Question

Calculate the depth in the ocean at which the pressure is three times atmospheric pressure. Atmospheric pressure is 1.013 × 105 Pa. The acceleration of gravity is 9.81 m/s^2 and the density of sea water is 1025 kg/m^3. Answer in units of m.

Answer 1

To calculate the depth in the ocean at which the pressure is three times atmospheric pressure, we can use the following formula for pressure due to a column of fluid:

\[
P = P_0 + \rho g h
\]

where:
- \( P \) is the total pressure at depth \( h \),
- \( P_0 \) is the atmospheric pressure,
- \( \rho \) is the density of the fluid (sea water in this case),
- \( g \) is the acceleration due to gravity, and
- \( h \) is the depth.

We need to find the depth at which \( P = 3 P_0 \).

1. Calculate \( 3 P_0 \):

\[
3 P_0 = 3 \times 1.013 \times 10^5 \, \text{Pa} = 3.039 \times 10^5 \, \text{Pa}
\]

2. Set up the equation:

\[
3.039 \times 10^5 = 1.013 \times 10^5 + \rho g h
\]

3. Now, rearrange it to find \( h \):

\[
\rho g h = 3.039 \times 10^5 - 1.013 \times 10^5
\]

\[
\rho g h = 2.026 \times 10^5 \, \text{Pa}
\]

4. Substitute the values for \( \rho \) and \( g \):

\[
1025 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times h = 2.026 \times 10^5
\]

5. Calculate \( \rho g \):

\[
1025 \times 9.81 = 10039.25 \,\text{N/m}^3
\]

6. Substitute this back into the equation:

\[
10039.25 \, h = 2.026 \times 10^5
\]

7. Solve for \( h \):

\[
h = \frac{2.026 \times 10^5}{10039.25} \approx 20.17 \, \text{m}
\]

Thus, the depth in the ocean at which the pressure is three times atmospheric pressure is approximately:

\[
\boxed{20.17 \, \text{m}}
\]