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A tube of length h, closed at one end and open at the other, is filled with air at atmospheric pressure. The tube is pushed vertically downward into a mercury tank until the open end is submerged to a depth of 2h below the mercury surface. Mercury is observed to rise a height x inside the tube. If the atmospheric pressure equals h (in terms of mercury column height), assuming isothermal compression of the trapped air, which equation correctly relates x and h?

1. (3h - x)(h - x) = h²
2. (2h - x)(h - x) = h²
3. (2h - x)(h + x) = h²
4. (2h)(h - x) = h²

Correct answer: (2h - x)(h - x) = h²

Solution

Initially the air in the tube has length h at pressure h (atm = h mercury). After insertion, the open end is at depth 2h; mercury rises x, so air column = h - x. The pressure of mercury at the open end of the tube's air column is: atmospheric + depth of open end - x risen = h + (2h - x) at the bottom... Careful: mercury outside at the level of the closed (top) end is at depth (2h - (h - x)) = h + x below the surface, so pressure at air-mercury interface inside = h + (h + x)? Let me redo: closed end is at top. Tube length h. Open end inserted to depth 2h. Mercury rises x. Air column = h - x from closed (top) end. The mercury inside is at height x from the open end (bottom). The mercury surface is at depth 0 (reference). The open end of the tube is at depth 2h. The mercury interface inside the tube (top of mercury inside) is at depth 2h - x from the surface. So pressure at that interface = atmospheric + (2h - x) in mercury units = h + (2h - x). By Boyle's law: h*(h) = (h + 2h - x)*(h - x), giving h² = (3h - x)(h - x). This matches option A.

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