Question

A bubble rises from the bottom of a pond to its surface by increasing its radius by 3 times its value when it was at the bottom. Calculate the depth of the pond

Answer 1

Answer ⇒ 260 m.

Explanation ⇒

Let the depth of the pond be h m.

Radius of the bubble at depth = r

Pressure = P + hρg

Also, Radius of the bubble at surface = 3r

Pressure = P  [P = Atmospheric pressure.]

V = 4/3 π (3r)³

In this case,

Pressure is inversely proportional to volume at constant temperature, that is an condition of Boyle's law.

∴ P₁V₁ = P₂V₂

∴ (P + hρg) × 4/3 πr³ = P × 4/3 π(3r)³

∴ (P + hρg) = P × 27

∴ 26P = hρg

∴ 26 × 10⁵ = h × 1000 × 10

∴ h = 260 m.

Hence, depth is 260 m.

Hope it helps.

Answer 2

Since, we know that the volume of the bubble is inversely proportional to its pressure inside.

              Pressure * volume will be constant at same temperature.

Now when the bubble raises to the top, the pressure created on the bubble due to the water in that level will decrease leading to its expansion.

The bubble's volume will be 27 times at surface with respect to volume at the bottom of lake.  So at the bottom the pressure will be 27 times that at the surface.  Since, the radius is thrice so the volume will be 3^3 times.

 

Pressure at surface  =  1 atm = water density * H * g.

Pressure at bottom  = 27 * 1 atm = 27 atm  = density * 27 * H * g

                                  = density * depth * g + 1 atm at the surface.  

Which on solving we get the depth of the lake to be = 27 * H - 1 * H = 26 * H.