Question

A pipe A is connected with other pipes B and C as shown in the figure. The areas of cross-section of A, B and C are respectively \( \alpha \), \( \frac{\alpha}{2} \) and \( \frac{\alpha}{4} \). If the velocities of flow of water through A and B are 10 m/sec and 6 m/sec, respectively, then velocity of flow, \( V_c \) along C is

• A. 21 m/sec
• B. 12 m/sec
• C. 28 m/sec
• D. 18 m/sec

Hint:

The continuity equation is a direct consequence of the conservation of mass. For branching pipes, the total incoming volume flow rate must equal the sum of the outgoing volume flow rates: \( Q_{\text{in}} = \sum Q_{\text{out}} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem shows a main pipe branching into two smaller pipes. We need to find the velocity of flow \( V_c \) in the branch pipe \( C \) using the given areas of cross-section and flow velocities in pipes \( A \) and \( B \).

Step 2: Key Formula or Approach:
For an incompressible and steady fluid flow, we apply the Equation of Continuity (mass/volume flow rate conservation):
\[ A_a V_a = A_b V_b + A_c V_c \]

Step 3: Detailed Explanation:
Given parameters:
- Area of cross-section of A, \( A_a = \alpha \)
- Velocity of flow in A, \( V_a = 10 \text{ m/s} \)
- Area of cross-section of B, \( A_b = \frac{\alpha}{2} \)
- Velocity of flow in B, \( V_b = 6 \text{ m/s} \)
- Area of cross-section of C, \( A_c = \frac{\alpha}{4} \)
- Velocity of flow in C, \( V_c \)
Applying the equation of continuity:
\[ \alpha (10) = \left(\frac{\alpha}{2}\right) (6) + \left(\frac{\alpha}{4}\right) V_c \]
Since \( \alpha \) is common to all terms, we can divide the entire equation by \( \alpha \):
\[ 10 = 3 + \frac{V_c}{4} \]
\[ 7 = \frac{V_c}{4} \]
\[ V_c = 28 \text{ m/s} \]

Step 4: Final Answer:
The velocity of flow along pipe C is 28 m/sec.