Question

4) (idea of a crush depth) A bottle of wine falls overboard. The cork has a radius of 1.5 cm. To move the cork you'd need a force of 356.98 Newtons.

(a) At what depth would the water pressure crush the cork into the bottle?

(b) Suppose it is a bottle of champagne so that its contents are under a pressure of 73.456 PSI (gauge) from all the 'bubbly'. At what depth during its descent into the ocean would it succumb?

Answer 1

Part a:

Pressure required to crush the cork = Force /Area = F / (π * r²) = 356.98 / (π * 0.015²) = 505023 N/m²

As we know that pressure of water at depth h will be "ρgh" where ρ = 1000 kg/m³

ρgh = 505023

1000 * 9.8 * h = 505023

h = 505023 /1000/9.8 = 51.53 m

PART B:

It is given that Champagne will have a pressure of 73.456 psi = 73.456 * 6894.7 N/m² = 506461 N/m²

Water pressure must be able to overcome this pressure as well as put extra pressure of 505023 N/m² s that cork can be crushed

ρgh = 505023 + 506461 = 1011484

h = 1011484 / 9.8/1000 = 103.2 m