When a large bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is

Options

A. H
B. 2H
C. 7H
D. 8H

Answer

To solve the problem step by step, we will analyze the situation involving the bubble rising in the lake and apply the principles of hydrostatics and the ideal gas law.

### Step 1: Understanding the Problem
We have a bubble at the bottom of a lake with an initial radius \( r \). As it rises to the surface, its radius doubles to \( 2r \). We need to find the depth \( H \) of the lake, given that the atmospheric pressure is equal to the pressure exerted by a column of water of height \( H \).

### Step 2: Define the Pressures
1. **At the surface of the lake**: The pressure \( P_0 \) (atmospheric pressure) is given by:
\[ P_0 = \rho g H \]
where \( \rho \) is the density of water, \( g \) is the acceleration due to gravity, and \( H \) is the height of the water column.

2. **At the bottom of the lake**: The pressure \( P' \) at the depth \( x \) is given by:
\[ P' = P_0 + \rho g x = \rho g H + \rho g x \]

### Step 3: Apply the Ideal Gas Law
Using the ideal gas law for isothermal processes, we have:
\[ P_1 V_1 = P_2 V_2 \]
where:
- \( P_1 = P_0 \)
- \( V_1 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \)
- \( P_2 = P' = \rho g H + \rho g x \)
- \( V_2 = \frac{4}{3} \pi r^3 \)

### Step 4: Set Up the Equation
Substituting the values into the equation:
\[ P_0 \cdot \frac{32}{3} \pi r^3 = (\rho g H + \rho g x) \cdot \frac{4}{3} \pi r^3 \]
We can cancel \( \frac{4}{3} \pi r^3 \) from both sides:
\[ P_0 \cdot 8 = \rho g H + \rho g x \]

### Step 5: Substitute \( P_0 \)
Substituting \( P_0 = \rho g H \) into the equation:
\[ \rho g H \cdot 8 = \rho g H + \rho g x \]

### Step 6: Simplify the Equation
Dividing through by \( \rho g \) (assuming \( \rho g \neq 0 \)):
\[ 8H = H + x \]
Rearranging gives:
\[ x = 8H - H = 7H \]

### Step 7: Conclusion
The depth of the lake \( H \) is equal to the distance \( x \) that the bubble rises:
\[ \text{Depth of the lake} = 7H \]

Thus, the depth of the lake is \( 7H \).