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. `sqrt3`
B. `sqrt2`
C. `(2-sqrt2)/sqrt2`
D. `(2-sqrt3)/sqrt3`

Answer

To solve the problem, we need to find the ratio \( \frac{t_1}{t_2} \) where \( t_1 \) is the time taken to empty one-fourth of the tank and \( t_2 \) is the time taken to empty the remaining three-fourths of the tank.

### Step 1: Understanding the problem
We have a tank filled with water, and there is a hole at the bottom. The speed of efflux of water through the hole can be described by Torricelli's theorem, which states that the speed \( v \) of efflux 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.

### Step 2: Setting up the equations
We can express the rate of change of height of water in the tank as:
\[ \frac{dh}{dt} = -\sqrt{2gh} \]
The negative sign indicates that the height of the water is decreasing.

### Step 3: Separating variables
We can separate the variables to integrate:
\[ \frac{dh}{\sqrt{h}} = -\sqrt{2g} dt \]

### Step 4: Integrating for \( t_1 \)
For the first part, we want to find \( t_1 \) when the height changes from \( h \) to \( \frac{3h}{4} \):
\[ \int_h^{\frac{3h}{4}} \frac{dh}{\sqrt{h}} = -\sqrt{2g} \int_0^{t_1} dt \]
The left-hand side becomes:
\[ \left[ 2\sqrt{h} \right]_h^{\frac{3h}{4}} = 2\sqrt{\frac{3h}{4}} - 2\sqrt{h} = 2\left( \frac{\sqrt{3}}{2}\sqrt{h} - \sqrt{h} \right) = 2\left( \frac{\sqrt{3} - 2}{2}\sqrt{h} \right) \]
Thus, we have:
\[ 2\left( \frac{\sqrt{3} - 2}{2}\sqrt{h} \right) = -\sqrt{2g} t_1 \]
This simplifies to:
\[ t_1 = -\frac{\sqrt{h}(\sqrt{3} - 2)}{\sqrt{2g}} \]

### Step 5: Integrating for \( t_2 \)
For the second part, we want to find \( t_2 \) when the height changes from \( \frac{3h}{4} \) to \( 0 \):
\[ \int_{\frac{3h}{4}}^0 \frac{dh}{\sqrt{h}} = -\sqrt{2g} \int_0^{t_2} dt \]
The left-hand side becomes:
\[ \left[ 2\sqrt{h} \right]_{\frac{3h}{4}}^0 = 0 - 2\sqrt{\frac{3h}{4}} = -\sqrt{3h} \]
Thus, we have:
\[ -\sqrt{3h} = -\sqrt{2g} t_2 \]
This simplifies to:
\[ t_2 = \frac{\sqrt{3h}}{\sqrt{2g}} \]

### Step 6: Finding the ratio \( \frac{t_1}{t_2} \)
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{-\frac{\sqrt{h}(\sqrt{3} - 2)}{\sqrt{2g}}}{\frac{\sqrt{3h}}{\sqrt{2g}}} \]
The \( \sqrt{2g} \) cancels out:
\[ \frac{t_1}{t_2} = -\frac{\sqrt{h}(\sqrt{3} - 2)}{\sqrt{3h}} = -\frac{\sqrt{3} - 2}{\sqrt{3}} \]

### Final Result
Thus, the ratio \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = 2 - \sqrt{3} \]