A tank is filled with a liquid upto a height H, A small hole is made at the bottom of this tank Let `t_(1)` be the time taken to empty first half of the tank and `t_(2)` time taken to empty rest half of the tank then find `(t_(1))/(t_(2))`

Explanation

To solve the problem of finding the ratio \( \frac{t_1}{t_2} \) where \( t_1 \) is the time taken to empty the first half of the tank and \( t_2 \) is the time taken to empty the second half, we can follow these steps:

### Step 1: Understand the relationship between time and height
The time \( t \) taken to empty a tank is proportional to the square root of the height of the liquid in the tank. This can be expressed mathematically as:
\[ t \propto \sqrt{h} \]
where \( h \) is the height of the liquid.

### Step 2: Define the total time to empty the tank
Let \( T \) be the total time to empty the entire tank from height \( H \). Thus, we can write:
\[ T = k \sqrt{H} \]
where \( k \) is a constant of proportionality.

### Step 3: Calculate \( t_1 \) for the first half
To find \( t_1 \), the time taken to empty the first half of the tank (from height \( H \) to \( H/2 \)), we can express it as:
\[ t_1 = k \sqrt{H} - k \sqrt{\frac{H}{2}} = k \left( \sqrt{H} - \sqrt{\frac{H}{2}} \right) \]
This simplifies to:
\[ t_1 = k \sqrt{H} \left( 1 - \frac{1}{\sqrt{2}} \right) \]

### Step 4: Calculate \( t_2 \) for the second half
For \( t_2 \), the time taken to empty the second half of the tank (from height \( H/2 \) to \( 0 \)), we have:
\[ t_2 = k \sqrt{\frac{H}{2}} = k \frac{\sqrt{H}}{\sqrt{2}} \]

### Step 5: Find the ratio \( \frac{t_1}{t_2} \)
Now, we can find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{k \sqrt{H} \left( 1 - \frac{1}{\sqrt{2}} \right)}{k \frac{\sqrt{H}}{\sqrt{2}}} \]
The \( k \) and \( \sqrt{H} \) terms cancel out:
\[ \frac{t_1}{t_2} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \left( 1 - \frac{1}{\sqrt{2}} \right) \sqrt{2} \]

### Step 6: Simplify the expression
Now we can simplify:
\[ \frac{t_1}{t_2} = \sqrt{2} - 1 \]
Using the approximate value of \( \sqrt{2} \approx 1.414 \):
\[ \frac{t_1}{t_2} \approx 1.414 - 1 = 0.414 \]

### Final Answer
Thus, the ratio \( \frac{t_1}{t_2} \) is approximately:
\[ \frac{t_1}{t_2} \approx 0.414 \]