The human lungs can function satisfactorily up to a limit where the pressure difference between the outside and inside of the lungs is one -twentieth of an atmosphere . If a diver uses a snorkel for breathing , how for below the water can she swim ? Assume the diver is in salt water whose density is `1025 kg//m^(3)` .

Options

A. 0.25 m
B. 0.35 m
C. 0.50 m
D. 0.20 m

Answer

To solve the problem of how deep a diver can swim while using a snorkel, we need to analyze the pressure differences involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the pressure difference limit
The problem states that the human lungs can function satisfactorily up to a pressure difference of one-twentieth of an atmosphere.

\[ \Delta P = \frac{1}{20} \times P_{\text{atm}} \]

Where \( P_{\text{atm}} \) is the atmospheric pressure, approximately \( 1.013 \times 10^5 \, \text{Pa} \).

### Step 2: Calculate the maximum pressure difference in Pascals
Substituting the value of atmospheric pressure into the equation gives:

\[ \Delta P = \frac{1}{20} \times 1.013 \times 10^5 \, \text{Pa} = \frac{1.013 \times 10^5}{20} \, \text{Pa} \]

Calculating this:

\[ \Delta P = 5065 \, \text{Pa} \]

### Step 3: Relate pressure difference to depth
As the diver goes deeper into the water, the pressure increases due to the weight of the water above. The pressure at a depth \( h \) in a fluid is given by:

\[ P = \rho g h \]

Where:
- \( \rho \) is the density of the salt water (given as \( 1025 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the depth in meters.

### Step 4: Set up the equation for pressure difference
The pressure difference that the diver experiences as they go deeper should equal the maximum pressure difference that their lungs can handle:

\[ \rho g h = 5065 \, \text{Pa} \]

### Step 5: Substitute known values into the equation
Substituting the values of \( \rho \) and \( g \):

\[ 1025 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times h = 5065 \, \text{Pa} \]

### Step 6: Solve for \( h \)
Rearranging the equation to solve for \( h \):

\[ h = \frac{5065 \, \text{Pa}}{1025 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2} \]

Calculating the denominator:

\[ 1025 \times 9.8 \approx 10045 \, \text{kg/(m}^2\text{s}^2\text{)} \]

Now substituting back into the equation for \( h \):

\[ h = \frac{5065}{10045} \approx 0.504 \, \text{m} \]

### Step 7: Final result
Thus, the diver can swim approximately:

\[ h \approx 0.5 \, \text{m} \]

### Conclusion
The maximum depth the diver can swim while using a snorkel is approximately **0.5 meters**.

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