Question

When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes $\dfrac{5r}{4}$. Taking the atmosphere pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):
A. 10.5 m
B. 8.7 m
C. 11.2 m
D. 9.5 m

Answer 1

- Hint: Try to understand the concept of pressure at a depth under water. Find the expression for both volume and pressure at the bottom and at the surface of the lake. Then at constant temperature try to find the solution with the help of P-V relation.

Complete step-by-step solution:
At the bottom of the lake the radius of the air bubble is r.
At the surface of the lake the radius of the bubble will be $\dfrac{5r}{4}$.
Atmospheric pressure is given as, ${{P}_{atm}}=10m$ height of water column.
Let the depth of the water column is h meter.
Volume of the bubble at the bottom of the lake, ${{V}_{1}}=\dfrac{4}{3}\pi {{r}^{3}}=V$
Volume of the bubble at the surface is, ${{V}_{2}}=\dfrac{4}{3}\pi {{\left( \dfrac{5r}{4} \right)}^{3}}=\dfrac{125}{64}V$
Now the atmospheric pressure at the surface is, ${{P}_{2}}=10m$
Atmospheric pressure at the bottom will be, $\begin{aligned} & {{P}_{1}}={{P}_{atm}}+\text{ pressure at the bottom of water of height h} \\\ & {{P}_{1}}=(10+h)m \\\ \end{aligned}$
We have represented the pressure in m because the atmospheric pressure is given terms of meter.
Now. The temperature of the system is constant. So, we can write,
$PV\to \text{constant}$
So,
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$
Putting the values of the quantities in the above equation, we get,
$\begin{aligned} & \left( 10+h \right)V=10\times \dfrac{125}{64}V \\\ & \left( 10+h \right)=10\times \dfrac{125}{64} \\\ & h=10\times \dfrac{61}{64} \\\ & h=9.53m \\\ & h\simeq 9.5m \\\ \end{aligned}$
So, the depth of the lake is 9.5 m.
The correct option is (D)

Note: Hydrostatic pressure can be defined as the pressure that is exerted by a fluid at equilibrium to an object at a depth due to the gravitational force.
It can be given as,
$p=\rho gh$
Where g is the acceleration due to gravity, h is the height and $\rho$ is the density of the fluid.In the solution, we took a pressure unit as m because in the question we have the pressure in the unit of depth of water.