A drop of oil 00320 Nm has a radius of 001 mm The drop is

A drop of oil ( = 0.0320 N/m) has a radius of 0.01 mm. The drop is located a distance of 2.55 m below the surface of fresh water. The atmospheric pressure above the water is 1.01 × 105 Pa.
a) What is the absolute pressure in the water at this depth?
b) Determine the absolute pressure inside the oil drop.

Solution

The formula that gives the P(N/m^2) pressure on an object submerged in a fluid is:

P = r * g * h

where

P(fluid) = 1.03*1000*9.81*2.55= 0.2576 *100000 N/m^2

P_total = P_atmosphere + P_fluid = (1.01 x 10⁵) + (0.2576 x = 10⁵ ) Pa = 1.2676 x 10 ² kPa (kilo Pascals)

absolute pressure inside the oil drop

Area of dropet= pi*r^2= 3.14 *0.01^2= 0.000314

density= 0.032 /0.000314= 101.91 kg/m^3

P(Oil) = 101.91*9.81*2.55= 2549 N/m^2