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Properties-of-fluid

The pressure outside droplet of water of diameter 0.04 mm is 10.32 N/`(cm)^2`. Calculate pressure within droplet if surface tension of water is 0.0725 N/m.

SOLUTION:

Given,

Surface Tension,`sigma=0.0725`N/m

Diameter of air  bubble,`d=0.04`mm

Since an air bubble has only one surface,

`P*(pi*d^2)/4=sigma*pi*d`


or,`P=(4*sigma)/d=7250`N/`m^2`


or, `P=0.725`N/`(cm)^2`


Thus, the pressure inside the droplet = P + pressure outside the droplet (atmospheric)

or, `P_(i n)=0.725+10.32`


or, `P_(i n)=11.045` N/`(cm)^2`

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Properties of fluid

Fluid-Mechanics