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A water drop is divided into 8 equal droplets. The pressure difference between inner and outer side of big drop will be

A water drop is divided into 8 equal droplets. The pressure difference between inner and outer side of big drop will be

Grade:11

2 Answers

Vikas TU
14149 Points
7 years ago
For small 8 droplets individuals the Pressure difference is:
Pi – Po = 2T/r 
where r is the radii of the bubble.
For one big drop formed by all the 8 equal droplets the pressure difference would be:
Pi’ – Po’ = 2T/R
Now from Volume conservation we get,
V1 + V2 + -------+V8 = V
8V8 = V
8*(4pir^3/3) = (4piR^3/3)
R = 2r.......(1)
Put eqn. (1) in big drop pressure differnece, we get,
Pi’ – Po’ = 2T/2r = > T/r
where T is the Tension generated.
christy
29 Points
5 years ago
Volume of the big drop = Volume of 8 droplets  i.e., 4/3\piR3 = 8 × 4/3\pir3 
  • \therefore r = R/2
  1. For smaller drop \DeltaPs = 2T/r = 2T/R/2 = 4T/R
  2. For bigger drop \DeltaPb = 2T/R = ½ \DeltaPs
  3.  

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