A) \[\frac{\rho L}{T}\]
B) \[\sqrt{\frac{T}{\rho L}}\]
C) \[\frac{T}{\rho L}\]
D) \[\frac{2T}{\rho L}\]
Correct Answer: D
Solution :
Surface energy of the drop \[=(4\pi {{R}^{2}})T\] Let\[(\delta m)\]mass evaporate. \[(\delta m)L=8\pi R.\text{ }\delta RT\] \[m=\rho \frac{4}{3}\pi {{R}^{3}}\]\[\Rightarrow \]\[\frac{\delta m}{\delta R}=4\pi {{R}^{2}}\rho \] \[\therefore \] \[\rho 4\pi {{R}^{2}}L=8\pi R.T\] \[\Rightarrow \] \[R=\frac{2T}{L\rho }\]
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