Question

Scuba divers are admonished not to rise faster than their air bubbles when rising to the surface. This rule helps them avoid the rapid pressure changes that cause the bends. Air bubbles of 1.0 mm radius are rising from a scuba diver to the surface of the sea. Assume a water temperature of 20 'C.

a) If the viscosity of the water is 1.0 x 10-3 Pas, what is the terminal velocity of the bubbles?

b) What is the largest rate of pressure change tolerable for the diver according to this rule?

Answer 1

As per the question,

The radius of the bubble is $(r)=1.0mm =1\times 10^{-3}m$

water temperature $(T)=20 ^\circ c$

Viscosity of water $(\eta) =1.0\times 10^{-3}Pas$

We know that density of water at $20 ^\circ c$ is $(\rho)=998 kg/m^3$

density of air bubble at $20 ^\circ c$ is$(\sigma)=1.194 kg/m^3$

gravitational acceleration $(g)=9.8 m/sec^2$

a) We know that terminal velocity $(v)=\dfrac{2r^2(\rho -\sigma)g}{9\eta}=\dfrac{2\times (1\times 10^{-3})^2(998-1.194)\times 9.8}{9\times1.0\times 10^{-3}} m/sec$

$=2170.0\times 10^{-3} m/sec = 2.170 m/sec$

b)

When we fixed the radius and increase the pressure,there will be a force per unit area exerted on the surface of the drop which will result in decreasing the volume of drop,and when the volume decreases the velocity will decrease