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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
one year ago

Vikas TU
6869 Points
							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/RNow from Volume conservation we get,V1 + V2 + -------+V8 = V8V8 = V8*(4pir^3/3) = (4piR^3/3)R = 2r.......(1)Put eqn. (1) in big drop pressure differnece, we get,Pi’ – Po’ = 2T/2r = > T/rwhere T is the Tension generated.

one year ago
christy
29 Points
							Volume of the big drop = Volume of 8 droplets  i.e., 4/3$\pi$R3 = 8 × 4/3$\pi$r3 	$\therefore$ r = R/2	For smaller drop $\Delta$Ps = 2T/r = 2T/R/2 = 4T/R	For bigger drop $\Delta$Pb = 2T/R = ½ $\Delta$Ps

3 months ago
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### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions