A tank full of water has a small hole at its bottom. Let `t_(1)` be the time taken to empty the first half of the tank and `t_(2)` be the time needed to empty the rest half of the tank, then

Options

A. `t_(1)=t_(2)`
B. `t_(1) gt t_(2)`
C. `t_(1) lt t_(2)`
D. `t_(1)=0.523 t_(2)`

Explanation

To solve the problem of determining the relationship between the time taken to empty the first half of the tank (T1) and the time taken to empty the second half of the tank (T2), we can follow these steps:

### Step 1: Understand the scenario
We have a tank full of water with a small hole at the bottom. The water will flow out of the hole due to gravity. The height of the water column above the hole will determine the pressure and, consequently, the velocity of efflux.

### Step 2: Apply Torricelli's Law
According to Torricelli's Law, the velocity of efflux (v) of a fluid under the influence of gravity through a hole is given by:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the water column above the hole.

### Step 3: Analyze the first half of the tank (T1)
When the tank is full, the height of the water column is \( H \). The velocity of efflux when the tank is full is:
\[ v_1 = \sqrt{2gH} \]
Using this velocity, we can find the time \( T_1 \) to empty the first half of the tank. The flow rate (Q) can be expressed as:
\[ Q = A \cdot v_1 \]
where \( A \) is the area of the hole.

The volume of the first half of the tank is \( V_1 = \frac{1}{2}A_t H \) (where \( A_t \) is the cross-sectional area of the tank). The time taken to empty this volume is:
\[ T_1 = \frac{V_1}{Q} = \frac{\frac{1}{2}A_t H}{A \cdot \sqrt{2gH}} \]

### Step 4: Analyze the second half of the tank (T2)
Once the first half is emptied, the height of the water column is now \( \frac{H}{2} \). The velocity of efflux at this height is:
\[ v_2 = \sqrt{2g \cdot \frac{H}{2}} = \sqrt{gH} \]
Using this velocity, we can find the time \( T_2 \) to empty the second half of the tank. The volume of the second half of the tank is the same as the first half:
\[ V_2 = \frac{1}{2}A_t H \]
The time taken to empty this volume is:
\[ T_2 = \frac{V_2}{Q} = \frac{\frac{1}{2}A_t H}{A \cdot \sqrt{gH}} \]

### Step 5: Compare T1 and T2
Now we can compare \( T_1 \) and \( T_2 \):
- For \( T_1 \):
\[ T_1 = \frac{\frac{1}{2}A_t H}{A \cdot \sqrt{2gH}} = \frac{A_t H}{2A \sqrt{2gH}} \]
- For \( T_2 \):
\[ T_2 = \frac{\frac{1}{2}A_t H}{A \cdot \sqrt{gH}} = \frac{A_t H}{2A \sqrt{gH}} \]

### Step 6: Simplifying the expressions
Now we can see that:
\[ T_1 = \frac{A_t H}{2A \sqrt{2gH}} \]
\[ T_2 = \frac{A_t H}{2A \sqrt{gH}} \]

Since \( \sqrt{2gH} > \sqrt{gH} \), it follows that:
\[ T_1 < T_2 \]

### Conclusion
Thus, we conclude that:
\[ T_2 > T_1 \]