We would apply the equation of continuity. Recall, it states that in case of flow of an incompressible fluid, the rate of flow at any time is constant at all points. The volume of fluid entering a section of flow in a period of time equals the volume of fluid that leaves at that same period of time. This means that
A1V1 = A2V2
where
A1 and A2 are the areas of the inlet and outlet
V1 and V2 are the velocities at inlet and outlet
AV is the volume flow rate in m^3/s
From the information given,
radius of inlet = 1 cm
Recall,
1 cm = 0.01 m
thus, radius of inlet = 0.01
Also,
Area of circle = pi x radius^2
Area of inlet = 3.14 x 0.01^2 = 0.000314
V1 = 3.84
A1V1 = 0.000314 x 3.84 = 0.00121
inner radius of outlet = 0.2 cm
Converting to meters, inner radius = 0.2 x 0.01 = 0.002
Area = 3.14 x 0.002^2 = 0.00001256
A2V2 = 0.00001256V2
Thus,
0.00121 = 0.00001256V2
V2 = 0.00121/0.00001256 = 96.33
The outlet velocity is 96.33 m/s