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 7H.

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

Step 1: Define the pressure and volume at the bottom of the lake
Let the depth of the lake be $d$ and the initial radius of the bubble at the bottom be $r$.
The atmospheric pressure is equal to the pressure of a water column of height $H$:
$P_0=\rho gH$
where $\rho$ is the density of water and $g$ is the acceleration due to gravity.

The total pressure at the bottom of the lake ($P_1$) is the sum of atmospheric pressure and the pressure due to the water column of depth $d$:
$P_1=P_0+\rho gd=\rho gH+\rho gd=\rho g(H+d)$

The initial volume of the bubble ($V_1$) is:
$V_1=\frac{4}{3}\pi r^3$

Step 2: Define the pressure and volume at the surface of the lake
At the surface, the pressure ($P_2$) is equal to the atmospheric pressure:
$P_2=P_0=\rho gH$

Since the radius of the bubble doubles at the surface, the new radius is $2r$. The final volume ($V_2$) becomes:
$V_2=\frac{4}{3}\pi {\left(2r\right)}^3=8·\left(\frac{4}{3}\pi r^3\right)=8V_1$

Step 3: Apply Boyle's Law
For an isothermal process:
$P_1{V}_1=P_2{V}_2$

Substitute the values of pressure and volume into the equation:
$\rho g(H+d)·V_1=\rho gH·\left(8V_1\right)$

Divide both sides by $\rho g{V}_1$ to simplify:
$H+d=8H$

Solving for the depth of the lake $d$:
$d=8H-H$
$d=7H$

Thus, the depth of the lake is 7H.