Question

Water flows through a horizontal pipe at a speed 'V'. Internal diameter of the pipe is 'd'. If the water is emerging at a speed '\(V_1\)' then the diameter of the nozzle is

• A. \(\text{d}\sqrt{\frac{\text{V}}{\text{V}_1}}\)
• B. \(\text{d}\sqrt{\frac{\text{V}_1}{\text{V}}}\)
• C. \(\frac{\text{dV}}{\text{V}_1}\)
• D. \(\frac{\text{dV}_1}{\text{V}}\)

Hint:

Flow rate A v is the same at the pipe and nozzle.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For an incompressible liquid the equation of continuity is \(A_1v_1 = A_2v_2\). The area of a circular section is \(\dfrac{\pi d^2}{4}\).

Step 2: Apply:
\[ \frac{\pi d^2}{4}V = \frac{\pi d_1^2}{4}V_1 \]

Step 3: Solve:
\[ d_1^2 = d^2\frac{V}{V_1} \Rightarrow d_1 = d\sqrt{\frac{V}{V_1}} \]

Step 4: Check:
Option (A) matches. Since the nozzle makes the water faster (\(V_1 > V\)), \(d_1 < d\), which makes sense. Option (B) has the ratio inverted, and (C) and (D) forget the square root.

Final Answer:
Area times speed is constant, so d1 = d root of V over V1. \[ \boxed{\text{(A) }d\sqrt{\dfrac{V}{V_1}}} \]