A large number of liquid drops each of radius 'a' coalesce to form a single spherical drop of radius 'b'. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of big drop will be:
A
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
A
Energy released = ( 
A) × σ { σ = surface tension}
Let us say n no. of small drops coalesced.
⇒ n. 
= 
⇒ b = a. n 1/3 ⇒ n = 
A = 4 π b 2 – n.4 π a 2 {this is –ve, hence energy is released}
= 4 π a 2 (n 2/3 – n)
⇒ U = 4 π a 2 T (n – n 2/3 ). = 4 π a 2 T 
This U converts to K.E.
Hence 
V 2 = 4 π a 2 T 
V = 
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