A 10.0 cm column of air is trapped by a column of Hg, 4.00 cm long in capillary tube of uniform bore when the tube is held horizontally in a room at 1 atm. Length of the air column when the tube is held vertically with the open end up is

When open end up, the weight of mercury represents an increase in the pressure on the gas of 4.00 cm Hg = 4omm Hg

$\therefore \text{ Final pressure }=760+40=800\mathrm{mmHg}$

${\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2\Rightarrow {\mathrm{p}}_1\times \mathrm{A}\times {\mathrm{I}}_1={\mathrm{p}}_2\times \mathrm{A}\times {\mathrm{l}}_2$

where, A is the area of cross-section (which is constant) and l₁, l₂ length

$\therefore {\mathrm{p}}_1\times 1_1={\mathrm{p}}_2\times {\mathrm{I}}_2$

$760\times 10=800\times {\mathrm{l}}_2\Rightarrow {\mathrm{l}}_2=9.50\mathrm{cm}$