Water flowing through a pipe suddenly comes to a section of pipe where the pipe diameter decreases to 86% of its previous value. if the speed of the water in the larger section of the pipe was 66 m/s, what is its speed in this smaller section?

In all the sections of the pipe, volumetric flow rate is the same. Therefore,
Q1 = Q2

Q = VA, V = velocity, A = Cross-sectional area.

Then,
V1A1 = V2A2

But,
A = πD^2/4
Then,
V1·πD1^2/4 = V2·πD2^2/4
V1·D1^2 = V2·D2^2

But
D2/D1 = 86% = 0.86, or D1/D2 = 1/0.86
Therefore,
V2 = (V1D1^2)/D2^2 = V1·(D1/D2)^2 = 66*(1/0.86)^2 = 89.24 m/s

The speed in the smaller section is 89.24 m/s.

2. they have negligible mass.