The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to h meter of mercury column and relative density of mercury is then the depth of lake in metre is -

Answer

To solve the problem, we need to apply the principles of gas laws and hydrostatics. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- The air bubble rises from the bottom of a lake to the surface.
- The temperature of the air bubble remains constant during this process.
- The diameter of the bubble doubles as it rises, which means its radius also doubles.
- The pressure at the surface of the lake is given as \( P_0 = h \) meters of mercury (Hg) column, where the relative density of mercury is given.

### Step 2: Use the Ideal Gas Law
Since the temperature is constant, we can use Boyle's Law, which states that for a given mass of gas 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 of the lake.
- \( V_2 \) is the volume of the bubble at the surface.

### Step 3: Calculate the Volumes
Let the initial radius of the bubble be \( r \). Therefore, the initial volume \( V_1 \) is:
\[ V_1 = \frac{4}{3} \pi r^3 \]
When the bubble rises, its radius doubles to \( 2r \), so the new volume \( V_2 \) is:
\[ V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \cdot \frac{4}{3} \pi r^3 = 8 V_1 \]

### Step 4: Set Up the Pressure Equation
Now, substituting the volumes into Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \implies P_1 \left(\frac{4}{3} \pi r^3\right) = P_2 \left(8 \cdot \frac{4}{3} \pi r^3\right) \]
This simplifies to:
\[ P_1 = 8 P_2 \]

### Step 5: Relate Pressures to Depth
At the surface of the lake, the pressure \( P_2 \) is equal to the atmospheric pressure \( P_0 \):
\[ P_2 = P_0 = \rho_{Hg} g h \]
At the bottom of the lake, the pressure \( P_1 \) can be expressed as:
\[ P_1 = P_0 + \rho_{lw} g y \]
Where \( y \) is the depth of the lake and \( \rho_{lw} \) is the density of lake water.

### Step 6: Substitute and Solve for Depth
Substituting \( P_2 \) into the equation \( P_1 = 8 P_2 \):
\[ P_1 = 8 \rho_{Hg} g h \]
Now equate the two expressions for \( P_1 \):
\[ 8 \rho_{Hg} g h = \rho_{lw} g y \]
Cancel \( g \) from both sides:
\[ 8 \rho_{Hg} h = \rho_{lw} y \]
Now, solving for \( y \):
\[ y = \frac{8 \rho_{Hg} h}{\rho_{lw}} \]

### Final Answer
Thus, the depth of the lake \( y \) in meters is:
\[ y = \frac{8 \rho_{Hg} h}{\rho_{lw}} \]