Question

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

• A. 20.2 cm
• B. 23.8 cm
• C. 25.6 cm
• D. 32.1 cm

Answer 1

Answer: (B)

In the horizontal position. $(P_{atm}=76cm$ of Hg)

$P_1{v}_1=(76cm)(15cm)A$
$A$ is cross-sectional area of the tube.
In the vertical position if we assume that $h$ cm of $Hg$ flows out then,
$P_2{V}_2=[76-(76-h)]\cdot (15+9+h)A$
$=h(24+h)A$

From Boyle’s law, $P_1{V}_1=P_2{V}_2$
$\Rightarrow 76\times 15=h(24+h)$
$\Rightarrow h^2+24h-1140=0$
$\Rightarrow h=\frac{-24\pm \sqrt{(24{)}^2+4(1140)}}{2}$
$=23.8$ or $-47.8$
$h$ can't be negative, so $23.8cm$ of $Hg$ flows out.