A tank has a hole at its bottom. The time needed to empty the tank from level `h_(1)` to `h_(2)` will be proportional to

Options

A. `h_(1)+h_(2)`
B. `sqrt(h_(1))+sqrt(h_(2))`
C. `h_(1)-h_(2)`
D. `sqrt(h_(1))-sqrt(h_(2))`

Explanation

To solve the problem of determining how the time needed to empty a tank from level \( h_1 \) to \( h_2 \) is proportional, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a tank with a hole at the bottom. We need to find the relationship between the time taken to empty the tank from height \( h_1 \) to height \( h_2 \).

2. **Using Torricelli's Law**:
The efflux speed \( v \) of fluid flowing out of the hole can be described by Torricelli's Law:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of the fluid above the hole.

3. **Volume Flow Rate**:
The volume flow rate \( Q \) can be expressed as:
\[ Q = A v = A \sqrt{2gh} \]
where \( A \) is the cross-sectional area of the hole.

4. **Relating Height Change to Time**:
The volume flow rate can also be related to the change in height over time. If the area of the tank is \( a \) and the height decreases by \( dh \) in time \( dt \), we have:
\[ A \sqrt{2gh} = a \frac{dh}{dt} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{dh}{dt} = -\frac{A}{a} \sqrt{2gh} \]
The negative sign indicates that the height is decreasing over time.

6. **Separating Variables**:
We can separate the variables to integrate:
\[ \frac{dh}{\sqrt{h}} = -\frac{A \sqrt{2g}}{a} dt \]

7. **Integrating**:
Integrating both sides, we get:
\[ \int_{h_1}^{h_2} \frac{dh}{\sqrt{h}} = -\frac{A \sqrt{2g}}{a} \int_{0}^{t} dt \]
The left side integrates to:
\[ 2(\sqrt{h_2} - \sqrt{h_1}) = -\frac{A \sqrt{2g}}{a} t \]

8. **Finding Time**:
Rearranging gives us the time \( t \):
\[ t = -\frac{2a}{A \sqrt{2g}} (\sqrt{h_2} - \sqrt{h_1}) \]
This shows that the time \( t \) is proportional to:
\[ t \propto \sqrt{h_1} - \sqrt{h_2} \]

9. **Conclusion**:
Therefore, the time needed to empty the tank from level \( h_1 \) to \( h_2 \) is proportional to:
\[ t \propto \sqrt{h_1} - \sqrt{h_2} \]