A bubble of gas released at the bottom of a lake increases to four times its original volume when it reaches the surface. Assuming that atmospheric pressure is equivalent to the pressure exerted by a column of water 10 m high, what is the depth of the lake?

Options

A. 20 m
B. 10 m
C. 30 m
D. 40 m

Explanation

To solve the problem step-by-step, we will use the principles of gas laws and hydrostatics.

### Step 1: Understand the problem
We have a bubble of gas that expands to four times its original volume when it rises from the bottom of a lake to the surface. We need to find the depth of the lake, given that the atmospheric pressure at the surface is equivalent to the pressure exerted by a column of water 10 meters high.

### Step 2: Define the variables
Let:
- \( V_1 \) = original volume of the gas bubble at the bottom of the lake
- \( V_2 \) = volume of the gas bubble at the surface
- \( P_1 \) = pressure at the bottom of the lake
- \( P_2 \) = pressure at the surface of the lake

From the problem, we know that:
\[ V_2 = 4V_1 \]

### Step 3: Apply Boyle's Law
Since the temperature is constant, we can apply Boyle's Law, which states that:
\[ P_1 V_1 = P_2 V_2 \]

Substituting \( V_2 \) into the equation:
\[ P_1 V_1 = P_2 (4V_1) \]

### Step 4: Simplify the equation
We can cancel \( V_1 \) from both sides (assuming \( V_1 \neq 0 \)):
\[ P_1 = 4P_2 \]

### Step 5: Relate pressures to depth
Using the hydrostatic pressure formula, we know that the pressure at a depth \( h \) in a fluid is given by:
\[ P = \rho g h \]
where:
- \( \rho \) = density of water
- \( g \) = acceleration due to gravity
- \( h \) = depth of the lake

At the surface (where the pressure is \( P_2 \)):
\[ P_2 = \rho g (10 \, \text{m}) \]

### Step 6: Substitute \( P_2 \) into the equation for \( P_1 \)
Now substituting \( P_2 \) into the equation for \( P_1 \):
\[ P_1 = 4P_2 = 4(\rho g (10)) = 40 \rho g \]

### Step 7: Use the hydrostatic pressure difference
According to hydrostatics, the difference in pressure between the bottom and the surface is:
\[ P_1 - P_2 = \rho g h \]

Substituting the values of \( P_1 \) and \( P_2 \):
\[ 40 \rho g - 10 \rho g = \rho g h \]

### Step 8: Simplify to find \( h \)
This simplifies to:
\[ 30 \rho g = \rho g h \]

Now, we can cancel \( \rho g \) from both sides (assuming \( \rho g \neq 0 \)):
\[ 30 = h \]

### Conclusion
Thus, the depth of the lake \( h \) is:
\[ h = 30 \, \text{meters} \]

### Final Answer
The depth of the lake is **30 meters**.

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