Question

From a waterfall, water is falling at the rate of \[100{\text{ }}kg/s\] on the blades of a turbine. If the height of fall is \[100{\text{ }}m\], the power delivered to the turbine is
$(A)100KW$
$(B)10KW$
$(C)1KW$
$(D)1000KW$

Answer 1

Hint: The definition of power has to be known. The relationship between the work done by the water and the delivered power by it has to be used to solve the problem. The work done can be found by the force exerting from the mass of the water that falls on the turbine blades. Note that the mass can be determined from the rate of the mass of the falling water.

Formula used:
The power, $P = \dfrac{W}{t}$
Where, $W$ is work done by water in time $t$
$W = mgh$
$m$ is the mass of water, $g$ is the acceleration due to gravity, and $h$ is the height from where the water is falling.

Complete step-by-step solution:
The water is falling during a waterfall at a given rate of its mass. Since the waterfalls on the turbine with some force, the force exerting from the water will be for its weight.
Therefore, the force will be $mg$. Where $W = mgh$
$m$ is the mass of water, $g$ is the acceleration due to gravity. Gravitational acceleration is included here due to the waterfall from a certain height $h$.
So, the work done will be $W = mgh$, in time $t$
Here is the problem: the power delivered to the turbine by this waterfall is needed. Power is defined by the rate of change of the work done. So, The power, $P = \dfrac{W}{t}$
Now given the rate of mass i.e. $\dfrac{m}{t} = 100kg/s$
So, the mass of the water can be taken $m = 100t{\text{ }}kg$, the height $h = 100m$ and, $g = 10m/{s^2}$
$\therefore W = 100t \times 10 \times 100$
$ \Rightarrow W = {10^5}t$
So, The power,
$P = \dfrac{{{{10}^5}t}}{t}$
$ \Rightarrow P = 100KW$
Therefore, the power delivered by the water to the turbine is $ P = 100KW$.

Note: The rate of mass is defined as the mass of a liquid flowing per unit of time. If the rate of mass is written by, \[\dfrac{m}{t} = \rho \] .
\[\rho \]is called the density.
Now from the formula of the power, we get
$P = \dfrac{{mgh}}{t}$[since, $W = mgh$]
$ \Rightarrow P = \dfrac{m}{t} \times gh$
$ \Rightarrow P = \rho gh$
This is the relation between the power and mass rate.
This relation is also used in determining the pressure exerted by a liquid of density \[\rho \]in time $t$.