Question

A person decides to use his bathtub water to generate electric power to run a 40W bulb. The bathtub is located at a height of 10m from the ground and it holds 200 litres of water. He installs a water-driven wheel generator on the ground. At what rate should the water drain from the bathtub to light the bulb? How long can he keep the bulb on, if the bathtub was full initially? The efficiency of the generator is 90%. Take $g = 9.8m{s^{ - 2}}$.

Answer 1

Hint: To solve this problem, we have to first calculate the work done by the falling of the water on the water-driven generator on the ground. The total mass of the falling water can be calculated by considering the density of water is $1000kg{m^{ - 3}}$.

Formula Used:
Work done,
$W=mgh$
where m= mass of the water, $g = 9.8m{s^{ - 2}}$ , and h= height.

Complete step by step solution:
We have to calculate the work done by the falling water to get the rate at which the water is to be drained.
Given, in the question, Output power = 40 W, height = 10 m, Volume of water= 200l, Efficiency = 90%, $g = 9.8m{s^{ - 2}}$.
We know that the density of water is ${\rho _w} = 1000kg{m^{ - 3}}$.
Again, Efficiency can be written as, $\eta = \dfrac{{{P_o}}}{{{P_i}}}$,
where ${P_o}$= Output power and ${P_i}$= input power.
Substituting, we get –
$90\% = \dfrac{{40}}{{{P_i}}}$
$ \Rightarrow \dfrac{{90}}{{100}} = \dfrac{{40}}{{{P_i}}}$
$ \Rightarrow {P_i} = \dfrac{{40 \times 100}}{{90}} = \dfrac{{400}}{9}W$
Now, considering if $d$ kg of water falls from height $h$ , the work done can be given by,
$W = dgh$.
The mass of water can be written as, $d = {\rho _w}V$,
where V= volume of the water.
Thus,
$W = {\rho _w}Vgh$.
The power can be obtained by differentiating work with respect to time.
${P_i} = \dfrac{{dW}}{{dt}}$
$ \Rightarrow {P_i} = \dfrac{{d{\rho _w}Vgh}}{{dt}}$
$ \Rightarrow {P_i} = {\rho _w}gh\left( {\dfrac{{dV}}{{dt}}} \right)$
Substituting the value of power input, ${P_i}$
${\rho _w}gh\left( {\dfrac{{dV}}{{dt}}} \right) = \dfrac{{400}}{9}$
$ \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{{400}}{{9{\rho _w}gh}} = \dfrac{{400}}{{9 \times 1000 \times 9.8 \times 10}} = 0.000453{m^3}{s^{ - 1}}$
By expressing the volume rate in litres, we get –
$\dfrac{{dV}}{{dt}} = 0.000453 \times 1000l{s^{ - 1}} = 0.453l{s^{ - 1}}$.
Thus, the rate at which the water has to be drained is $0.453l{s^{ - 1}}$
Now, Time for which the bulb will glow if the tank was initially full can be given as,
Time, $T = \dfrac{V}{{\dfrac{{dV}}{{dt}}}} = \dfrac{{200}}{{0.453}} = 441.5s$

Thus, if the tank was initially full with 200l of water then the bulb will glow till 441.5s.

Note: Here, it is mentioned that the efficiency is 90%. The efficiency of an ideal machine is 100% which means there is zero power loss from the input to output. However, it is impossible to obtain an ideal machine with 100% where there is zero loss in power due to friction. Thus, the machines are benchmarked on the basis of this quantity efficiency.