A large tank filled with water to a height `h` is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to `(h)/(2)` and from `(h)/(2)` to zero is

Options

A. `sqrt(2)`
B. `1/(sqrt(2))`
C. `sqrt(2) - 1`
D. `1/(sqrt(2) - 1)`

Answer

To solve the problem of finding the ratio of times taken for the water level in a tank to fall from height \( h \) to \( \frac{h}{2} \) and from \( \frac{h}{2} \) to \( 0 \), we can follow these steps:

### Step 1: Understand the flow of water
The speed of the water flowing out of the hole at the bottom of the tank is given by Torricelli's law, which states that the speed \( v \) of efflux of a fluid under the force of gravity is given by:
\[ v = \sqrt{2gy} \]
where \( y \) is the height of the water above the hole.

### Step 2: Set up the volume flow rate
Let \( A \) be the cross-sectional area of the tank and \( a \) be the area of the hole. The volume flow rate \( \frac{dV}{dt} \) can be expressed as:
\[ \frac{dV}{dt} = a \cdot v = a \cdot \sqrt{2gy} \]
Since the volume \( V \) of water in the tank is given by \( V = A \cdot y \), we can relate the change in volume to the change in height:
\[ \frac{dV}{dt} = A \frac{dy}{dt} \]

### Step 3: Relate the two equations
Equating the two expressions for \( \frac{dV}{dt} \):
\[ A \frac{dy}{dt} = a \sqrt{2gy} \]
Rearranging gives:
\[ \frac{dy}{dt} = \frac{a}{A} \sqrt{2gy} \]

### Step 4: Separate variables and integrate
We can separate the variables to integrate:
\[ \frac{dy}{\sqrt{y}} = \frac{a}{A} \sqrt{2g} dt \]
Integrating both sides:
\[ \int \frac{dy}{\sqrt{y}} = \frac{a}{A} \sqrt{2g} \int dt \]
The left side integrates to \( 2\sqrt{y} \), and the right side integrates to \( \frac{a}{A} \sqrt{2g} t \):
\[ 2\sqrt{y} = \frac{a}{A} \sqrt{2g} t + C \]

### Step 5: Find the time taken for each segment
1. **From \( h \) to \( \frac{h}{2} \)**:
- Initial height \( y_i = h \), final height \( y_f = \frac{h}{2} \):
\[ 2\sqrt{\frac{h}{2}} - 2\sqrt{h} = \frac{a}{A} \sqrt{2g} t_1 \]
This simplifies to:
\[ 2\left(\sqrt{\frac{h}{2}} - \sqrt{h}\right) = \frac{a}{A} \sqrt{2g} t_1 \]

2. **From \( \frac{h}{2} \) to \( 0 \)**:
- Initial height \( y_i = \frac{h}{2} \), final height \( y_f = 0 \):
\[ 2\sqrt{0} - 2\sqrt{\frac{h}{2}} = \frac{a}{A} \sqrt{2g} t_2 \]
This simplifies to:
\[ -2\sqrt{\frac{h}{2}} = \frac{a}{A} \sqrt{2g} t_2 \]

### Step 6: Calculate the ratio of times
Now we can find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{2\left(\sqrt{\frac{h}{2}} - \sqrt{h}\right)}{-2\sqrt{\frac{h}{2}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{h}{2}} - \sqrt{h}}{-\sqrt{\frac{h}{2}}} \]
After simplification, we find:
\[ \frac{t_1}{t_2} = \sqrt{2} - 1 \]

### Final Answer
The ratio of times taken for the level of water to fall from \( h \) to \( \frac{h}{2} \) and from \( \frac{h}{2} \) to \( 0 \) is:
\[ \sqrt{2} - 1 \]