Question

Problem 18.87 Drops of an oil that has a mass density of 700
kg/m3 are released from a broken undersea oil pipe that is 1200 m
below the ocean surface. Use 1000 kg/m3 for the mass density of
ocean water. **Part A** If the drops have an average
diameter of 100 *μ*m, how long do they take to rise to the
surface? Hint: The drops quickly reach terminal speed and
experience a viscous drag force of magnitude
*F**d*wo=(0.0200Pa⋅s)*R**v*, where
*R* is the radius of the drops and *v* is their
speed. **Express your answer to two significant digits and
include the appropriate units. Part B** After a chemical
dispersant is applied to the oil to reduce the average drop
diameter to 2.0 *μ*m how long do the drops take to rise to
the surface? Express your answer with the appropriate units.

Answer #1

**Part A: **Given that the density of
oil drop _{o}
= 700 kg/m^{3}, density of water _{w}
= 1000 kg/m^{3}, the drag force is F_{d} = (0.02
Pa.s)Rv and radius of oil drop is R = (100/2) _{}m =
50 _{}m.
Now the velocity of oil drop can be calculated by balancing the
forces on oil drop as shown below,

Thus the time taken to reach to the surface is,

**Part B:** When the radius of oil drop is R =
(2.0/2)= 1.0 _{}m
then the velocity is,

Thus the time taken is,

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