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 tube.$A_{1} v_{1} = A_{2} v_{2}$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 $\frac{v_{1}}{v_{2}} = \left(\frac{r_{2}}{r_{1}}\right)^{2}$.Given the ratio of radii $\frac{r_{1}}{r_{2}} = \frac{3}{2}$,we have $\frac{r_{2}}{r_{1}} = \frac{2}{3}$.Substituting this into the equation,$\frac{v_{1}}{v_{2}} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$.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 ends of the pipe is $3:2$. Then the ratio of velocities at the entry and exit of the liquid is

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