A large tank is filled with water to a height `H`. A small hole is made at the base of the tank. It takes `T_(1)` time to decrease the height of water to `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` time to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` is

Options

A. 2
B. 3
C. 4
D. `2sqrt(2)`

Explanation

To solve the problem step by step, we need to analyze the time taken for the water to drain from the tank through a hole at the base.

### Step 1: Understand the problem
We have a tank filled with water to a height \( H \). A hole is made at the base, and we need to find the value of \( \eta \) given that it takes time \( T_1 \) to decrease the height from \( H \) to \( \frac{H}{\eta} \) and time \( T_2 \) to drain the remaining water, with the condition that \( T_1 = T_2 \).

### Step 2: Write the formula for time taken
The time taken \( T \) to drain water from a height \( h_1 \) to \( h_2 \) through a hole can be expressed as:
\[ T = \frac{A}{a} \sqrt{\frac{2}{g}} \left( \sqrt{h_1} - \sqrt{h_2} \right) \]
where:
- \( A \) is the cross-sectional area of the tank,
- \( a \) is the cross-sectional area of the hole,
- \( g \) is the acceleration due to gravity.

### Step 3: Calculate \( T_1 \)
For the first part, where the height decreases from \( H \) to \( \frac{H}{\eta} \):
\[ T_1 = \frac{A}{a} \sqrt{\frac{2}{g}} \left( \sqrt{H} - \sqrt{\frac{H}{\eta}} \right) \]
This simplifies to:
\[ T_1 = \frac{A}{a} \sqrt{\frac{2}{g}} \left( \sqrt{H} - \sqrt{H} \cdot \frac{1}{\sqrt{\eta}} \right) = \frac{A}{a} \sqrt{\frac{2H}{g}} \left( 1 - \frac{1}{\sqrt{\eta}} \right) \]

### Step 4: Calculate \( T_2 \)
For the second part, where the height decreases from \( \frac{H}{\eta} \) to \( 0 \):
\[ T_2 = \frac{A}{a} \sqrt{\frac{2}{g}} \left( \sqrt{\frac{H}{\eta}} - 0 \right) = \frac{A}{a} \sqrt{\frac{2H}{g}} \cdot \frac{1}{\sqrt{\eta}} \]

### Step 5: Set \( T_1 = T_2 \)
Since \( T_1 = T_2 \), we can equate the two expressions:
\[ \frac{A}{a} \sqrt{\frac{2H}{g}} \left( 1 - \frac{1}{\sqrt{\eta}} \right) = \frac{A}{a} \sqrt{\frac{2H}{g}} \cdot \frac{1}{\sqrt{\eta}} \]
We can cancel \( \frac{A}{a} \sqrt{\frac{2H}{g}} \) from both sides (assuming \( A \) and \( a \) are not zero):
\[ 1 - \frac{1}{\sqrt{\eta}} = \frac{1}{\sqrt{\eta}} \]

### Step 6: Solve for \( \eta \)
Rearranging gives:
\[ 1 = 2 \cdot \frac{1}{\sqrt{\eta}} \]
This leads to:
\[ \sqrt{\eta} = 2 \quad \Rightarrow \quad \eta = 4 \]

### Final Answer
Thus, the value of \( \eta \) is \( 4 \).

---