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Question

(A) According to the equation of continuity for an incompressible fluid, the product of the cross-sectional area $(A)$ and the velocity $(v)$ remains

Vedclass · Accepted Answer

$d\sqrt{\frac{V}{V_1}}$

Vedclass · Answer

(A) According to the equation of continuity for an incompressible fluid, the product of the cross-sectional area $(A)$ and the velocity $(v)$ remains constant throughout the flow, i.e.,$A_1V_1 = A_2V_2$.Let the diameter of the pipe be $d$ and the velocity be $V$. The cross-sectional area of the pipe is $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Let the diameter of the nozzle be $d_1$ and the velocity of emerging water be $V_1$. The cross-sectional area of the nozzle is $A_1 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}$.Applying the equation of continuity: $A \cdot V = A_1 \cdot V_1$.Substituting the areas: $\frac{\pi d^2}{4} \cdot V = \frac{\pi d_1^2}{4} \cdot V_1$.Simplifying the equation: $d^2 V = d_1^2 V_1$.Solving for $d_1$: $d_1^2 = d^2 \cdot \frac{V}{V_1}$.Taking the square root on both sides: $d_1 = d \sqrt{\frac{V}{V_1}}$.

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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