Answer
(a) less than
(b) $\Delta P=-\frac{15}{2}\rho v^2$
Work Step by Step
(a) We know that the pressure in the narrow region is less than the initial pressure because the velocity of the water is greater at this point.
(b) We can find the required change in pressure as follows:
$P_2-P_1=\frac{1}{2}\rho [v_1^2-(4v_1)^2]$
We plug in the known values to obtain:
$P_2-P_1=\frac{1}{2}\rho v^2(1-16)$
$P_2-P_1=\Delta P=-\frac{15}{2}\rho v^2$
