A $1.0$ HP motor pumps out water from a well of depth $20 m$ and fills a water tank of volume $2238$ litres at a height of $10 m$ from the ground. The running time of the motor to fill the empty water tank is $(g = 10 m s^{- 2})$

5

minutes

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10

minutes

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15

minutes

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20

minutes

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Solution

The correct option is C $15$ minutes
Power = P = 1 HP = 746 W
Depth = d = 20 m
H = 10 + 20 = 30 m
Volume = V = 2238 l
Mass of water = m = $m = d e n s i t y \times v o l u m e = 1 k g / m^{3} \times 2238 m^{3} = 2238 k g$
$t = \frac{p o t e n t i a l E n e r g y}{P} = \frac{m g h}{P} = \frac{2238 k g \times 9.8 m / s^{2} \times 30 m}{746 W} = 900 s = \frac{900}{60} m i n = 15 m i n$

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A 1.0 HP motor pumps out water from a well of depth 20m and fills a water tank of volume 2238 litres at a height of 10m from the ground. The running time of the motor to fill the empty water tank is (g=10ms−2)

The correct option is C 15 minutesPower = P = 1 HP = 746 WDepth = d = 20 mH = 10 + 20 = 30 mVolume = V = 2238 lMass of water = m = m=density×volume=1kg/m3×2238m3=2238kgt=potentialEnergyP=mghP=2238kg×9.8m/s2×30m746W=900s=90060min=15min

A $1.0$ HP motor pumps out water from a well of depth $20 m$ and fills a water tank of volume $2238$ litres at a height of $10 m$ from the ground. The running time of the motor to fill the empty water tank is $(g = 10 m s^{- 2})$