An air bubble rises from the bottom of a deep lake. The radius of the air bubble near the surface is r its radius was (Atmospheric pressure = 10 m of water
and assume temperature is constant)
(A) 12 at depth 30 m
(B) 2 at depth 70 m
(C) 13 at depth 140 m
(D) 13 is depth 160 m
Answers
Answer:
When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 45r. Taking the atmospheric pressure to be equal to 10m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):
Radius of the air bubble will be 24 at the depth of 70m
Explanation:
P1 = Pressure under water at required depth = 7 atm
P2 = Pressure at surface of water = 1 atm.
V1 = Volume of bubble at original depth.
V2 = Volume of bubble at surface of water.
If original radius of bubble is r, then the new radius will be 2r.
When temperature is constant, we know that P1V1 = P2V2
P1(4/3πr³) = P2(4/3)π(2r)³
P1(4/3)πr³ = P2(4/3)π.8r³
P1 = 8P2
We know that P2 is 1 atm.
P1 = 7atm = P1 + hρg
1atm + h(ρg) = 8 atm
h(ρg) =7 atm = P
Therefore h = 70m as 10 m column of water produces 1atm of pressure.
Option B is the answer