1
Two water drops each of radius 'r' coalesce to form a bigger drop. If 'T' is the surface tension, the surface energy released in this process is :
- 4 r2 T [2 - 21/3 ]
- 4 r2 T[1 + 2]
- 4 r2 T [2 - 22/3 ]
- 4 r2 T[ 2 - 1]
Answer
(3) 4 r2 T [2 - 22/3 ]
Solution
Volume is conserved: 43 R3 = 2 43 r3
R = 21/3 r
Initial surface area = 2 4 r2
Final surface area = 4 R2 = 4 r2 22/3
Energy released = T (decrease in area)
= 4 r2 T [2 - 22/3 ]
Answer: (3)
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