The volume of the air bubble increases by 10%, when it rises from bottom to the surface of the water pond. If the temperature of the water is constant, find the depth of the pond. (1 atm = 10m of water column)

Explanation

Volume of the air bubble at the bottom of the pond `(V_(1))=V`
Volume of the air bubble at the surface of the pond
`V_(2)=Vxx110/100`
Pressure at the bottom of the pond `(P_(1))=(h+H)`cm of water column
where .h. is the depth of the pond and .H. is the atmospheric pressure
Pressure on the surface of the pond `(P_(2))` = H cm of water column
From Boyle.s law, `P_(1)V_(1)=P_(2)V_(2)` or
`(h+H)V=H(Vxx110/100)," "h+H=Hxx110/100`
`h=Hxx110/100-H=Hxx10/100=H/10`
`therefore` Depth of the pond = H/10 cm
= `1000/10=100cm=1m`