Question

A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time \[T\] . How long will it take to empty the tank if it is half filled?

Answer 1

Hint:To solve this above question, we will apply Bernoulli’s theorem to determine how long it will take to empty the tank if it is half filled and we will integrate the obtained equation to get the required solution.

Formula used:
$v = \sqrt {2gh} $
Where, $g$ is the acceleration due to gravity and $h$ is the height.

Complete step by step answer:
Let us assume the actual height of the tank is $H$ and its cross sectional area is $A$. Now let us consider that at some moment of time the height of water level is $h$ and the velocity of water emerging through the cross section of the area is $a$ at the bottom of the tank be $v$.As the surface of water and the orifice are in open atmosphere, then with Bernoulli's equation we can say that,
$v = \sqrt {2gh} $

And now, let $dh$ is the decrease in level of water during infinitesimally small time intervals $dt$. When the level of water is at the height of $h$. So, we can say that the rate of decrease in the volume of water will be $ - A\dfrac{{dg}}{{dt}}$.And the rate of flow of water at this moment through the orifice is given by,
$v \times a = a\sqrt {2gh} $
Then we can say that with the principle of continuity the both be same i.e.,
$a\sqrt {2gh} = - A\dfrac{{dg}}{{dt}} \\
\Rightarrow dt = - A\dfrac{{dg}}{{a\sqrt {2g} }} \times {h^{\dfrac{{ - 1}}{2}}}dh \\ $
Now, we can integrate the above equation to get the time taken to empty the tank. If the tank is filled to the brim, then the height of the water level will be $H$ and the time required to empty the tank is $T$ .
i.e.,
$T = \int\limits_0^{{T_0}} {dt} = - A\dfrac{{dg}}{{a\sqrt {2g} }} \times \int\limits_H^0 {{h^{\dfrac{{ - 1}}{2}}}dh} \\
\Rightarrow T = \dfrac{{ - A}}{{a\sqrt {2g} }}\left[ {\dfrac{{{h^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right]_H^0 \\
\Rightarrow T = \dfrac{{A\sqrt {2H} }}{{a\sqrt {2g} }} \\ $
And according to the question if the tank is half filled with water and the time to empty be $T'$ then
\[T' = \dfrac{{A\sqrt {2H} }}{{a\sqrt {2g} }} \\
\Rightarrow \dfrac{{T'}}{T} = \dfrac{1}{{\sqrt 2 }} \\
\therefore T' = \dfrac{T}{{\sqrt 2 }} \\ \]
So, time taken to empty the tank if it is half filled is \[\dfrac{T}{{\sqrt 2 }}\].

Note:The rule of conservation of energy is used to derive Bernoulli's equation. If you don't remember Bernoulli's efflux equation, you can apply the law of conservation of energy, which states that potential energy must be transferred to kinetic energy.