One end of a cylinderical pipe has a radius of 2 cm. Water comes out at 10 m/s. The rate at which mass is leaving the pipe is

Options

A. 3.14 kg/s
B. 10 kg/s
C. 12.56 kg/s
D. 9.56 kg/s

Answer

To find the rate at which mass is leaving the pipe, we can use the formula derived from the continuity equation in fluid dynamics. The mass flow rate (ṁ) can be calculated using the formula:

\[ \dot{m} = \rho \cdot A \cdot V \]

Where:
- \(\dot{m}\) is the mass flow rate (in kg/s),
- \(\rho\) is the density of the fluid (for water, \(\rho \approx 1000 \, \text{kg/m}^3\)),
- \(A\) is the cross-sectional area of the pipe (in m²),
- \(V\) is the velocity of the fluid (in m/s).

### Step 1: Calculate the cross-sectional area (A) of the pipe.
The radius of the pipe is given as 2 cm, which we need to convert to meters for consistency in units:

\[ r = 2 \, \text{cm} = 0.02 \, \text{m} \]

The area \(A\) of a circle is given by the formula:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 0.00125664 \, \text{m}^2 \]

### Step 2: Use the velocity (V) of the water.
The velocity of water coming out of the pipe is given as:

\[ V = 10 \, \text{m/s} \]

### Step 3: Use the density (\(\rho\)) of water.
The density of water is approximately:

\[ \rho = 1000 \, \text{kg/m}^3 \]

### Step 4: Calculate the mass flow rate (\(\dot{m}\)).
Now, we can substitute the values into the mass flow rate formula:

\[ \dot{m} = \rho \cdot A \cdot V \]

Substituting the known values:

\[ \dot{m} = 1000 \, \text{kg/m}^3 \cdot 0.00125664 \, \text{m}^2 \cdot 10 \, \text{m/s} \]

Calculating this gives:

\[ \dot{m} = 1000 \cdot 0.00125664 \cdot 10 = 12.5664 \, \text{kg/s} \]

Rounding this to two decimal places, we find:

\[ \dot{m} \approx 12.56 \, \text{kg/s} \]

### Conclusion:
The rate at which mass is leaving the pipe is approximately **12.56 kg/s**.

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