Question

The ratio of the diameters of certain air bubbles at the bottom and at the surface 1:2. What is the depth of the lake? (1 atmosphere = 10 m depth of water)
(A) 80 m
(B) 60 m
(C) 35 m
(D) 70 m

Answer 1

Hint: In this question we have to apply the PV formula for the two radius that are given. Once we put the values in the formula we can compare the two values of two radii of the air bubble. This will lead to the development of an equation. Once the equation is formed we have to put the value of the 1 atmosphere that is mentioned to obtain the answer.

Complete step by step answer:
Let us consider that the depth of the lake is H m.
The relation between the diameters of the air bubble at the bottom and at the surface is given as ${r_2} = 2{r_1}$.
Here the radius of the air bubble at the bottom is ${r_1}$ and the radius of the air bubble at the surface is ${r_2}$.
Now we have to apply the formula of ${P_1}{V_1} = {P_2}{V_2}$.
Now we have to put the values mentioned in the question in the above equation. So this becomes:
$
   \Rightarrow ({P_0} + H\rho g)\dfrac{4}{3}\pi r_1^3 = {P_0}\dfrac{4}{3}\pi r_2^3 \\
   \Rightarrow ({P_0} + H\rho g)\dfrac{4}{3}\pi r_1^3 = {P_0}{(2{r_1})^3} \\
   \Rightarrow {P_0} + H\rho g = 8{P_0} \\
   \Rightarrow {P_0} = \dfrac{H}{7}\rho g \\
 $
 It is given that 1 atmosphere = 10 m depth of water. Hence we have to put the value in the equation to obtain that:
$
  \dfrac{H}{7} = 10m \\
   \Rightarrow H = 70m \\
 $
Hence the depth of the lake is 70 m.

So the correct answer is option D.

Note: For a better understanding we should know that if an air bubble which is supposed to have a radius of r rises from the surface of suppose a lake, in this case, then the pressure of the atmosphere will be equal to the height of about 10 m from that of the water column.
We should also remember that the shape of an air bubble is considered to be a sphere. So in that case while the radius of the air bubble is taken into consideration, we should be implementing the formula of a sphere.