Answer
The speed of the water through the point where the pipe tapers is $3.2~m/s$
Work Step by Step
We can find the speed $v_2$ that the water moves through the point where the pipe tapers:
$A_2~v_2 = A_1~v_1$
$v_2 = \frac{A_1~v_1}{A_2}$
$v_2 = \frac{\pi~r_1^2~v_1}{\pi~r_2^2}$
$v_2 = \frac{r_1^2~v_1}{r_2^2}$
$v_2 = \frac{(0.010~m)^2~(0.20~m/s)}{(0.0025~m)^2}$
$v_2 = 3.2~m/s$
The speed of the water through the point where the pipe tapers is $3.2~m/s$.
