Question

A basketball is pressurized to a gauge pressure of PG = 55 kPa when at the surface of a swimming pool. (Patm = 101 kPa). The ball is then submerged in the pool of water which has a density rho = 1000 kg/m³. Assume the ball does not change in mass, temperature, or volume as it is submerged. Solve the pressure equation for the depth (in meters) at which the pressure difference between the inside and outside of the ball will become zero. At this depth the pressure inside the basketball is the same as the pressure outside the ball. y=?

Answer 1

To find the depth at which the pressure inside a basketball is equal to the external water pressure, you set the total pressure to the sum of the gauge pressure and atmospheric pressure and solve for depth, which results in a depth of 5.61 meters.

Step-by-step explanation:

We need to calculate the depth at which the pressure inside a basketball equals the external pressure in water. According to Pascal's Principle, the total pressure at a certain depth in a fluid is the sum of the atmospheric pressure and the pressure due to the column of water above that depth. We have the gauge pressure PG = 55 kPa, the atmospheric pressure Patm = 101 kPa, the density of water ρ = 1000 kg/m³, and the acceleration due to gravity g = 9.81 m/s².

The pressure in water at a depth of h meters is given by P = Patm + ρgh. To find the depth at which the pressure difference between inside and outside of the basketball becomes zero, we set P equal to the sum of PG and Patm and solve for h:

P = Patm + PG = Patm + ρgh

Therefore, h = (Patm + PG - Patm) / (ρg) = PG / (ρg).

Plugging in the given values:

h = (55,000 Pa) / (1000 kg/m³ * 9.81 m/s²) = 5.61 m

Thus, the basketball would experience equal pressure inside and out at a depth of 5.61 meters.