A cylindrical vessel of cross section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is:

Options

A. `(2A)/(pi a^(2)) sqrt(h/g)`
B. `(sqrt2A)/(pia^(2)) sqrt(h/g)`
C. `(2sqrt2A)/(pi a^(2)) sqrt(h/g)`
D. `(A)/(sqrt2pia^(2)) sqrt(h/g)`

Explanation

To find the time taken for a cylindrical vessel filled with water to empty through a hole at the bottom, we can follow these steps:

### Step 1: Define the Variables
- Let \( A \) be the cross-sectional area of the cylindrical vessel.
- Let \( a \) be the radius of the hole at the bottom.
- Let \( h \) be the initial height of the water in the vessel.
- Let \( y \) be the height of the water at any time \( t \).

### Step 2: Write the Volume Discharge Equation
The volume of water that discharges through the hole in a small time \( dt \) can be expressed as:
\[ dV = A \, dy \]
where \( dy \) is the change in height of the water.

### Step 3: Use Torricelli's Law
According to Torricelli's law, the velocity \( V \) of water flowing out of the hole is given by:
\[ V = \sqrt{2gy} \]
where \( g \) is the acceleration due to gravity.

### Step 4: Write the Volume Flow Rate
The volume flow rate \( dV \) through the hole can also be expressed as:
\[ dV = \text{Area of the hole} \times \text{Velocity} \times dt = \pi a^2 \sqrt{2gy} \, dt \]

### Step 5: Set the Two Volume Equations Equal
Since both expressions for \( dV \) represent the same volume of water discharged, we can set them equal:
\[ A \, dy = \pi a^2 \sqrt{2gy} \, dt \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ \frac{dy}{\sqrt{y}} = -\frac{\pi a^2}{A \sqrt{2g}} \, dt \]

### Step 7: Integrate Both Sides
Integrate both sides:
\[ \int_{h}^{0} \frac{dy}{\sqrt{y}} = -\frac{\pi a^2}{A \sqrt{2g}} \int_{0}^{t} dt \]
The left side integrates to:
\[ -2(\sqrt{y}) \bigg|_{h}^{0} = -2(0 - \sqrt{h}) = 2\sqrt{h} \]
The right side integrates to:
\[ -\frac{\pi a^2}{A \sqrt{2g}} t \]

### Step 8: Set the Integrals Equal
Setting the results of the integrals equal gives:
\[ 2\sqrt{h} = -\frac{\pi a^2}{A \sqrt{2g}} t \]

### Step 9: Solve for Time \( t \)
Rearranging for \( t \):
\[ t = \frac{2A\sqrt{h}}{\pi a^2 \sqrt{2g}} \]

### Final Result
Thus, the time taken to empty the vessel is:
\[ t = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}} \]