Water is descending in a vertical pipe of diameter 7 cm and open to the atmosphere. At a lower point the water flows into a smaller pipe of diameter .84 cm. At a certain instant the depth of the water just above the narrowing point is 30 cm and the water is moving at 135 cm/s. How fast will water be going as it exits a small hole just below the narrowing point?

To find the speed of the water as it exits the small hole below the narrowing point, we can use the principle of conservation of mass, which states that the mass flow rate of a fluid is constant at any point in a continuous flow.

To solve this problem, we can use the equation of continuity, which relates the speed of the fluid to the cross-sectional area of the pipe.

The equation of continuity can be written as:

A_1 * v_1 = A_2 * v_2

where A_1 and A_2 are the cross-sectional areas of the pipe at points 1 and 2 respectively, and v_1 and v_2 are the corresponding speeds of the water at those points.

Given:
Diameter of the vertical pipe (A_1) = 7 cm
Diameter of the smaller pipe (A_2) = 0.84 cm
Depth just above the narrowing point = 30 cm (this would be the height of the water column above point 1)

To find the area of the pipes, we need to calculate their radii:

Radius of the vertical pipe (r_1) = Diameter / 2 = 7 / 2 = 3.5 cm = 0.035 m
Radius of the smaller pipe (r_2) = Diameter / 2 = 0.84 / 2 = 0.42 cm = 0.0042 m

The cross-sectional areas of the pipes can be calculated using the formula for the area of a circle:

A_1 = π * r_1^2
A_2 = π * r_2^2

Now, substitute the values into the equation of continuity:

A_1 * v_1 = A_2 * v_2

Let's consider the point just above the narrowing point as point 1, and the point just below the narrowing point (where the small hole is) as point 2.

We have:
v_1 = 135 cm/s (as given)
A_1 = π * (0.035)^2
A_2 = π * (0.0042)^2

Simplify the equation:

π * (0.035)^2 * 135 = π * (0.0042)^2 * v_2

Now, solve for v_2:

v_2 = (π * (0.035)^2 * 135) / (π * (0.0042)^2)

Simplify further:

v_2 = (0.035)^2 * 135 / (0.0042)^2

Evaluate the expression:

v_2 ≈ 10,036.59 cm/s

So, the water will be moving at a speed of approximately 10,036.59 cm/s as it exits the small hole just below the narrowing point.

Note: If you want to convert the answer to meters per second (m/s), you need to divide the value by 100:
v_2 ≈ 100.37 m/s