Water Pressure Calculation at Sea Bottom

The pressure exerted by a column of fluid is known as hydrostatic pressure. This pressure depends on the density of the fluid, the acceleration due to gravity, and the depth of the fluid column.

The formula for hydrostatic pressure \(P\) is given by:

\[P = \rho g h\]

Where:

• \(\rho\) (rho) is the density of the fluid.
• \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
• \(h\) is the depth of the fluid column.

Identifying Given Values

In this problem, we are given:

• Depth of the sea bottom, \(h = 6500 \, \text{m}\).
• Specific gravity of sea water = 1.03.

We know that the density of pure water is approximately \(1000 \, \text{kg/m}^3\). The specific gravity of a substance is the ratio of its density to the density of a reference substance (usually water for liquids).

So, the density of sea water \(\rho_{\text{seawater}}\) can be calculated as:

\[\rho_{\text{seawater}} = \text{Specific gravity} \times \text{Density of pure water}\]

\[\rho_{\text{seawater}} = 1.03 \times 1000 \, \text{kg/m}^3 = 1030 \, \text{kg/m}^3\]

Step-by-Step Pressure Calculation

Now we can substitute the values of \(\rho\), \(g\), and \(h\) into the pressure formula. We'll use \(g = 9.8 \, \text{m/s}^2\).

\[P = \rho g h\]

\[P = (1030 \, \text{kg/m}^3) \times (9.8 \, \text{m/s}^2) \times (6500 \, \text{m})\]

Calculating the product:

\[P = 1030 \times 9.8 \times 6500 \, \text{Pa}\]

\[P = 65321000 \, \text{Pa}\]

The pressure is usually expressed in Pascals (Pa). The options are given in MegaPascals (MPa). To convert Pascals to MegaPascals, we divide by \(10^6\).

\[1 \, \text{MPa} = 10^6 \, \text{Pa}\]

So, \(65321000 \, \text{Pa}\) is equal to:

\[P = \frac{65321000}{10^6} \, \text{MPa}\]

\[P = 65.321 \, \text{MPa}\]

Comparing with Options

Let's compare the calculated pressure with the given options:

• 70 MPa
• 65 MPa
• 75 MPa
• 60 MPa

The calculated pressure of approximately \(65.321 \, \text{MPa}\) is closest to 65 MPa.

Therefore, the water pressure on the sea bottom at a depth of 6500 m, assuming the specific gravity of sea water is 1.03, is approximately 65 MPa.