Question

An in-compressible fluid flows steadily through a horizontal cylindrical pipe. The pipe has radius '\(3R\)' at point A and '\(1.5R\)' at point B, further along the flow direction at the same level. If the velocity at point A is '\(v\)', then that at point B is

• A. \(v\)
• B. \(2v\)
• C. \(3v\)
• D. \(4v\)

Hint:

Use the equation of continuity A1 v1 = A2 v2.

Answer 1

The Correct Option is D

Step 1: Continuity Equation:
For an incompressible fluid in steady flow, \(A_1v_1=A_2v_2\).

Step 2: Areas:
\(A_A=\pi(3R)^2=9\pi R^2\) and \(A_B=\pi(1.5R)^2=2.25\pi R^2\).

Step 3: Solve:
\[ v_B=\frac{A_A}{A_B}v=\frac{9}{2.25}v=4v \]

Step 4: Check the Options:
The radius ratio is 2, but speed depends on the area ratio, which is \(2^2=4\). Option (B) \(2v\) comes from using the radius ratio directly. So (D) is correct.

Final Answer:
The speed at B is \(4v\), option (D). \[ \boxed{\text{(D) } 4v} \]