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

Question

Dr. Abubacker · Accepted Answer

(c) Work done $W = mgh + \mathrm{\Delta}pV$
=$Vpgh + \mathrm{\Delta}pV$
Given V=$4m^{3},\ p = 10^{3}kg^{- 2},g = 10ms^{- 2},$
$h = 20m,\mathrm{\Delta}p = \left( 2 	imes 10^{5} - 1 	imes 10^{5} ight)Nm^{- 2}$
$W = 4 	imes 10^{3} 	imes 10 	imes 20 + (2 	imes 10^{5} - 1 	imes 10^{5}) 	imes 4$
$8 	imes 10^{5} + 4 	imes 10^{5} = 12 	imes 10^{5}J$

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

Explanation

Reviewed by Dr. Abubacker

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