Water is poured in a tank through a cylindrical tube of area of cross-section $A$ and ejecting water at a constant speed $4 \mathrm{~m} / \mathrm{s}$. The tank contains a hole of area $\dfrac{A}{2}$ at bottom. Level of water in the tank will not go up beyond

Let $A_{in} = A$ be the area of the inlet tube and $v_{in} = 4 \text{ m/s}$ be the velocity of water entering the tank. Let $A_{out} = \dfrac{A}{2}$ be the area of the hole at the bottom of the tank.

The level of water in the tank stops rising when the rate of water entering the tank equals the rate of water exiting the tank through the hole. This is the condition for steady state.

The rate of volume flow entering the tank is $Q_{in} = A_{in} \times v_{in} = A \times 4 = 4A$.

The rate of volume flow exiting the tank through the hole is $Q_{out} = A_{out} \times v_{out}$, where $v_{out}$ is the velocity of efflux given by Torricelli's law as $v_{out} = \sqrt{2gh}$, with $h$ being the height of the water level.

Equating the flow rates: $4A = \dfrac{A}{2} \times \sqrt{2gh}$.

Dividing both sides by $A$: $4 = \dfrac{1}{2} \sqrt{2gh} \Rightarrow 8 = \sqrt{2gh}$.

Squaring both sides: $64 = 2gh$.

Using $g = 10 \text{ m/s}^2$: $64 = 2 \times 10 \times h \Rightarrow 64 = 20h$.

$h = \dfrac{64}{20} = 3.2 \text{ m}$.

Answer: $3.2 \mathrm{~m}$