Question

When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 5r 4 . Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension an

Answer 1

Let us assuming the Original radius of the Air bubble to be R.

This means it volume will be,

V = 4/3 πR³,

Now, After it has been rises to the Height, its volume becomes,

R' = 5R/4

V' = 4/3 π × (5R/4)³

V' = 4/3 π × 125R³/64

V' = V × 125/64

Now, Atmospheric Pressure (P) = 1.013 × 10⁵ Pa or 1 atm.

Now, At the depth, Pressure will be,

P' = P + hρg

P ' = 1.013 × 10⁵ + h × 1000 × 9.8

∴ P' = 101300 + 9800h

Now, Using Boyle's law,

PV = P'V'

∴ 1.013 × 10⁵ × V =  (101300 + 9800h) × V × 125/64

10130 × 64/125 =  (101300 + 9800h)

5186.56 = 101300 + 9800h

-9800h = 101300 - 5186.56

h = -9.8 m.

Thus, depth of the lake is 9.8 m.

Hope it helps.