The pressure inside a water droplet is higher than the surroundings, this is related to surface tension. If the surface tension of a spherical droplet is 73mN/m and has a diameter of 2mm, then what is the gauge pressure rise (Pg) between droplet and the surroundings (to within 10%)?

Gauge pressure rise = 292 Pa

Explanation:

We are given;

Diameter of the water droplet is d = 2mm = 0.002m

Radius = 0.002/2 = 0.001

Surface tension; σ = 73 mNm = 73 x 10^(-3)m

So we have to find the pressure rise.

Now, pressure rise is given by the formula ;

ΔP = P_inside - P_atmosphere = 4σ/r

Where r is radius.

Thus, plugging in the relevant values, we have;

ΔP = (4 x 73 x 10^(-3))/0.001 = 292 pa

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west