A pressure measurement device fitted on the surface of a submarine, located at a depth H below the surface of an ocean, reads an absolute pressure of 4.2 MPa. The density of sea water is 1050 kg/m³ , the atmospheric pressure is 101 kPa, and the acceleration due to gravity is 9.8 m/s² . The depth H is ___________ m (round off to the nearest integer).

The correct answer is 398.347 (or 398 when rounded to the nearest integer).

1. Identify the given parameters:
Based on the problem description and the provided diagram showing a submarine at depth H, we have:
- Absolute pressure at depth H: $P=4.2\text{MPa}=4.2\times {10}^6\text{Pa}$
- Atmospheric pressure at the ocean surface: $P_{\text{atm}}=101\text{kPa}=101\times {10}^3\text{Pa}$
- Density of seawater: ${\rho }_{\text{sea}}=1050{\text{kg/m}}^3$
- Acceleration due to gravity: $g=9.8{\text{m/s}}^2$

2. Hydrostatic pressure equation:
The absolute pressure at a depth H below the liquid surface is the sum of the atmospheric pressure at the surface and the pressure exerted by the fluid column above the submarine:
$P=P_{\text{atm}}+{\rho }_{\text{sea}}gH$

3. Substitute the values and solve for H:
Substituting the given values into the equation yields:
$4.2\times {10}^6=101\times {10}^3+1050\times 9.8\times H$

Simplify the terms:
$\mathrm{4,200,000}=\mathrm{101,000}+\mathrm{10,290}H$

Isolate the term containing H:
$\mathrm{10,290}H=\mathrm{4,200,000}-\mathrm{101,000}$
$\mathrm{10,290}H=\mathrm{4,099,000}$

Solve for H:
$H=\frac{\mathrm{4,099,000}}{\mathrm{10,290}}\approx 398.3479\text{m}$

Rounding this calculation to three decimal places matches the provided correct answer of 398.347 m (which rounds to 398 m when rounded to the nearest integer).