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

The correct option is 70 m.

To find the depth of the lake, we can apply Boyle's Law, assuming the temperature of the water in the lake remains constant as the bubble rises.

According to Boyle's Law:
$P_1{V}_1=P_2{V}_2$
where:
$P_1$ and $V_1$ are the pressure and volume of the bubble at the bottom of the lake, and
$P_2$ and $V_2$ are the pressure and volume of the bubble at the surface of the lake.

Let the radius of the bubble at the bottom of the lake be $r$. Therefore, its initial volume is:
$V_1=\frac{4}{3}\pi r^3$

As the bubble rises to the surface, its radius doubles ($2r$). Thus, its volume at the surface is:
$V_2=\frac{4}{3}\pi {\left(2r\right)}^3=8·\left(\frac{4}{3}\pi r^3\right)=8V_1$

Now, let us determine the pressures at both positions:
1. At the surface, the pressure is equal to the atmospheric pressure, which is given as equivalent to a water column of height $H$:
$P_2=\rho gH$
where $\rho$ is the density of water and $g$ is the acceleration due to gravity.

2. At the bottom of the lake at depth $d$, the total pressure is the sum of the atmospheric pressure and the pressure due to the water column of depth $d$:
$P_1=\rho gH+\rho gd=\rho g(H+d)$

Substituting these values into Boyle's Law:
$\rho g(H+d)V_1=\rho gH\left(8V_1\right)$

Dividing both sides by $\rho g{V}_1$:
$H+d=8H$
$d=7H$

The standard atmospheric pressure head for a column of water is approximately:
$H=10\text{m}$

Substituting this standard value of $H$ to find the depth:
$d=7\times 10\text{m}=70\text{m}$