Question

A uniform tube closed at one end contains a pallet of mercury $10\, cm$ long. When the tube is kept vertically with the closed end upward, the length of the air column trapped is $20 \,cm$. Find the length (in $cm$ ) of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure $=75\, cm$ of mercury.

Answer 1

For air column,
$P_{1} =P-\frac{ mg }{ A } $
$\Rightarrow P _{1} =\rho g (75-10)=65\, cm $ of $ Hg$
$V _{1} =20\, A , T _{1}= T $
$P _{2} = P +\frac{ mg }{ A } $
$\Rightarrow P _{2} =\rho g (75+10)$
$P _{2} =85\, cm $ of $ Hg $
$V _{2} = hA , T _{2}= T $
$P _{1} V _{1} = P _{2} V _{2}$
$\Rightarrow 65 \times 20 \,A =85 \times Ha $
$h =15.294 \,cm$