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

Answer

To solve the problem, we will use the principles of fluid dynamics and energy conservation. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a large open vessel with a hole at the bottom. When the vessel is filled with water to a height \( h \), it takes time \( t \) to empty. We need to find the time taken to empty the vessel when it is filled to a height \( 4h \).

### Step 2: Use Torricelli’s Law
According to Torricelli’s Law, the speed \( v \) of fluid flowing out of an orifice under the influence of gravity is given by:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the water column above the hole.

### Step 3: Calculate the time to empty the vessel for height \( h \)
When the water height is \( h \), the speed of water flowing out is:
\[ v = \sqrt{2gh} \]
The volume of water flowing out can be expressed in terms of the area of the hole \( A \) and the time \( t \):
\[ \text{Volume} = A \cdot v \cdot t \]
The volume of water in the vessel when filled to height \( h \) is:
\[ \text{Volume} = A \cdot h \]
Setting these equal gives:
\[ A \cdot h = A \cdot v \cdot t \]
Cancelling \( A \) from both sides (assuming \( A \neq 0 \)):
\[ h = v \cdot t \]
Substituting \( v = \sqrt{2gh} \):
\[ h = \sqrt{2gh} \cdot t \]

### Step 4: Solve for time \( t \)
Rearranging gives:
\[ t = \frac{h}{\sqrt{2gh}} = \frac{h}{\sqrt{2g}} \cdot \frac{1}{\sqrt{h}} = \frac{\sqrt{h}}{\sqrt{2g}} \]

### Step 5: Calculate the time to empty the vessel for height \( 4h \)
Now, when the height is \( 4h \), the speed of water flowing out becomes:
\[ v' = \sqrt{2g(4h)} = 2\sqrt{2gh} \]
The volume of water when filled to height \( 4h \) is:
\[ \text{Volume} = A \cdot 4h \]
Setting the volume equal to the outflow gives:
\[ 4h = v' \cdot t' \]
Substituting \( v' \):
\[ 4h = (2\sqrt{2gh}) \cdot t' \]
Cancelling \( 2 \) from both sides:
\[ 2h = \sqrt{2gh} \cdot t' \]
Solving for \( t' \):
\[ t' = \frac{2h}{\sqrt{2gh}} = 2 \cdot \frac{\sqrt{h}}{\sqrt{2g}} = 2t \]

### Conclusion
Thus, the time taken to empty the vessel when filled to a height \( 4h \) is:
\[ t' = 2t \]

### Final Answer
The time taken to empty the vessel when filled to a height \( 4h \) is \( 2t \).

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