Question

From a waterfall, water is falling down at the rate of 100kg/s on the blades of the turbine. If the height of the fall is 100m, then the power delivered to the turbine is approximately equal to
A. 100kW
B. 10kW
C. 1kW
D. 1000kW

Answer 1

Hint: Power is the rate of work done with respect to time. i.e. $P=\dfrac{W}{t}$. Work done is given as W=Fd. Calculate the work done on m mass of water in time t. here, m=100t. Then to find the power delivered divide the work done with time t.

Formula used:
$P=\dfrac{W}{t}$
W=Fd

Complete answer:
Let us first understand what power. Power is defined as the rate of work done with respect to time. In other words, power is the amount of work done in one unit of time.
If the rate of work with respect to time is constant, then the power delivered is constant. Then if W amount of work is done in a time interval t, then the power delivered is equal to $P=\dfrac{W}{t}$.
When a force is applied on a body and the body displaces, we say that a work is done on the body. If the directions of force (F) and the body’s displacement (d) are the same then the work is equal to W = Fd.
Here, water is falling from a height of 100m. Suppose a water of mass m falls down on the blades of a turbine in time t.
In this time, earth will exert a gravitational force on this mass of water, which is equal to F = mg and is directed downwards. Due to this the displacement of the water is d=100m downwards. Let the value of g be 10$m{{s}^{-2}}$.
Hence, the work done is equal to $W=mgd=m(10)(100)=1000m$.
It is given that the flow of water is 100kg/s. This means that 100kg mass of water falls down in 1 second. Therefore, in time the mass of water falling down will be 100t.
This means that m=100t.
Therefore, $W=1000\times 100t={{10}^{5}}t$
Therefore, the power delivered is equal to $P=\dfrac{W}{t}=\dfrac{{{10}^{5}}t}{t}={{10}^{^{5}}}W=100kW$.

So, the correct answer is “Option A”.

Note:
Mass per unit time is called mass rate. Let it be denoted by $\rho $. If mass m flows in time t, then $\rho =\dfrac{m}{t}$.
In the above question, we got that power is equal to $P=\dfrac{mgd}{t}$. Substitute the value of $\dfrac{m}{t}$.
$\Rightarrow P=\rho gd$ …. (1).
Hence, whenever we face a similar problem we can directly use the formula (1).