An incompressible fluid flows steadily throug | vedclass

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

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$9V$

Vedclass · Answer

(D) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of the fluid remains constant at all points along the flow: $A_1 V_1 = A_2 V_2$.Here,the cross-sectional area of the pipe is $A = \pi r^2$.At point $A$,the radius is $R$,so the area $A_A = \pi R^2$ and velocity $V_A = V$.At point $B$,the radius is $\frac{R}{3}$,so the area $A_B = \pi \left(\frac{R}{3}ight)^2 = \frac{\pi R^2}{9}$.Using the continuity equation: $(\pi R^2) 	imes V = \left(\frac{\pi R^2}{9}ight) 	imes V_B$.Canceling $\pi R^2$ from both sides,we get: $V = \frac{V_B}{9}$.Therefore,the velocity at point $B$ is $V_B = 9V$.

An incompressible fluid flows steadily through a cylindrical pipe having radius $R$ at point $A$ and $\frac{R}{3}$ at point $B$ further along the direction of flow of liquid. If the velocity at point $A$ is $V$,then the velocity at point $B$ is:

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