A basketball is pressurized to a gauge pressure of PG = 75 kPa when at the surfa

Question

A basketball is pressurized to a gauge pressure of PG = 75 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 = 1000 kg/m3. Assume the ball does not change in mass, temperature, or volume as it is submerged. Write an equation for the pressure difference P between the inside and outside of the ball when it is submerged a distance y below the surface of the water. 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. At what depth, in meters, would the pressure difference between the inside and outside of the ball be zero if the ball were submerged in mercury ( = 13,500 kg/m3) instead of in water?

Explanation / Answer

PG = 75 kpa

Gauge pressure at a depth 'y' below the liquid surface, Ppool = gd

So, pressure difference inside and outside the ball,

P = PG - gy

For P = 0,

PG - gy = 0

=> y = PG/g

For water, y = (75 * 103) / (1000 * 9.81) = 7.64 m

For mercury, y = (75 * 103) / (13500 * 9.81) = 0.566 m

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