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 solve the problem of finding the rate at which mass is leaving the cylindrical pipe, we can follow these steps:

### Step 1: Identify the given values
- Radius of the pipe (r) = 2 cm = 0.02 m (convert to meters)
- Velocity of water (v) = 10 m/s
- Density of water (ρ) = 1000 kg/m³

### Step 2: Calculate the cross-sectional area (A) of the pipe
The area of a circle is given by the formula:
\[ A = \pi r^2 \]

Substituting the radius:
\[ A = \pi (0.02)^2 \]
\[ A = \pi (0.0004) \]
\[ A = 0.0004\pi \, \text{m}^2 \]

### Step 3: Use the formula for mass flow rate
The mass flow rate (dm/dt) can be calculated using the formula:
\[ \frac{dm}{dt} = \rho \cdot A \cdot v \]

### Step 4: Substitute the values into the mass flow rate formula
Now substituting the values we have:
\[ \frac{dm}{dt} = 1000 \cdot (0.0004\pi) \cdot 10 \]

### Step 5: Simplify the expression
Calculating the expression:
\[ \frac{dm}{dt} = 1000 \cdot 0.0004 \cdot 10 \cdot \pi \]
\[ \frac{dm}{dt} = 4\pi \, \text{kg/s} \]

### Step 6: Calculate the numerical value
Using the value of π (approximately 3.14):
\[ \frac{dm}{dt} = 4 \cdot 3.14 \]
\[ \frac{dm}{dt} \approx 12.56 \, \text{kg/s} \]

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

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