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

The correct option is 4 : 9.

To find the ratio of the velocities of water at the entry and exit of the tube, we can use the equation of continuity for the steady flow of an incompressible fluid.

According to the equation of continuity, the mass flow rate of a fluid through any cross-section of a tube remains constant. For an incompressible fluid like water, this means the volume flow rate is conserved:
$A_1{v}_1=A_2{v}_2$
where:
• $A_1$ and $A_2$ are the cross-sectional areas at the entry and exit ends, respectively.
• $v_1$ and $v_2$ are the velocities of water at the entry and exit ends, respectively.

For a circular tube of radius $r$, the cross-sectional area is given by $A=\pi r^2$. Substituting this into the continuity equation:
$\pi r_1^2{v}_1=\pi r_2^2{v}_2$
Dividing both sides by $\pi$, we get:
$r_1^2{v}_1=r_2^2{v}_2$
Rearranging the terms to find the ratio of velocities ($v_1:v_2$):
$\frac{v_1}{v_2}={\left(\frac{r_2}{r_1}\right)}^2$

Given that the ratio of the radius at the entry ($r_1$) to the exit ($r_2$) is 3 : 2:
$\frac{r_1}{r_2}=\frac{3}{2}\Rightarrow \frac{r_2}{r_1}=\frac{2}{3}$

Substituting this ratio into our velocity equation:
$\frac{v_1}{v_2}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$
Thus, the ratio of velocities at entry and exit is 4 : 9.