Water enters a pipe of diameter 3.0 cm with a velocity of 3.0 m/s . The water encounters a construction where its velocity is 12 m/s . What is the diameter of the constricted portion of the pipe ?

Options

A. 0.33 cm
B. 0.75 cm
C. 1.0 cm
D. 1.5 cm

Answer

To solve the problem of finding the diameter of the constricted portion of the pipe, we will use the principle of conservation of mass for incompressible fluids, known as the equation of continuity.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the initial section of the pipe (D1) = 3.0 cm
- Velocity of water in the initial section (V1) = 3.0 m/s
- Velocity of water in the constricted section (V2) = 12.0 m/s

2. **Convert Diameter to Radius:**
- The radius (R1) of the initial section can be calculated as:
\[ R1 = \frac{D1}{2} = \frac{3.0 \, \text{cm}}{2} = 1.5 \, \text{cm} \]

3. **Calculate the Cross-Sectional Area of the Initial Section (A1):**
- The cross-sectional area (A1) can be calculated using the formula for the area of a circle:
\[ A1 = \pi R1^2 = \pi (1.5 \, \text{cm})^2 = \pi (2.25 \, \text{cm}^2) \approx 7.07 \, \text{cm}^2 \]

4. **Apply the Equation of Continuity:**
- According to the equation of continuity, the mass flow rate must be constant:
\[ A1 \cdot V1 = A2 \cdot V2 \]
- Here, A2 is the cross-sectional area of the constricted section.

5. **Express A2 in Terms of D2:**
- The area of the constricted section (A2) can be expressed as:
\[ A2 = \pi R2^2 = \pi \left(\frac{D2}{2}\right)^2 = \frac{\pi D2^2}{4} \]

6. **Substitute Known Values into the Equation:**
- Substitute A1, V1, A2, and V2 into the equation:
\[ A1 \cdot V1 = A2 \cdot V2 \]
\[ (7.07 \, \text{cm}^2) \cdot (3.0 \, \text{m/s}) = \left(\frac{\pi D2^2}{4}\right) \cdot (12.0 \, \text{m/s}) \]

7. **Solve for D2:**
- Rearranging the equation gives:
\[ D2^2 = \frac{(7.07 \cdot 3.0)}{(12.0)} \cdot \frac{4}{\pi} \]
- Calculate the left-hand side:
\[ D2^2 = \frac{21.21}{12.0} \cdot \frac{4}{\pi} \approx \frac{21.21 \cdot 4}{37.7} \approx \frac{84.84}{37.7} \approx 2.25 \, \text{cm}^2 \]
- Taking the square root:
\[ D2 = \sqrt{2.25} = 1.5 \, \text{cm} \]

### Final Answer:
The diameter of the constricted portion of the pipe (D2) is **1.5 cm**.