A horizontal pipe has a part with a diameter 1/3 of the rest of the pipe. If v is the speed of the fluid in

the wider section, then the speed in the narrower section is:

To find the speed of the fluid in the narrower section of the pipe, we can use the principle of continuity. According to this principle, the mass flow rate of an incompressible fluid remains constant as it flows through a pipe of varying cross-sectional area.

The cross-sectional area of a pipe is directly proportional to the square of its diameter. So, if the diameter of the wider section of the pipe is D, then the diameter of the narrower section is (1/3)D.

Using the principle of continuity, we can write the equation:

A1 * v1 = A2 * v2

where A1 and v1 are the cross-sectional area and velocity of the wider section, and A2 and v2 are the cross-sectional area and velocity of the narrower section.

We know that the ratio of the areas is equal to the square of the ratio of the diameters:

A1/A2 = (D/((1/3)D))^2 = (3/1)^2 = 9

So, A1 = 9A2

Substituting this into the continuity equation, we get:

9A2 * v1 = A2 * v2

Simplifying, we find:

v2 = 9v1

Therefore, the speed in the narrower section of the pipe is 9 times the speed in the wider section.