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An ideal liquid flows through two pipes: AC is inclined and BD is horizontal, connected by two vertical tubes (heights h1 and h2). The flow is streamlined in both pipes, and no fluid flows through the vertical connections AB and CD. If the speeds of the liquid at A, B and C are 2 m/s, 4 m/s and 4 m/s respectively, what is the speed at D?
- 4 m/s
- sqrt(14) m/s
- sqrt(28) m/s
- None
Correct answer: sqrt(28) m/s
Solution
Apply Bernoulli along the vertical tube AB (connecting the inclined pipe at A to the horizontal pipe at B) and along CD. Because the pressure differences along the two vertical tubes are linked by Bernoulli's equation applied to the whole streamline pattern, the speeds satisfy vD² = vB² + vC² - vA². Numerically vD² = 16 + 16 - 4 = 28, so vD = sqrt(28) m/s.
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