Question

A uniform tube closed at one end, contains a pellet 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 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

Case 1 : Net pressure on air in volume

$V=P_{atm}-h\rho g$

= $75\times {\rho }_{Hg}\times g-{10}_{{\rho }_{Hg}}\times g$

= ${\rho }_{Hg}\times g\times 65$

Case 2: Net pressure on air in volume 'V'

V = $P_{atm}+\rho Hg\times g\times h$

$P_1{V}_1=P_2{V}_2$

= ${\rho }_{Hg}\times g\times 65\times A\times 20$

= $({\rho }_{Hg}\times g\times 75+{\rho }_{Hg}\times g\times 10)A\times h$

$\Rightarrow$ 65+20 = 85 h

$\Rightarrow h=\frac{65\times 20}{85}$ = 15.2 cm

= 15 cm