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Q.

Assume that liquid drop evaporates partially by decreasing its surface energy so that its temperature remains unchanged. The surface tension of liquid drop is T, density of liquid is ρ and L is latent heat of vapouration of liquid. What should be the minimum radius of the drop for this to be possible?

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a

T/ρL

b

3T/ρL

c

ρL/T

d

2T/ρL

answer is D.

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Detailed Solution

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Let r be the minimum radius of a drop of liquid. Therefore its surface energy, U = 4πr2T . If a thin layer of liquid of thickness t evaporates, then necessary latent heat to be supplied = ( 4πr2t).ρ.L.
Decrease in surface energy = 4πr2T - 4π(r - t)2.T 8πrt.T(r>>t)
Temperature of the drop will not change if, (8πrt)T = 4πr2t.ρ.L. ⇒ r = 2T/ρ.L

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