Question

An apparatus produces water droplets of diameter 70 m. If the coefficient of surface tension of water in air is 0.07 n/m, the excess pressure in these droplets is kpa

Answer 1

Explanation:

pressure = 2*surface tension/radius

= 2*70/0.07 = 2*10*10^2

= 2000 N/m^2

Answer 2

GIVEN:

•   water droplets of diameter 70 m.
•   the coefficient of surface tension of water in air is 0.07N/m

TO FIND:

•   excess pressure in these droplets

SOLUTION:

   S = 0.07 N/m

    r = 70 m

Excess pressure inside the water droplet  ΔP = 2S/r

ΔP =  2×0.07 / 70

     = 0.14/70

     =0.002 N/m²

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