A cylindrical tank has a small hole at bottom. At `t = 0`, a tap starts to supply water into the tank at a constant rate `beta m^3//s`.
(a) Find the maximum level of water `H_(max)` in the tank ?
(b) At what time level of water becomes `h(h lt H_(max))` Given `a` , area of hole, `A` : area of tank.

Explanation

To solve the problem, we will break it down into two parts as per the question.

### Part (a): Finding the Maximum Level of Water \( H_{\text{max}} \)

1. **Understanding the inflow and outflow**:
The inflow rate of water into the tank is given by \( \beta \) (in \( m^3/s \)). The outflow rate through the hole at the bottom of the tank can be expressed using Torricelli's law, which states that the speed \( v \) of efflux of fluid under gravity through a hole is given by:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the water above the hole and \( g \) is the acceleration due to gravity.

2. **Outflow rate**:
The volume flow rate \( Q_{\text{out}} \) through the hole can be expressed as:
\[ Q_{\text{out}} = A v = A \sqrt{2gh} \]
where \( A \) is the area of the hole.

3. **Setting up the equilibrium**:
At maximum height \( H_{\text{max}} \), the inflow rate equals the outflow rate:
\[ \beta = A \sqrt{2gH_{\text{max}}} \]

4. **Solving for \( H_{\text{max}} \)**:
Rearranging the equation gives:
\[ \sqrt{2gH_{\text{max}}} = \frac{\beta}{A} \]
Squaring both sides:
\[ 2gH_{\text{max}} = \left(\frac{\beta}{A}\right)^2 \]
Finally, solving for \( H_{\text{max}} \):
\[ H_{\text{max}} = \frac{\beta^2}{2gA^2} \]

### Part (b): Finding the Time When Water Level is \( h \) (where \( h < H_{\text{max}} \))

1. **Setting up the differential equation**:
The change in height \( h \) with respect to time \( t \) can be expressed as:
\[ A \frac{dh}{dt} = \beta - A \sqrt{2gh} \]
Rearranging gives:
\[ A \frac{dh}{dt} = \beta - A \sqrt{2gh} \]

2. **Separating variables**:
Rearranging the equation to separate variables:
\[ \frac{dh}{\beta - A \sqrt{2gh}} = \frac{dt}{A} \]

3. **Integrating both sides**:
Integrating the left side with respect to \( h \) and the right side with respect to \( t \):
\[ \int \frac{dh}{\beta - A \sqrt{2gh}} = \int \frac{dt}{A} \]

4. **Solving the integral**:
The left-hand side requires substitution and can be solved to yield a logarithmic function. After integration and simplification, we can express \( t \) in terms of \( h \):
\[ t = \frac{A}{\beta} \ln\left(\frac{\beta - A \sqrt{2gh}}{\beta - A \sqrt{2gH_{\text{max}}}}\right) \]

### Final Answers:
- **(a)** The maximum level of water \( H_{\text{max}} \) in the tank is:
\[ H_{\text{max}} = \frac{\beta^2}{2gA^2} \]

- **(b)** The time \( t \) when the water level is \( h \) is given by:
\[ t = \frac{A}{\beta} \ln\left(\frac{\beta - A \sqrt{2gh}}{\beta - A \sqrt{2gH_{\text{max}}}}\right) \]