

Work done = change in gravitational potential energy + change in pressure energy

Work done = mgh + V∆p=Vdgh+ V∆p; where d is density of water

Work done = 4000*10*20+ 4 ( 2*10^5 - 10^5)= 8*10^5+4*10^5 = 12*10^5 joule 

Posted by

shubham.krishnan

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