In a hydroelectric power station, the height of the dam is $10m$. How many $kg$ of water must fall per second on the blades of a turbine in order to generate $1MW$of electric power?
A) ${10^3}kg{s^{ - 1}}$
B) ${10^4}kg{s^{ - 1}}$
C) ${10^5}kg{s^{ - 1}}$
D) ${10^6}kg{s^{ - 1}}$

Hint: To solve the given problem we have to consider the work done. The work done can be explained as the product of the force and the displacement. The gravitational potential energy of water will be converted into electrical energy.

Complete step by step answer:
Given, the height of the dam, $h = 10m$
Power generated,
$ \Rightarrow P = 1MW$
 $ \Rightarrow 1 \times {10^6}W$
Let us consider $m$ to be the amount of water fall per second on the blades of the turbine.
Power is a ratio of work done by time taken. That is,
${\text{power = }}\dfrac{{{\text{work done}}}}{{{\text{time}}}} = \dfrac{W}{t}$ ……………. (1)
We know that the product of force and displacement is known as work completed. That is $W = F.s$ Equation (1) then becomes,
$ \Rightarrow P = \dfrac{{F.s}}{t}$
 \[ \Rightarrow \dfrac{{mgh}}{t}\]
Where the height of the dam, which is $h$, is $s$. $Mg$ refers to the weight of water.
We can calculate the mass of water falling per second from the above equation, which is given by,
$ \Rightarrow \dfrac{m}{t} = \dfrac{P}{{gh}}$
By replacing the values we get,
$ \Rightarrow \dfrac{m}{t} = \dfrac{{1 \times {{10}^6}}}{{10 \times 10}}$
By using the arithmetic multiplication, we can multiply the denominator as the base number is same. We get,
$ \Rightarrow \dfrac{m}{t} = \dfrac{{1 \times {{10}^6}}}{{{{10}^2}}}$
 $\therefore {10^4}kg{s^{ - 1}}$
Thus, ${10^4}kg$ of water must fall per second on the blades to generate $1MW$ of electric power.
$\therefore $ Correct option is (B).

Note: Power is defined as the rate of time at which work is carried out or energy is transferred. If $W$ is the sum of work completed in a time $t$, then the ratio of the work done $W$ to the total time t gives the average power $P$.
That is, power$P = \dfrac{W}{t}$
Here, the SI unit of work done is joule and time in second.
Therefore, unit of power will be,
Joule per second or $J{s^{ - 1}}$ which can be written as \[Watt\left( W \right)\].
Different types of units of power.
If one $joule$ per second of work is completed, the output is one $watt$.
\[{\text{1watt}}\left( {\text{W}} \right){\text{ = 1}}\dfrac{{{\text{joule}}}}{{{\text{second}}}}\]
\[1kilowatt\left( {kW} \right){\text{ }} = 1000watt\]
\[1megawatt\left( {MW} \right) = {10^6}watt\]
\[1horsepower\left( {HP} \right) = 746watts\]
Power is also referred to as the dot product of force and velocity.Power which has only magnitude and no direction.