. If n number of little droplets of water each of radius 'r' and surface tension 's' coalesce to form bigger drop then pressure inside bigger drop is ?
(po is atmosphere pressure).
Answers
The pressure inside the bigger drop is V = √6S(R - r/ Rr)
Explanation:
Let n be the number of little droplets which coalesce to form single drop. Then,
Volume of n little droplets = Volume of single drop
or n × 4/3 πr³ = 4/3 πR³ or nr³ = R³
Decrease in surface area = n × 4πr² - 4πR²
= 4π [ nr² - R²] = 4π [ nr³/r - R²]
= 4π [ R³/r - R²] = 4πR³ [ 1/r - 1/R]
{ ∴ nr³ = R³ }
The energy released,
E = Surface tension × decrease in surface area
= 4πSR³ [ 1/r - 1/R]
The mass of bigger drop,
M = 4/3 πR³ × 1 = 4/3πR³
E = 4/3 πSR³ × 3 [ 1/r - 1/R]
E = 3SM [ 1/r - 1/R] { ∴ M = 4/3 πR³ }
KE of bigger droplets = Energy released
K.E = 1/2 MV² = 3SM [ 1/r - 1/R]
V = √6S(R - r/ Rr)
Hence the pressure inside the bigger drop is V = √6S(R - r/ Rr)
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