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Question

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of flow remains const

Vedclass · Accepted Answer

$d \sqrt{\frac{V}{V_{1}}}$

Vedclass · Answer

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of flow remains constant at any point along the pipe: $A_1 V_1 = A_2 V_2$.Here,$A_1$ is the area of the pipe,$V$ is the velocity in the pipe,$A_2$ is the area of the nozzle,and $V_1$ is the velocity at the nozzle.Since the area $A = \pi (d/2)^2$,we have $A \propto d^2$.Substituting this into the continuity equation: $d^2 V = d_2^2 V_1$.Solving for the nozzle diameter $d_2$: $d_2^2 = d^2 \frac{V}{V_1}$.Taking the square root on both sides: $d_2 = d \sqrt{\frac{V}{V_1}}$.

Water flows through a horizontal pipe at a speed $V$. The internal diameter of the pipe is $d$. If the water is emerging from a nozzle at a speed $V_{1}$,then the diameter of the nozzle is

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