Q.

A large number of droplets, each of radius a, coalesce, to form a bigger drop of radius b. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is (σ surface tension, ρ  density)

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a

σρ1a−1b1/2

b

2σρ1a−1b1/2

c

6σρ1a−1b1/2

d

3σρ1a−1b1/2

answer is C.

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Detailed Solution

Let say number of droplets are n. Energy released =n×4πa2−4πb2σ Now, n×43πa3=43πb3  or  n=b3a3Therefore, energy released is=b3a3×4πa2−4πb2σ=4πb2ba−1σ Now, K.E of big drop=energy released⇒1243πb3ρv2=4πb2ba−1σ⇒v=6σρ1a−1b1/2
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