A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water

Options

A. 9 min
B. 7 min
C. 5 min
D. 3 min

Explanation

To solve the problem step by step, we can use the principles of fluid dynamics, specifically Torricelli's law, which relates the time taken to empty a vessel through an orifice to the height of the liquid above the orifice.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a rectangular vessel that takes 10 minutes to empty when full (height = H).
- We need to find the time taken to empty the vessel when it is half-filled (height = H/2).

2. **Use the Formula for Time to Empty**:
- The time \( T \) taken to empty a vessel through an orifice is given by:
\[ T = k \cdot \frac{A}{A_0} \sqrt{\frac{2h}{g}} \]
where:
- \( k \) is a constant,
- \( A \) is the cross-sectional area of the vessel,
- \( A_0 \) is the area of the orifice,
- \( h \) is the height of the liquid,
- \( g \) is the acceleration due to gravity.

3. **Set Up the Equations**:
- For the full vessel (height = H):
\[ T_1 = k \cdot \frac{A}{A_0} \sqrt{\frac{2H}{g}} = 10 \text{ minutes} \]
- For the half-filled vessel (height = H/2):
\[ T_2 = k \cdot \frac{A}{A_0} \sqrt{\frac{2(H/2)}{g}} = k \cdot \frac{A}{A_0} \sqrt{\frac{H}{g}} \]

4. **Relate the Two Times**:
- We can relate \( T_1 \) and \( T_2 \):
\[ \frac{T_1}{T_2} = \frac{\sqrt{2H}}{\sqrt{H/2}} = \frac{\sqrt{2H}}{\sqrt{H}/\sqrt{2}} = \frac{\sqrt{2H} \cdot \sqrt{2}}{\sqrt{H}} = \frac{2\sqrt{H}}{\sqrt{H}} = 2 \]
- Thus, we have:
\[ T_2 = \frac{T_1}{2} \]

5. **Calculate \( T_2 \)**:
- Since \( T_1 = 10 \) minutes:
\[ T_2 = \frac{10}{2} = 5 \text{ minutes} \]

### Final Answer:
The time taken to empty the vessel when it is half-filled with water is **5 minutes**.