A water pump rated 600 W has an efficiency of | vedclass

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(A) The effective power $P_p$ of the pump used to lift the water is given by the product of its efficiency $\eta$ and its rated power $P_{	ext{rated}

Vedclass · Accepted Answer

$1.14$

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(A) The effective power $P_p$ of the pump used to lift the water is given by the product of its efficiency $\eta$ and its rated power $P_{	ext{rated}}$.$P_p = \eta 	imes P_{	ext{rated}} = 0.95 	imes 600 W = 570 W$.The power required to lift a volume $V$ of water to a height $h$ in time $t$ is given by $P_p = \frac{mgh}{t} = \frac{ho V gh}{t}$,where $ho$ is the density of water.Given: $ho = 1000 kg/m^3$,$g = 10 m/s^2$,$h = 60 m$,and $t = 20 	ext{ minutes} = 20 	imes 60 s = 1200 s$.Substituting the values: $570 = \frac{1000 	imes V 	imes 10 	imes 60}{1200}$.$570 = \frac{600000 	imes V}{1200} = 500 	imes V$.$V = \frac{570}{500} = 1.14 m^3$.

$A$ water pump rated $600 W$ has an efficiency of $95 \%$. If it is used to raise water through a vertical distance of $60 m$,what is the volume of water drawn in $20$ minutes (in $m^3$)? [Use density of water $= 1000 kg/m^3$,$g = 10 m/s^2$]

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