If D1 = 0.1 m, D2 = 0.2 m, and V1 = 3.0 m/s, what is V2?

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Multiple Choice

If D1 = 0.1 m, D2 = 0.2 m, and V1 = 3.0 m/s, what is V2?

Explanation:
The key idea is that for a steady, incompressible flow, the volume flow rate must be the same through both sections: A1 V1 = A2 V2. For a circular pipe, the cross-sectional area is proportional to the square of the diameter, since A ∝ D^2 (A = πD^2/4). Here the diameter doubles from 0.1 m to 0.2 m, so the area increases by a factor of 4. That means the velocity must decrease by the same factor to keep A1 V1 equal to A2 V2. So V2 = V1 × (A1/A2) = 3.0 m/s × (1/4) = 0.75 m/s. Since the diameter doubles, the velocity drops to one quarter of the original, which aligns with the 0.75 m/s result.

The key idea is that for a steady, incompressible flow, the volume flow rate must be the same through both sections: A1 V1 = A2 V2. For a circular pipe, the cross-sectional area is proportional to the square of the diameter, since A ∝ D^2 (A = πD^2/4). Here the diameter doubles from 0.1 m to 0.2 m, so the area increases by a factor of 4. That means the velocity must decrease by the same factor to keep A1 V1 equal to A2 V2. So V2 = V1 × (A1/A2) = 3.0 m/s × (1/4) = 0.75 m/s.

Since the diameter doubles, the velocity drops to one quarter of the original, which aligns with the 0.75 m/s result.

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