There is a hole at the bottom of a large open vessel. If water is filled upto a height h, it flows out in time t. if water is filled to a height 4h, it will flow out in time

Options

A. 4 t
B. t/4
C. t/2
D. 2t

Answer

To solve the problem, we can use Torricelli's law, which states that the speed of efflux of a fluid under the force of gravity through a hole is given by the equation:

\[ v = \sqrt{2gh} \]

where:
- \( v \) is the speed of efflux,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the fluid above the hole.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a large open vessel with a hole at the bottom. We need to find the time taken for water to flow out when the height of water is increased from \( h \) to \( 4h \).

2. **Flow Rate**:
The flow rate \( Q \) through the hole can be expressed as:
\[ Q = A \cdot v \]
where \( A \) is the area of the hole and \( v \) is the velocity of the water flowing out.

3. **Applying Torricelli's Law**:
For a height \( h \):
\[ v_h = \sqrt{2gh} \]
For a height \( 4h \):
\[ v_{4h} = \sqrt{2g(4h)} = \sqrt{8gh} = 2\sqrt{2gh} \]

4. **Volume of Water**:
The volume of water \( V \) in the vessel when filled to height \( h \) is:
\[ V_h = A \cdot h \]
The volume when filled to height \( 4h \) is:
\[ V_{4h} = A \cdot 4h \]

5. **Time to Empty the Vessel**:
The time \( t \) taken to empty the water from the height \( h \) is given by:
\[ t = \frac{V_h}{Q} = \frac{A \cdot h}{A \cdot v_h} = \frac{h}{v_h} = \frac{h}{\sqrt{2gh}} = \frac{\sqrt{h}}{\sqrt{2g}} \]

For the height \( 4h \):
\[ t_{4h} = \frac{V_{4h}}{Q} = \frac{A \cdot 4h}{A \cdot v_{4h}} = \frac{4h}{2\sqrt{2gh}} = \frac{2\sqrt{h}}{\sqrt{2g}} \]

6. **Relating the Times**:
We can see that:
\[ t_{4h} = 2 \cdot t \]
Therefore, if the water is filled to a height of \( 4h \), the time taken to flow out will be \( 2t \).

### Final Answer:
If the water is filled to a height of \( 4h \), it will flow out in time \( 2t \).