Diving into the sea, the scuba diver took a balloon with him. The air pressure in the ball was 100 kPa.

Diving into the sea, the scuba diver took a balloon with him. The air pressure in the ball was 100 kPa. To what depth did the scuba diver dive if the air pressure in the ball increased to 250 kPa?

The hydrostatic pressure of water on a ball can be expressed in the formula:
P = ρ * g * h, where P is the pressure created by water on the ball (P = P2 – P1 = 250 kPa – 100 kPa = 150 kPa = 150 * 10 ^ 3 Pa), ρ is the density of seawater (we take ρ = 1025 kg / m ^ 3), g is the acceleration of gravity (we take g = 10 m / s ^ 2), h is the depth of the scuba diver’s immersion (m).
Let us express and calculate the depth of the scuba diver’s dive:
h = P / (ρ * g) = (150 * 10 ^ 3) / (1025 * 10) = 14.63 m.
Answer: The diving depth of the scuba diver is 14.63 m.