A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in `t_(1)` seconds and the remaining three-fourths of the tank is emptied in `t_(2)` seconds. Then the ratio `(t_(1))/(t_(2))` is

Options

A. `sqrt(3)`
B. `sqrt(2)`
C. `(1)/(sqrt(2))`
D. `(2)/(sqrt(3))-1`

Explanation

To solve the problem, we need to determine the ratio \( \frac{t_1}{t_2} \) based on the time taken to empty different portions of the tank through a small hole at the bottom.

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have a tank full of water with a small hole at the bottom.
- One-fourth of the tank is emptied in \( t_1 \) seconds.
- The remaining three-fourths of the tank is emptied in \( t_2 \) seconds.

2. **Using Torricelli's Law**:
- The velocity \( v \) of the water flowing out of the hole is given by Torricelli's theorem:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the water above the hole.

3. **Setting Up the Equations**:
- Let the total height of the tank be \( H \).
- When one-fourth of the tank is emptied, the height of the water left in the tank is \( \frac{3H}{4} \).
- When three-fourths of the tank is emptied, the height of the water left in the tank is \( 0 \).

4. **Deriving the Time for \( t_1 \)**:
- The volume flow rate \( Q \) through the hole is given by:
\[ Q = A \cdot v = A \cdot \sqrt{2gh} \]
where \( A \) is the area of the hole.
- The change in height \( dh \) over time \( dt \) can be expressed as:
\[ A \cdot \sqrt{2gh} = -A \cdot \frac{dh}{dt} \]
- Rearranging gives:
\[ \frac{dh}{\sqrt{h}} = -k \, dt \]
where \( k \) is a constant.

5. **Integrating for \( t_1 \)**:
- Integrating from \( h = H \) to \( h = \frac{3H}{4} \):
\[ \int_H^{\frac{3H}{4}} \frac{dh}{\sqrt{h}} = -k \int_0^{t_1} dt \]
- The left side integrates to:
\[ 2(\sqrt{h}) \bigg|_H^{\frac{3H}{4}} = 2\left(\sqrt{\frac{3H}{4}} - \sqrt{H}\right) = 2\left(\frac{\sqrt{3H}}{2} - \sqrt{H}\right) = \sqrt{3H} - 2\sqrt{H} \]
- Thus, we have:
\[ t_1 = k \left(\sqrt{3H} - 2\sqrt{H}\right) \]

6. **Deriving the Time for \( t_2 \)**:
- Now, integrating from \( h = \frac{3H}{4} \) to \( h = 0 \):
\[ \int_{\frac{3H}{4}}^{0} \frac{dh}{\sqrt{h}} = -k \int_{t_1}^{t_2} dt \]
- The left side integrates to:
\[ 2(\sqrt{h}) \bigg|_{\frac{3H}{4}}^{0} = 0 - 2\left(\frac{\sqrt{3H}}{2}\right) = -\sqrt{3H} \]
- Thus, we have:
\[ t_2 = k \sqrt{3H} \]

7. **Finding the Ratio \( \frac{t_1}{t_2} \)**:
- Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{k(\sqrt{3H} - 2\sqrt{H})}{k\sqrt{3H}} = \frac{\sqrt{3H} - 2\sqrt{H}}{\sqrt{3H}} = 1 - \frac{2\sqrt{H}}{\sqrt{3H}} = 1 - \frac{2}{\sqrt{3}} \]
- Simplifying gives:
\[ \frac{t_1}{t_2} = \frac{\sqrt{H}}{\sqrt{3H}} - 1 = \frac{2}{\sqrt{3}} - 1 \]

8. **Final Result**:
- The final ratio \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = \frac{2}{\sqrt{3}} - 1 \]
- Thus, the answer is \( \frac{2}{\sqrt{3}} - 1 \).