A 10 text{ H.P.} motor pumps out water from a | vedclass

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(C) The total height through which the water is raised is $H = 20 	ext{ m} + 10 	ext{ m} = 30 	ext{ m}$.Volume of water $V = 22380 	ext{ litres} =

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The total height through which the water is raised is $H = 20 	ext{ m} + 10 	ext{ m} = 30 	ext{ m}$.Volume of water $V = 22380 	ext{ litres} = 22380 	imes 10^{-3} 	ext{ m}^3 = 22.38 	ext{ m}^3$.Density of water $ho = 10^3 	ext{ kg/m}^3$,so mass $m = V 	imes ho = 22.38 	imes 10^3 	ext{ kg} = 22380 	ext{ kg}$.Power of the motor $P = 10 	ext{ H.P.} = 10 	imes 746 	ext{ W} = 7460 	ext{ W}$.The work done to lift the water is $W = mgh = 22380 	imes 10 	imes 30 	ext{ J}$.Power $P = \frac{W}{t}$,so $t = \frac{mgh}{P} = \frac{22380 	imes 10 	imes 30}{7460} 	ext{ seconds}$.$t = \frac{22380 	imes 300}{7460} = 3 	imes 300 = 900 	ext{ seconds}$.Converting to minutes: $t = \frac{900}{60} = 15 	ext{ minutes}$.

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