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  4. A meter 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. When the tube is held vertically with the open end at the bottom, the length of mercury that flows out is

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Q. A meter 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. When the tube is held vertically with the open end at the bottom, the length of mercury that flows out is

Kinetic Theory

Solution:

In the horizontal position. $(P_{atm} = 76 \,cm$ of Hg)
image
$P_1v_1 = (76\, cm) (15 \,cm)A$
$A$ is cross-sectional area of the tube.
In the vertical position if we assume that $h$ cm of $Hg$ flows out then,
$P_2V_2 = [76 - (76 - h)] \cdot (15 + 9 + h)A$
$ = h(24 + h)A$
image
From Boyle’s law, $P_1V_1 = P_2V_2$
$\Rightarrow 76 \times 15 = h(24 + h)$
$\Rightarrow h^2 + 24h - 1140 = 0$
$\Rightarrow h = \frac{-24\pm \sqrt{(24)^2 + 4(1140)}}{2}$
$ = 23. 8$ or $- 47.8$
$h$ can't be negative, so $23.8\,cm$ of $Hg$ flows out.