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Q13.11 A metre-long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

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Initially, the pressure of the 15-cm-long air column is equal to the atmospheric pressure, P1 = 1 atm = 76 cm of Mercury

Let the cross-sectional area of the tube be x cm2

The initial volume of the air column, V1 = 15x cm3

Let's assume once the tube is held vertically, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P= 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V2 = (24 + y)x cm3

Since the temperature of the air column does not change,

\\P_{1}V_{1}=P_{2}V_{2}\\ 76\times 15x=y\times (24+y)x\\ 1140=y^{2}+24y\\ y^{2}+24y-1140=0

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it and the length of the air column becomes 47.8 cm

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