Question

When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is:-
A. H
B. 2H
C. 7H
D. 8H

Answer 1

The pressure at the surface of the lake will be ${{P}_{1}}=\rho gH$ and the pressure at the bottom will be ${{P}_{2}}={{P}_{1}}+\rho gd$. Then use the Boyle’s law which says that ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$. Calculate the volumes of the bubble at the bottom and at the surface and substitute in the equation to find d.

Formula used: $P=\rho gh$
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$

Complete step by step solution:
The pressure at the bottom of the lake will be more than the pressure at the surface. Therefor
Let the pressure at the bottom of the lake be ${{P}_{2}}$ and that at the top be ${{P}_{1}}$.
It is given that the atmospheric pressure is equal to the pressure of column of water height H. The pressure due to this water column will be equal to $\rho gH$, where $\rho$ is the density of water and g is acceleration due to gravity.
Therefore,
${{P}_{1}}=\rho gH$.
The pressure at the bottom of the lake will be equal to ${{P}_{2}}={{P}_{1}}+\rho gd$, where d is the depth of the lake.
Substitute the value of ${{P}_{1}}$.
$\Rightarrow {{P}_{2}}=\rho gH+\rho gd$.
Now, let us assume that the whole process takes place at constant temperature. Therefore, we can apply Boule’s law.
According to Boyle’s law, at constant temperature the product of pressure of the gas and its volume are constant.
Let the volumes of the bubble when its pressures are ${{P}_{1}}$ and ${{P}_{2}}$ be ${{V}_{1}}$ and ${{V}_{2}}$.
The radius of the bubble at the bottom be r. Its radius will be 2r at the surface.
Therefore,
${{V}_{1}}=\dfrac{4}{3}\pi {{(2r)}^{3}}$ and ${{V}_{2}}=\dfrac{4}{3}\pi {{r}^{3}}$.
According to Boyle’s law, ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$ …. (i).
Substitute the values in equation (i)
$\Rightarrow \rho gH\left( \dfrac{4}{3}\pi {{(2r)}^{3}} \right)=(\rho gH+\rho gd)\left( \dfrac{4}{3}\pi {{r}^{3}} \right)$
$\Rightarrow H{{(2r)}^{3}}=(H+d){{r}^{3}}$
$\Rightarrow 8H=(H+d)$
$\Rightarrow d=7H$
Therefore, the depth of the lake is equal to 7H.

So, the correct answer is “Option C”.

Note: Let us understand more clearly about what is happening to the bubble when it comes up to the surface.
A bubble is some volume of gas covered by water. Since the density of a gas is less than that of water, it rises up. As it rises up, the pressure exerted by the water reduces. Hence, its volume increases, which means its radius will increase.