Question

Assuming isothermal conditions, if the ratio of radius of a spherical bubble at the bottom and at the surface of the lake is $1:2$, then find the depth of the lake.
$($Take, $1\text{atm}=10\text{m}$ column of water, $g=10{\text{ms}}^{-2})$

• A. $45\text{m}$
• B. $90\text{m}$
• C. $70\text{m}$
• D. $30\text{m}$

Answer 1

The correct option is C $70\text{m}$

In isothermal condition,

$P_1{V}_1=P_2{V}_2......\left(1\right)$

$P_1$ and $V_1$ are the pressure and volume at the surface, respectively.

$P_2$ and $V_2$ are the pressure and volume at the bottom, respectively.

Let $P_0$ be the atmospheric pressure at the surface.

$(P_0=1\text{atm}={10}^5\text{Pa})$

So,
$P_1=P_0,\&P_2=P_0+\rho gh$

Here, density of water, $\rho =1000{\text{kg/m}}^3$

Also given that $R_1:R_2=2:1$

We know that,
$\frac{V_1}{V_2}=\frac{R_1^3}{R_2^3}=\frac{8}{1}$

$\therefore V_1=8V_2$

Putting in equation $\left(1\right)$, we get

$P_0\left(8V_2\right)=(P_0+\rho gh)\left(V_2\right)$

$7P_0=\rho gh$

$\Rightarrow h=\frac{7\times {10}^5}{{10}^3\times 10}(\because \rho ={10}^3{\text{kgm}}^{-3})$

$\therefore h=70\text{m}$

Hence, option $\left(\text{C}\right)$ is the correct answer.