To solve the problem step by step, we will use Boyle's Law and the relationship between pressure, volume, and radius of a bubble.
### Step 1: Understanding the problem
We have a bubble at the bottom of a lake at a depth \( h \). As the bubble rises to the surface, its radius increases three times. We need to find the depth \( h \) given that atmospheric pressure is equal to \( I \) meters of water column.
### Step 2: Applying Boyle's Law
Boyle's Law states that at constant temperature, the product of pressure and volume is constant:
\[
P_1 V_1 = P_2 V_2
\]
Where:
- \( P_1 \) is the pressure at the bottom of the lake,
- \( V_1 \) is the volume of the bubble at the bottom,
- \( P_2 \) is the pressure at the surface,
- \( V_2 \) is the volume of the bubble at the surface.
### Step 3: Determine the pressures
1. **Pressure at the surface (\( P_2 \))**: This is equal to the atmospheric pressure, which is given as \( P_0 = \rho g I \) (where \( \rho \) is the density of water, \( g \) is the acceleration due to gravity, and \( I \) is the height of the water column).
2. **Pressure at the bottom of the lake (\( P_1 \))**: This is the atmospheric pressure plus the pressure due to the water column above the bubble:
\[
P_1 = P_0 + \rho g h = \rho g I + \rho g h = \rho g (I + h)
\]
### Step 4: Calculate the volumes
The volume of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
- At the bottom, let the radius be \( R \). Thus, the volume \( V_1 \) is:
\[
V_1 = \frac{4}{3} \pi R^3
\]
- At the surface, the radius increases to \( 3R \). Thus, the volume \( V_2 \) is:
\[
V_2 = \frac{4}{3} \pi (3R)^3 = \frac{4}{3} \pi (27R^3) = 27 \left(\frac{4}{3} \pi R^3\right) = 27 V_1
\]
### Step 5: Substitute into Boyle's Law
Now substituting into Boyle's Law:
\[
P_1 V_1 = P_2 V_2
\]
Substituting the expressions for \( P_1 \), \( P_2 \), \( V_1 \), and \( V_2 \):
\[
(\rho g (I + h)) V_1 = (\rho g I) (27 V_1)
\]
We can cancel \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[
\rho g (I + h) = 27 \rho g I
\]
### Step 6: Simplify the equation
Dividing both sides by \( \rho g \) (assuming \( \rho g \neq 0 \)):
\[
I + h = 27 I
\]
Now, rearranging gives:
\[
h = 27 I - I = 26 I
\]
### Conclusion
Thus, the depth \( h \) is:
\[
h = 26 I
\]
