Working mechanism of straws and siphon

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A siphon operates by utilizing gravity and differential pressure, where the exit end is lower than the liquid source, allowing liquid to flow continuously once air is removed from the pipe. The process involves gravity pulling the liquid down, which maintains the differential pressure necessary for flow. In contrast, a straw relies on creating a temporary vacuum by sucking, which generates a pressure difference that allows liquid to rise. When sucking stops, the pressure difference is lost, halting the flow of liquid. Understanding these mechanisms clarifies the distinct functioning of siphons and straws.
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Homework Statement
What's the difference in the working mechanism between a straw and a siphon? Why do you only need to suck on the siphon once for continuous flow of the liquid whereas if you stop sucking on a straw then liquid stops flowing?
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Is it because for a siphon since the end where the liquid exits is lower than the container level so when the pipe is sucked all the air in the pipe is gone and so liquid moves due to differential pressure, and gravity pulls the liquid down, removing the air in the pipe as the liquid comes down, so this differential pressure is maintained so the liquid just keeps flowing down? Whereas for a straw when you stop sucking on the straw there is no longer a differential pressure to keep the liquid flowing?
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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