question_answer

If the radius of the opening of a dropper is \[r=5\times {{10}^{-4}}m\], density of liquid \[\rho ={{10}^{3}}kg\,{{m}^{-3}}\],\[g=10\,m{{s}^{-2}}\]and surface tension \[T=0.11N\,{{m}^{-1}}\], the radius of the drop when the drop detaches from the dropper is approximately

A) \[1.4\times m{{10}^{-3}}m\]    

B) \[3.3\times {{10}^{-3}}m\]

C) \[2.0\times {{10}^{-3}}m\]       

D) \[4.1\times {{10}^{-3}}m\]

Correct Answer: A

Solution :

[a] : The vertical force due to the surface tension on the drop \[=T2\pi r\sin \theta \] \[=T2\pi r\frac{r}{R}=\frac{2\pi {{r}^{2}}T}{R}\] When the drop detaches from the dropper, then \[\frac{2\pi {{r}^{2}}T}{R}=mg=\frac{4}{3}\pi {{R}^{3}}\rho g\]or\[{{R}^{4}}=\frac{3}{2}\frac{{{r}^{2}}T}{\rho g}\] or\[R=\left( \frac{3}{2}\frac{{{r}^{2}}T}{\rho g} \right)\] Substituting the given values, we get \[R={{\left( \frac{3\times {{\left( 5\times {{10}^{-4}} \right)}^{2}}\times 0.11}{2\times {{10}^{3}}\times 10} \right)}^{1/4}}=1.4\times {{10}^{-3}}m\]