Water is flowing through a tube of non-unifor | vedclass

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

Vedclass · Accepted Answer

$4:9$

Vedclass · Answer

(A) 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 pipe.$A_1 v_1 = A_2 v_2$Here,$A_1$ and $A_2$ are the cross-sectional areas at the entry and exit,and $v_1$ and $v_2$ are the velocities at the entry and exit,respectively.Since the cross-section is circular,$A = \pi r^2$,where $r$ is the radius.Therefore,$\pi r_1^2 v_1 = \pi r_2^2 v_2$.This simplifies to $\frac{v_1}{v_2} = \frac{r_2^2}{r_1^2} = \left( \frac{r_2}{r_1} \right)^2$.Given the ratio of radii at entry and exit is $r_1 : r_2 = 3 : 2$,we have $\frac{r_2}{r_1} = \frac{2}{3}$.Substituting this value: $\frac{v_1}{v_2} = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.Thus,the ratio of velocities at the entry and exit is $4:9$.

Water is flowing through a tube of non-uniform cross-section. The ratio of the radius at the entry and exit end of the pipe is $3:2$. Then the ratio of velocities at the entry and exit of the liquid is:

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