Question

The diameters of a bubble at the surface of a lake is $4mm$ and at the bottom of the lake is $1mm$. 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 $1gm/ml$ and $13.6gm/ml$ respectively]

• A. 120.5 m
• B. 200.4 m
• C. 184.5 m
• D. 153.03 m

Answer 1

Answer: (D)

153.03 m

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

So, we get that $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Case1- When the bubble is at surface.

The pressure is $P_1=1atm$

And the volume is $V_1=4\pi \times (4{)}^2$

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

The pressure will be $P2=1atm+\rho gh$ which will be $(1+\frac{1000\times 9.8\times h}{{10}^5})atm$

The volume is $V_2=4\pi 1^2$

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 $h=153.03m$