Water contained in a tank flows through an orifice of a diameter 2 cm under a constant pressure difference of 10 cm of water column. The rate of flow of water through the orifice is

Options

A. 44cc/s
B. 4.4cc/s
C. 440cc/s
D. 4400cc/s

Answer

To solve the problem of determining the rate of flow of water through an orifice, we can follow these steps:

### Step 1: Determine the velocity of water through the orifice
We can use Torricelli's theorem, which states that the velocity \( v \) of fluid flowing out of an orifice under the influence of gravity is given by the formula:
\[ v = \sqrt{2gh} \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the height of the water column above the orifice.

Given that the pressure difference corresponds to a height of 10 cm of water, we convert this to meters:
\[ h = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.10 \, \text{m} \]

Now substituting the values into the formula:
\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 0.10 \, \text{m}} = \sqrt{1.96} \approx 1.4 \, \text{m/s} \]

### Step 2: Calculate the area of the orifice
The area \( A \) of the orifice can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Given that the diameter of the orifice is 2 cm, we can find the radius:
\[ \text{Diameter} = 2 \, \text{cm} \Rightarrow \text{Radius} = \frac{2}{2} \, \text{cm} = 1 \, \text{cm} = 1 \times 10^{-2} \, \text{m} \]

Now substituting the radius into the area formula:
\[ A = \pi (1 \times 10^{-2})^2 = \pi \times 1 \times 10^{-4} \approx 3.14 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the rate of flow of water through the orifice
The rate of flow \( Q \) can be calculated using the formula:
\[ Q = v \times A \]
Substituting the values we calculated:
\[ Q = 1.4 \, \text{m/s} \times 3.14 \times 10^{-4} \, \text{m}^2 \approx 4.396 \times 10^{-4} \, \text{m}^3/\text{s} \]

### Step 4: Convert the flow rate to cubic centimeters per second
Since \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \):
\[ Q \approx 4.396 \times 10^{-4} \, \text{m}^3/\text{s} \times 10^6 \, \text{cm}^3/\text{m}^3 \approx 439.6 \, \text{cm}^3/\text{s} \]

Rounding this to two significant figures, we find:
\[ Q \approx 440 \, \text{cm}^3/\text{s} \]

### Conclusion
The rate of flow of water through the orifice is approximately \( 440 \, \text{cm}^3/\text{s} \).