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Question

# 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
6Tρ[1a1b]
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B
4Tρ[1a1b]
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C
8Tρ[1a1b]
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D
5Tρ[1a1b]
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Solution

## The correct option is A √6Tρ[1a−1b]Volume of n drops of radiusa= Volume of drop with radius b⇒n×43πa3=43πb3⇒na3=b3Work done =Surface Tension×Change in Area=TΔA ⇒12mv2=T[4πna2−4πb2]⇒12mv2=4πT[na2−b2]⇒12ρ×43πb3×v2=4πT[na2−b2]⇒v=√6Tρ[na2b3−b2b3]⇒v=√6Tρ[1a−1b]

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