Momentum and CollisionHard
Question
n drops of a liquid, each with surface energy E, joining to form a single drop
Options
A.some energy will be released in the process
B.some energy will be absorbed in the process
C.the energy released or absorbed will be E(n - n2/3)
D.the energy released or absorbed will be nE (22/3 - 1)
Solution
E = T 4πr2 ⇒ 
πR3 = n × 
⇒ n = 
⇒ R = n1/3r
Surface energy of big drop E′ = T4πR2 = T4πn2/3r2 = En2/3
Energy released = nE - E′ = nE-n2/3E = E(n-n2/3)
πR3 = n × ⇒ n = ⇒ R = n1/3rSurface energy of big drop E′ = T4πR2 = T4πn2/3r2 = En2/3
Energy released = nE - E′ = nE-n2/3E = E(n-n2/3)
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