Water drains out of a vessel filled with water upto height h through a hole of area A in t seconds.If the height of the water is 4 h then how much time will be required for the water to drain out ?[Assume `A lt lt A_(0)` (area of tank ) ]

Options

A. t seconds
B. 4 t second
C. 2 t seconds
D. `(t)/( 4)` seconds

Answer

To solve the problem of how long it takes for water to drain out of a vessel when the height of the water is increased from \( h \) to \( 4h \), we can use the principles of fluid dynamics and kinematics.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that water drains out of a vessel through a hole of area \( A \). The time taken to drain water from a height \( h \) is given as \( t \). We need to find the time taken \( t' \) to drain water from a height \( 4h \).

2. **Using Torricelli's Law**:
According to Torricelli's Law, the speed \( v \) of efflux of a fluid under gravity through an orifice is given by:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of the fluid above the hole.

3. **Finding the Time for Height \( h \)**:
For the initial height \( h \), the time \( t \) taken to drain the water can be expressed in terms of the height:
\[ t \propto \sqrt{h} \]
Therefore, we can write:
\[ t = k \sqrt{h} \]
where \( k \) is a constant of proportionality.

4. **Finding the Time for Height \( 4h \)**:
Now, for the height \( 4h \):
\[ t' \propto \sqrt{4h} = 2\sqrt{h} \]
Thus, we can express \( t' \) in terms of \( t \):
\[ t' = k \cdot 2\sqrt{h} = 2k\sqrt{h} = 2t \]

5. **Conclusion**:
Therefore, the time required to drain the water when the height is \( 4h \) is:
\[ t' = 2t \]

### Final Answer:
The time required for the water to drain out when the height is \( 4h \) is \( 2t \).

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