A cylindrical tank of height `H` is open at the top end and it has a radius `r`. Water is filled in it up to a height of `h`. The time taken to empty the tank through a hole of radius `r'` in its bottom is

Options

A. `sqrt((2h)/g)(r^(2))/r^('2)`
B. `sqrt((2H)/g)(r^(2))/(r^('2))`
C. `sqrt(h)H`
D. none of these

Explanation

To solve the problem of determining the time taken to empty a cylindrical tank through a hole at the bottom, we can follow these steps:

### Step 1: Understand the Setup
We have a cylindrical tank of height \( H \) and radius \( r \), filled with water up to a height \( h \). There is a hole at the bottom of the tank with radius \( r' \). We need to find the time taken to empty the tank through this hole.

### Step 2: Apply the Principle of Continuity
According to the principle of continuity, the flow rate of water out of the hole must equal the rate at which the water level in the tank decreases.

Let:
- \( A_1 = \pi r^2 \) (cross-sectional area of the tank)
- \( A_2 = \pi (r')^2 \) (cross-sectional area of the hole)

The flow rate out of the hole can be expressed as:
\[ A_2 v_2 = \pi (r')^2 v_2 \]
where \( v_2 \) is the velocity of water exiting the hole.

### Step 3: Apply Torricelli’s Law
The velocity \( v_2 \) of water exiting the hole can be derived from Torricelli’s law:
\[ v_2 = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of water above the hole.

### Step 4: Relate the Change in Height to Time
The rate of change of height of water in the tank can be expressed as:
\[ \frac{dh}{dt} = -\frac{A_2 v_2}{A_1} \]
Substituting for \( A_1 \) and \( A_2 \):
\[ \frac{dh}{dt} = -\frac{\pi (r')^2 \sqrt{2gh}}{\pi r^2} = -\frac{(r')^2 \sqrt{2gh}}{r^2} \]

### Step 5: Rearranging and Integrating
Rearranging gives:
\[ dt = -\frac{r^2}{(r')^2 \sqrt{2g}} \frac{dh}{\sqrt{h}} \]
Now, we integrate both sides. The limits for \( h \) will go from \( h \) to \( 0 \) and for \( t \) from \( 0 \) to \( T \) (the total time to empty the tank):
\[ T = -\frac{r^2}{(r')^2 \sqrt{2g}} \int_{h}^{0} \frac{dh}{\sqrt{h}} \]

### Step 6: Solve the Integral
The integral \( \int \frac{dh}{\sqrt{h}} \) evaluates to \( -2\sqrt{h} \). Thus:
\[ T = -\frac{r^2}{(r')^2 \sqrt{2g}} \left[-2\sqrt{h}\right]_{h}^{0} = \frac{2r^2}{(r')^2 \sqrt{2g}} \sqrt{h} \]

### Final Result
The time taken to empty the tank is given by:
\[ T = \frac{2r^2}{(r')^2 \sqrt{2g}} \sqrt{h} \]