JEE MAIN - Physics (2013 (Offline) - No. 22)
Assume that a drop of liquid evaporates by decreases 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.
$$\rho L/T$$
$$\sqrt {T/\rho L} $$
$$T/\rho L$$
$$2T/\rho L$$
Explanation
When radius is decrease by $$\Delta R,$$
$$4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]$$
$$ \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]$$
$$ \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,$$ [ $$\Delta R$$ is very small ]
$$ \Rightarrow R = {{2T} \over {\rho L}}$$
$$4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]$$
$$ \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]$$
$$ \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,$$ [ $$\Delta R$$ is very small ]
$$ \Rightarrow R = {{2T} \over {\rho L}}$$
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