Question

Water flows into a large tank with a flat bottom at the rate of ${10^{ - 4}}{m^3}{s^{ - 1}}$. Water is also leaking out of a hole of area $1c{m^2}$ at its bottom. If the height of the water in the tank remains steady, then this height is____?
(A) $4cm$
(B) $2.9cm$
(C) $1.7cm$
(D) $5.1cm$

Answer 1

Hint
We can find the velocity of the water leaking out from the bottom by equating the energy of the water at the top and bottom of the tank. Using that velocity we can calculate the height by equating the inflow rate and the outflow rate.
In this solution we will be using the following formula,
$\Rightarrow P.E. = mgh$ where $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is height.
and $K.E. = \dfrac{1}{2}m{v^2}$ where $v$ is the velocity of the body.

Complete step by step answer
According to the question, the height of the water in the tank will remain the same. For this to be possible, the water should be flowing in at the same rate as it is flowing out. So the rate of inflow should be equal to the rate of outflow of water.
Now in the question we are given that the water is flowing in at the rate of ${10^{ - 4}}{m^3}{s^{ - 1}}$. And the hole at the bottom of the tank has an area of $1c{m^2}$. This we can write as, ${10^{ - 4}}{m^2}$. So the rate of outflow of water will be the product of the area and the velocity with which the water is flowing out.
Hence we can equate the inflow and the outflow as,
$\Rightarrow {10^{ - 4}}{m^3}{s^{ - 1}} = {10^{ - 4}}{m^2} \times v$ where we have taken $v$ as the velocity of the water.
Now in the tank, the potential energy of the mass of the water at the top of the tank is equal to the kinetic energy for the water flow at the bottom of the tank.
So if we consider the mass of whole volume of water be $m$ and the height of the water in the tank is $h$, then we can write,
$\Rightarrow P.E. = mgh$ at the top of the water level and
$\Rightarrow K.E. = \dfrac{1}{2}m{v^2}$ at the bottom.
Equating them we get,
$\Rightarrow \dfrac{1}{2}m{v^2} = mgh$
Cancelling $m$ from both sides and taking all the terms except $v$ from the LHS to the RHS we get,
$\Rightarrow {v^2} = 2gh$
So taking square root on both sides we get,
$\Rightarrow v = \sqrt {2gh} $
So substituting this value to the first equation we get,
$\Rightarrow {10^{ - 4}}{m^3}{s^{ - 1}} = {10^{ - 4}}{m^2} \times \sqrt {2gh} $
We can cancel ${10^{ - 4}}$ from both sides of the equation to get,
$\Rightarrow \sqrt {2gh} = 1$
Squaring both sides,
$\Rightarrow 2gh = 1$
Taking all terms except $h$to the RHS we get
$\Rightarrow h = \dfrac{1}{{2g}}$
We can take the value of $g = 9.8m/{s^2}$
Hence we get
$\Rightarrow h = \dfrac{1}{{2 \times 9.8}}$
On doing the division, we get the height as, $h = 0.051m$
As the answer is in cm, so by converting we get
$\Rightarrow h = 5.1cm$
So the height of the water in the tank has to be $5.1cm$
Therefore, the correct answer is option (D).

Note
We can equate the kinetic energy of the mass of water at the bottom of the tank with the potential energy at the top of the tank because of the law of conservation of energy. According to law, energy can be changed from 1 form to another.