Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radiu - Physics MCQ JEE Main 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

1. Let the radius of the drop be R . Surface energy U_{s} = T (4 π R^{2}) .2. If the radius decreases by d R, the decrease in surface energy is d U_{s} = T (8 π R d R) .3. This energy is used to evaporate the mass d m = ρ (4 π R^{2} d R) .4. Heat required for evaporation d Q = d m \cdot L = ρ (4 π R^{2} d R) L .5. For this to be possible from surface energy alone: d U_{s} = d Q . T (8 π R d R) = ρ (4 π R^{2} d R) L 2 T = ρR L \Rightarrow R = \frac{2 T}{ρ L} .

Try it yourself before checking the explanation above.

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

\sqrt{T/\rho L}

T/\rho L

2T/\rho L

\rho L/T

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