**Answer:**

a) h = 1990 m, b) the height we can reach increases
, c) h = 2,329 m

**Explanation:**

.a) For this fluid mechanics problem let's use the pressure ratio

P = P₀ + ρ g y

P-P₀ = ρ g y

Where rho is the fluid density where it is contained, P₀ is the atmospheric pressure.

It does not indicate that the maximum pressure change before exporting is

ΔP = 2.32 * (0.1 P₀)

Let's look for this pressure

ΔP = 2.32 0.1 1.01 10⁵

ΔP = 0.234 10⁵ Pa.

This is the maximum pressure difference, let's look for the height where we reach this value, the container has an internal pressure of 2.02 10 Pa, which we must subtract from the initial difference

ΔP = ΔP - 2.02 10⁰ = 0.234 10⁵5 - 2.02

ΔP = 23398 Pa

ΔP = ρ g h

h = ΔP /ρ g

h = 23398 /1.20 9.8

h = 1990 m

b) If air density decreases the height from 1,225 at sea level to 0.3629 to 11000m, the height we can reach increases

Density must be functioned with height

c) Now we introduce the contain in the ocean, the density of sea water is

ρ = 1025 kg / m³

h = 23398 / 1025 9.8

h = 2,329 m