A pipe 2 cm in diameter has a constriction of diameter 1 cm. What it’s the velocity of flow at the constriction. If velocity of flow in the broader region of the pipe is 5 cm/s?

Options

A. 10 cm/s
B. 20 cm/s
C. 25 cm/s
D. 30 cm/s

Answer

To solve the problem, we can use the principle of continuity for fluid flow, which states that the product of the cross-sectional area and the velocity of flow must remain constant throughout a streamline flow. This can be expressed mathematically as:

\[ A_1 v_1 = A_2 v_2 \]

Where:
- \( A_1 \) = cross-sectional area at the broader region
- \( v_1 \) = velocity of flow at the broader region
- \( A_2 \) = cross-sectional area at the constriction
- \( v_2 \) = velocity of flow at the constriction

### Step 1: Calculate the cross-sectional areas

1. **Calculate \( A_1 \)** (Area at the broader region):
- The diameter \( d_1 = 2 \) cm.
- The radius \( r_1 = \frac{d_1}{2} = 1 \) cm.
- Area \( A_1 = \pi r_1^2 = \pi (1)^2 = \pi \) cm².

2. **Calculate \( A_2 \)** (Area at the constriction):
- The diameter \( d_2 = 1 \) cm.
- The radius \( r_2 = \frac{d_2}{2} = 0.5 \) cm.
- Area \( A_2 = \pi r_2^2 = \pi (0.5)^2 = \frac{\pi}{4} \) cm².

### Step 2: Use the continuity equation

Using the continuity equation:

\[ A_1 v_1 = A_2 v_2 \]

Substituting the values we have:

\[ \pi \cdot 5 = \frac{\pi}{4} v_2 \]

### Step 3: Solve for \( v_2 \)

1. Cancel \( \pi \) from both sides:

\[ 5 = \frac{1}{4} v_2 \]

2. Multiply both sides by 4:

\[ v_2 = 5 \cdot 4 = 20 \text{ cm/s} \]

### Conclusion

The velocity of flow at the constriction is \( v_2 = 20 \text{ cm/s} \).

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