An incompressible fluid flows steadily through a cylindrical pipe which has radius $2r$ at point $A$ and radius $r$ at point $B$ further along the flow direction. If the velocity at point $A$ is $v$,what is its velocity at point $B$?

Which fundamental law forms the basis of the equation of continuity?

Water flows through a horizontal pipe of varying cross-section at a rate of $0.314 \,m^3 \,s^{-1}$. The velocity of water at a point where the radius of the pipe is $10 \,cm$ is (in $\,m/s$)

$A$ tube of length $L$ is shown in the figure. The radius of the cross-section at point $(1)$ is $2 \ cm$ and at point $(2)$ is $1 \ cm$. If the velocity of water entering at point $(1)$ is $2 \ m/s$,then the velocity of water leaving at point $(2)$ will be: (in $m/s$)

The cylindrical tube of a spray pump has radius $R,$ one end of which has $n$ fine holes,each of radius $r.$ If the speed of the liquid in the tube is $v,$ the speed of the ejection of the liquid through the holes is

$A$ shower head has $25$ circular openings, each with radius $1 \,mm$. The shower head is connected to a pipe with radius $2 \,cm$. If the speed of the water in the pipe is $25 \,cm/s$, what is its speed as it exits the shower head openings (in $\,m/s$)?