A hydroelectric power station takes its water from a lake whose water level is 50m above the turbine. Assuming an overall efficiency of 40%, calculate the mass of water which must flow through the turbine each second to produce power output of 1MW. (g=10m${s^{ - 2}}$ )
A) 5000 kg
B) 2000 kg
C) 3333 kg
D) None of the above

Hint: In order to solve the question we must need to know the potential energy of the turbine which is mgh where, m is the mass, g is the gravity and h is the height. Mass is needed to find so we will organize values in terms of mass and with the help of the power formula which is work done per unit time we will get the mass of water.

Complete step by step answer:
Step 1:
We are given:
A hydroelectric power station takes its water from a lake whose water level is 50m above the turbine. This implies h=50m
Efficiency given is 40% which is equal to 0.4
We need to calculate the mass of water which must flow through the turbine each second to produce power output of 1MW.
This implies power P=1MW or $1 \times {10^6}$ W
Now coming to the solution:
Potential energy P.E=mgh where, m is the mass, g is the gravity and h is the height.
Substituting the value in potential energy we get P.E=m × 10 × 50, g=10m${s^{ - 2}}$
This equal to P=500m
We are given that overall efficiency is 40% so, $ \Rightarrow \left( {0.4} \right) \times 500m = 200m$, m is the mass (this is work done)
Now, we know power is work divide by time so, $P = \dfrac{{work}}{{time}}$
Substituting the value and time as 1sec we get $1 \times {10^6} = \dfrac{{200 \times m}}{{1\sec }}$
From here we can write m=$\dfrac{{{{10}^6}}}{{200}}kg$
This gives mass equal to 5000kg.
The mass of water which must flow through the turbine each second is 5000 Kg
Hence option A is correct.


Note: A turbine converts the potential and kinetic energy of a moving fluid (liquid or gas) to mechanical energy. In a turbine generator, a moving fluid, or air pushes a series of blades mounted on a shaft, which rotates the shaft connected to a generator. That’s why we have used here the formula of potential energy.