JEE Mains · Physics · STD 11 - 9.2 surface tension
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 \(\rho\) and \(L\) is its latent heat of vaporization
- A \(\frac{{2T}}{{\rho L}}\)
- B \(\;\frac{{\rho L}}{T}\)
- C \(\sqrt {\;\frac{T}{{\rho L}}} \)
- D \(\;\frac{T}{{\rho L}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{2T}}{{\rho L}}\)
Step-by-step Solution
Detailed explanation
When radius is decrease by \(\Delta R,\) \(4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]\) \( \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]\)…
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