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 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 `= 10 m`.

Answer

`5m`

Explanation

Let `V` be the total volume of the ballon and `M` be its mass. When it is floating on the surface of lake.
Weight of water displaced `=` weight of ballon
`implies 2/3vrhog=Mg` or `M=2/3Vrho`
Now let the depth of immersion be `h`, so that the new pressure is `(10+h)` of water. Applying Boyle' law
`P_(1)V_(1)=P_(2)V_(2)`
`implies10V=(10+h)V'impliesV'=(10V)/((10+h))`
let the balloon be at equilibrium at a depth `h` from the surface of the lake, then at equilibrium
`Mg=V'rhog`
`impliesm=V'rho`
`:. h=5m`