Question

The diameters of a bubble at the surface of
lake is 4 mm and at the bottom of the lake is
1 mm. If atmospheric pressure is 1 atm and
the temperature of the lake-water and the
atmosphere are equal, what is the depth of the
lake ?
(The density of lake-water and Hg are
1 gm/ml and 13.6 gm/ml respectively. Also
neglect the contribution of pressure due to
surface tension)​

Answer 1

Answer:

153.03 meter.

Explanation:

We know from the Charles Law that the pressure is directly proportional to the temperature at a given volume.

So, we get that P1V1/T1 = P2V2/T2

Case1- When the bubble is a t surface.

The pressure is P1 = 1 atm.

And the volume is V1 = 4π*(4)^2 .

Case2- When the bubble is at the bottom then we will get that.

The pressure will be P2 = 1 atm + pgh which will be 1 + (1000*9.8*h)/10^5 atm.

The volume is V2 = 4π*1^2.

Since, the temperature is same given in the question, so on substituting the values in the equation we will get.

16 = 1+ 0.098h

On solving we will get h = 153.03 meter.