A motor can pump 7560 text{ kg} of water per | vedclass

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(C) The mass of water pumped per hour is $m = 7560 	ext{ kg}$.The time taken is $t = 1 	ext{ hour} = 3600 	ext{ s}$.The depth of the well is $h = 1

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$3$

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(C) The mass of water pumped per hour is $m = 7560 	ext{ kg}$.The time taken is $t = 1 	ext{ hour} = 3600 	ext{ s}$.The depth of the well is $h = 100 	ext{ m}$.The acceleration due to gravity is $g = 10 	ext{ m s}^{-2}$.The useful work done (output energy) per second is the power output $P_{	ext{out}} = \frac{mgh}{t}$.$P_{	ext{out}} = \frac{7560 	imes 10 	imes 100}{3600} = \frac{7560000}{3600} = 2100 	ext{ W} = 2.1 	ext{ kW}$.Given the efficiency $\eta = 70 \% = 0.7$, the power input $P_{	ext{in}}$ is given by $\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}}$.$P_{	ext{in}} = \frac{P_{	ext{out}}}{\eta} = \frac{2.1 	ext{ kW}}{0.7} = 3 	ext{ kW}$.Therefore, the power of the pump is $3 	ext{ kW}$.

$A$ motor can pump $7560 \text{ kg}$ of water per hour from a well of depth $100 \text{ m}$. If the efficiency of the pump is $70 \%$, then the power of the pump is (Acceleration due to gravity $= 10 \text{ m s}^{-2}$) (in $\text{ kW}$)

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