Question

A pump is required to lift 800kg of water per minute from a 10 m deep well and eject it with speed of \[20\,m{s^{ - 1}}\]. The required power in watts of the pump will be
A. 6000
B. 4000
C. 5000
D. 8000

Answer 1

Hint: We need to find the energy required or the work done to lift 800kg of water to a height of 10m and throw it with a speed of \[20m{s^{ - 1}}\] to find the power of the pump. The work done in moving a particle under the conservative force is given by change in mechanical energy.

Formula used:
The work done in a system is given by the change in mechanical energy of the system,
\[W = (\dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v_i}^2) + (mg{h_f} - mg{h_i})\]
where, $m$ is the mass of the body, \[{v_f}\], \[{v_i}\] are final and initial velocities respectively and \[{h_f}\],\[{h_i}\] are the final and initial height of the body.

Complete step by step answer:
The pump is required to lift 800kg of water per minute from a 10 m deep well and eject it with speed of \[20\,m{s^{ - 1}}\]. Now, we know that the work done on moving a body under conservative force is given by the change in mechanical energy. So, here the change in mechanical energy is equal to the sum of change in kinetic energy and change in potential energy.

So, to move 800kg of water, the change in potential energy is equal to \[mgh\].
\[\text{Potential energy} = 800 \times 10 \times 10 = 80000\,J\].
To throw the water with a speed of \[20\,m{s^{ - 1}}\] the change in kinetic energy is equal to,
\[\text{kinetic energy} = \dfrac{1}{2}m{v^2}\]. (Since, the initial speed is zero.)
\[\Rightarrow \text{kinetic energy} = \dfrac{1}{2}800 \times {20^2}J\]
\[\Rightarrow \text{kinetic energy} = 400 \times 400J\]
\[\Rightarrow \text{kinetic energy} = 160000\,J\]

Therefore total energy required per minute by the pump is equal to,
\[\text{Total energy}= 160000 + 80000\,J\]
\[\Rightarrow \text{Total energy} = 240000\,J\]
So, this amount of energy is required per minute to eject water.
So, Power (P) of the pump must be,
\[P = \dfrac{{240000J}}{{1\min }}\]
\[\Rightarrow P =\dfrac{{240000J}}{{60s}}\]
\[\Rightarrow P = 4000\,J\,{s^{ - 1}}\]
\[\therefore P = 4000\,Watt\]
Hence, the required power of the pump must be, \[4000\,Watt\].

Hence, option B is the correct answer.

Note: When calculating the change in kinetic energy, note that the initial speed of the water is zero. So, change in kinetic energy is only the kinetic energy of water finally. Also, when calculating the power remember that the ejection rate is given in per minute so we have to convert it in per second to get the power in per second.