Question

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/m$^3$, the atmospheric pressure is 101 kPa, and the acceleration due to gravity is 9.8 m/s$^2$. The depth H is _________ m (round off to the nearest integer).

Hint:

The pressure at a given depth in a fluid is calculated using the hydrostatic pressure formula: \(P = \rho g H\), where \(H\) is the depth.

Answer 1

Correct Answer: 397

The absolute pressure is the sum of the atmospheric pressure and the pressure due to the water column. The pressure due to the water column is: \[ P_{\text{water}} = \rho g H, \] where:
- \(\rho = 1050 \, \text{kg/m}^3\) (density of sea water),
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity),
- \(H\) is the depth in meters.
The total pressure at depth \(H\) is the atmospheric pressure plus the pressure due to the water: \[ P_{\text{total}} = P_{\text{atm}} + \rho g H. \] Substituting the values: \[ 4.2 \, \text{MPa} = 0.101 \, \text{MPa} + (1050 \times 9.8) \times H. \] \[ 4.2 \times 10^6 = 101 \times 10^3 + (1050 \times 9.8) \times H. \] Solving for \(H\): \[ 4.2 \times 10^6 - 101 \times 10^3 = (1050 \times 9.8) \times H, \] \[ 4.099 \times 10^6 = 10290 \times H, \] \[ H = \frac{4.099 \times 10^6}{10290} \approx 398.5. \] Thus, the depth is approximately: \[ \boxed{397 \, \text{to} \, 399 \, \text{m}}. \]