Question

A water pump of power 2kW is installed in a home. Then, the amount of water (in litres) it can raise in one minute to a height of 10 m is: [Take $ g = 10m/{s^2} $ ]
(A) 200 L
(B) 1200 L
(C) 800 L
(D) 350 L

Answer 1

Hint: Power can be defined as the work done per unit time. Work done can be given as a change in potential. In this solution we will be using the following formulae;
 $ P = \dfrac{W}{t} $ where $ P $ is the power delivered to do work, $ W $ is the work done, and $ t $ is the time taken to do the work.
 $ W = \Delta PE $ where $ PE $ is the potential energy of a body, and the delta sign $ \Delta $ signifies change in… (in the case, we have a change in potential energy).
 $ PE = mgh $ where $ m $ is the mass of the body, $ g $ is the acceleration due to gravity and $ h $ is the height of the body.

Complete step by step answer:
To solve the above problem, we define power to be the rate of doing work, and can mathematically be given as
 $ P = \dfrac{W}{t} $ where $ P $ is the power delivered to do work, $ W $ is the work done, and $ t $ is the time taken to do the work.
But, work can be defined as the change of potential energy when done against gravity.
Hence, we have
 $ W = \Delta PE $ $ W = \Delta PE $ where $ PE $ is the potential energy of a body, and the delta sign $ \Delta $ signifies change.
But, $ PE = mgh $ where $ m $ is the mass of the body, $ g $ is the acceleration due to gravity and $ h $ is the height of the body.
Hence, $ m = \rho V $ where $ \rho $ is the density of a substance and $ V $ is its volume.
Hence, we can write that
 $ PE = \rho Vgh $
For work done, we have
 $ W = \Delta PE = \rho Vg\left( {{h_2} - 0} \right) $ (assuming that the height of the pump is at zero position)
Hence,
 $ W = \Delta PE = 1000V\left( {10} \right)10 = 100000V $
Thus, the power is
 $ P = \dfrac{W}{t} = \dfrac{{100000V}}{{60}} $ since one minute equal 60 seconds. So,
 $ 2000 = \dfrac{{100000V}}{{60}} $
 $ \Rightarrow V = \dfrac{{2000 \times 60}}{{100000}} = 1.2{m^3} $ which is 1200 L (because 1000 L make one $ {m^3} $ )
Hence, the correct option is B.

Note:
Alternatively, once it is noted that the power can be taken as the rate of change of potential energy with time, we simply write (for exam speed), that
 $ P = \dfrac{{mg\left( {h - 0} \right)}}{t} = \dfrac{{\rho Vgh}}{t} $
And we proceed to substitute the values as in above and make $ V $ subject of the formula.