A bubble is at the bottom of the lake of depth h. As the bubble comes to sea level, its radius increases three times. If atmospheric pressure is equal to `l` metre of water column, then h is equal to

Options

A. `26 l`
B. `l`
C. `25 l`
D. `30 l`

Explanation

To solve the problem, we need to analyze the situation of a bubble rising from the bottom of a lake to the surface and how its volume changes according to Boyle's Law. Here’s a step-by-step solution:

### Step 1: Understanding the Pressure at the Bottom of the Lake
When the bubble is at the bottom of the lake, the pressure acting on it is the sum of the atmospheric pressure and the pressure due to the water column above it.

- **Atmospheric Pressure (P0)**: Given as equivalent to `L` meters of water column, we can express this as:
\[ P_0 = \rho g L \]
where \( \rho \) is the density of water and \( g \) is the acceleration due to gravity.

- **Pressure due to Depth (h)**: The pressure due to the depth \( h \) of the water is:
\[ P_h = \rho g h \]

- **Total Pressure at the Bottom (P1)**: Therefore, the total pressure at the bottom of the lake is:
\[ P_1 = P_0 + P_h = \rho g L + \rho g h = \rho g (L + h) \]

### Step 2: Understanding the Volume Change of the Bubble
When the bubble rises to the surface, its radius increases three times.

- **Initial Volume (V1)**: The volume of the bubble at the bottom of the lake (with radius \( r_1 \)) is:
\[ V_1 = \frac{4}{3} \pi r_1^3 \]

- **Final Volume (V2)**: When the radius increases three times (to \( 3r_1 \)), the new volume is:
\[ V_2 = \frac{4}{3} \pi (3r_1)^3 = \frac{4}{3} \pi \cdot 27 r_1^3 = 27 V_1 \]

### Step 3: Applying Boyle's Law
According to Boyle's Law, the product of pressure and volume for a gas remains constant if the temperature is constant. Thus:
\[ P_1 V_1 = P_2 V_2 \]

- **Pressure at the Surface (P2)**: At the surface of the lake, the pressure is just the atmospheric pressure:
\[ P_2 = P_0 = \rho g L \]

### Step 4: Setting Up the Equation
Substituting the values into Boyle's Law:
\[ (\rho g (L + h)) V_1 = (\rho g L)(27 V_1) \]

### Step 5: Simplifying the Equation
Since \( V_1 \) appears on both sides, we can cancel it out:
\[ \rho g (L + h) = \rho g L \cdot 27 \]

### Step 6: Solving for h
Dividing both sides by \( \rho g \) (assuming \( \rho g \neq 0 \)):
\[ L + h = 27 L \]

Rearranging gives:
\[ h = 27L - L = 26L \]

### Final Answer
Thus, the depth \( h \) is:
\[ h = 26L \]

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