uniform vertical tube of 40cm is sealed at upper end is lowered into Mercury and length of air column become 35cm calculate depth of immersion of tube if atm is 77cm of mercury
V air after = 35/40 V original of air
so p air = 40/35 atm
40 - 35 = 5 cm of Hg in tube
5 cm hg = 5/77 atm of hg pressure
so total pressure at bottom of inside of tube = 40/35 atm + 5/77 atm
= 1.14 + 0.065 atm = 1.205 atm
outside the tube at the bottom
p = 1 atm + rho g h water
so
.205 atm = rho g * h
0.205 atm* 101325 N/m^2 / atm = 20772 pascals or N/m^2
so
20772 N/m^2 = 1000 Kg/m^3 * 9.81 m/s^2 * h
h = 0.126 meters
12.6 cm