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: Flow rate:
Volume per second $Q = \dfrac{\pi d^2}{4}V$ must be the same everywhere in the pipe.

Step 2: Ratio of diameters:
$\dfrac{d_1^2}{d^2} = \dfrac{V}{V_1}$.

Step 3: Result:
$d_1 = d\sqrt{V/V_1}$, option (A).

Final Answer:
The nozzle diameter is d root of V over V1. \[ \boxed{\text{(A) }d\sqrt{\dfrac{V}{V_1}}} \]