An air bubble of volume `1 cm^(3)` rises from the bottom of a lake `40 m` deep to the surface at a temperature of `12^(@)C`.The atmospheric pressure is `1 xx 10^(5)`Pa,the density of water is `1000 kg/m^(3)`and `g=10m/s^(2).There is no difference of the temperature of water at the depth of 40m and on the surface .The volume of air bubble when it reaches the surface will be:

Options

A. `3 cm^(3)`
B. `2 cm^(3)`
C. `4 cm^(3)`
D. `5 cm^(3)`

Answer

To solve the problem of the air bubble rising from the bottom of a lake, we can use the principle of conservation of pressure and volume, which is expressed by Boyle's Law. Let's break down the solution step by step.

### Step 1: Identify the given values
- Initial volume of the bubble, \( V_1 = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \)
- Depth of the lake, \( h = 40 \, \text{m} \)
- Atmospheric pressure, \( P_0 = 1 \times 10^5 \, \text{Pa} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the pressure at the depth of 40 m
The pressure at a depth in a fluid is given by the formula:
\[ P_1 = P_0 + \rho g h \]
Substituting the values:
\[ P_1 = 1 \times 10^5 \, \text{Pa} + (1000 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(40 \, \text{m}) \]
Calculating the second term:
\[ P_1 = 1 \times 10^5 \, \text{Pa} + 400000 \, \text{Pa} = 500000 \, \text{Pa} \]

### Step 3: Use Boyle's Law
According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_2 \) is the pressure at the surface (atmospheric pressure), which is \( P_0 = 1 \times 10^5 \, \text{Pa} \).
- \( V_2 \) is the volume of the bubble at the surface.

### Step 4: Rearranging the equation to find \( V_2 \)
Rearranging the equation gives:
\[ V_2 = \frac{P_1 V_1}{P_2} \]
Substituting the values we have:
\[ V_2 = \frac{(500000 \, \text{Pa})(1 \times 10^{-6} \, \text{m}^3)}{1 \times 10^5 \, \text{Pa}} \]

### Step 5: Calculate \( V_2 \)
Calculating \( V_2 \):
\[ V_2 = \frac{500000 \times 1 \times 10^{-6}}{1 \times 10^5} = \frac{500000 \times 10^{-6}}{10^5} = \frac{500000}{100000000} = 5 \times 10^{-3} \, \text{m}^3 \]
Converting \( V_2 \) back to cm³:
\[ V_2 = 5 \, \text{cm}^3 \]

### Final Answer
The volume of the air bubble when it reaches the surface will be:
\[ \boxed{5 \, \text{cm}^3} \]