Question

The top of the water tank is open to air and its water level is maintained. It is giving out $0.74{m^3}$ water per minute through a circular opening of $2cm$ radius in its wall. The depth of centre of the opening from the level of water in the tank is close to:
A) $9.6m$
B) $4.8m$
C) $2.9m$
D) $6.0m$

Answer 1

In this question, first find out the amount of water going in and the amount of water giving out from the principle of continuity equation and from there find out the depth of the centre the opening from the level of water in the tank.

Complete step by step solution:
It is given in the question that the top of the water tank is open and its water level is maintained. Now there is a circular opening of $2cm$ radius in its wall.
Hence, $r = 2cm = 2 \times {10^{ - 2}}m$
We know the continuity equation which states that in flow volume is equal to the out flow volume. That means the amount of water is going in and the amount of water which is coming in is equal.
Now the amount of water which is giving out per minute is $\dfrac{{0.74}}{{60}} = 0.012{m^3}.$

Hence this amount of water is equal to the amount of which is going through the circular opening.
Thus, the amount of water is going through the circular opening is
$\Rightarrow \pi {r^2} \times \sqrt {2gh}$ So, $0.012 = \pi {r^2}\sqrt {2gh}$
$\Rightarrow h = \left( {\dfrac{{0.012}}{{\pi \times {{\left( {2 \times {{10}^{ - 2}}} \right)}^2}}}} \right) \times \dfrac{1}{{2 \times 9.8}} = 4.8m$
Here $g$ is gravity, $h$ is the depth of centre of the opening from the level of water in the tank.
So, the depth of the centre of the opening from the level of water in the tank is $4.8m$.

Hence option (B) is correct.

Note: Continuity equation is a local form of conservation laws. A continuity equation is a mathematical representation of a fluid move by a continuous flow.