Question

An expansible balloon filled with air is floating on the top surface of a lake, with 2/3 of its volume is submerged in water. How deep (in m) should it be sunk in water without changing the temperature of air in it, in order that it is just in equilibrium, neither sinking nor rising further? Height of water barometer $=10m$

Answer 1

Le the volume of bubble when floating with $2/3$ volume submerged be $V$.

Thus $mg=\frac{2}{3}V{\rho }_{w}g$

When it is completely submerged and is in equilibrium,

$mg=v{\rho }_{w}g$

$⟹v=\frac{2}{3}V$

Since the temperature of the air inside the water remains the same, from Boyle's Law,

$PV=pv$

where p,v are the pressure inside and volume of the bubble inside water at equilibrium.

Thus $p=\frac{3}{2}P$

$P=10mofwater$

Thus $p=15mofwater$

Thus the depth inside the water=$15-10=5mofwater$