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. \( \dfrac{V}{V_1} \)
• B. \( d\sqrt{\dfrac{V_1}{V}} \)
• C. \( d\sqrt{\dfrac{V}{V_1}} \)
• D. \( \dfrac{dV_1}{V} \)

Hint:

For incompressible fluids, area is inversely proportional to velocity.

Answer 1

The Correct Option is C

Step 1: Apply equation of continuity.
\[ A_1 V = A_2 V_1 \]

Step 2: Write areas in terms of diameters.
\[ \frac{\pi d^2}{4}V = \frac{\pi d_1^2}{4}V_1 \]

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