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Q.

Assume that liquid drop evaporates by decreasing in 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 vapourisation of liquid. What should be the minimum radius of the drop for this to be possible

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a

$\frac{T}{ρL}$

b

$\frac{3 T}{ρL}$

c

$\frac{ρL}{T}$

d

$\frac{2 T}{ρL}$

answer is D.

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Detailed Solution

$\begin{array}{l} For evaportation to begin with the the rate of loss of surface energy \\ must be equal to the rate of energy required for evaporation \\ w = 4 {πR}^{2} T; \\ d (Lρ \frac{4}{3} {πR}^{3}) = d (4 {πR}^{2} T) \\ \Rightarrow L ρ \frac{4}{3} π (3 R^{2}) = 4 π T (2 R); \\ R = \frac{2 T}{ρ L} \end{array}$

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