Consider a submarine cruising 30 m below the free surface of seawater whose density is 1025 kg/m3. What is the increase in the pressure exerted on the submarine when it dives to a depth of 110 m below the free surface?

804.42 kPa

Explanation:

Data provided in the question:

Initial depth of the submarine, d = 30 m

Density of water, ρ = 1025 kg/m³

Final depth of the submarine, D = 110 m

Now,

Pressure due to water = ρgh

here,

g is the acceleration due to gravity = 9.8 m/s²

h = height of water above

therefore,

Increase in pressure = ρgD - ρgd

= ρg (D - d)

= 1025 * 9.81 * (110 - 30)

= 804,420 Pa

= 804,420 * 10⁻³ kPa

= 804.42 kPa