A) \[3:2\]
B) \[4:9\]
C) \[9:4\]
D) \[27:8\]
Correct Answer: D
Solution :
As, terminal velocity is given by \[v=\frac{2(\rho -\sigma )\,9{{r}^{2}}}{9\,\eta }\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{v}\propto {{r}^{2}}\] \[\therefore \,\,\,\,\,\,\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\,\,\,\,\,\Rightarrow \,\,\,\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{4}\,\,\,\,\Rightarrow \,\,\frac{{{r}_{1}}}{{{r}_{2}}}\,\,=\,\,\frac{3}{2}\,\] Now, \[\frac{Volume\,\,of\,\,1st\,\,drop}{Volume\,\,of\,\,2nd\,\,drop}=\frac{4/3\,\pi r_{1}^{3}}{4/3\,\pi r_{2}^{3}}\] \[=\,\,\,\,{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}={{\left( \frac{3}{2} \right)}^{3}}=\frac{27}{8}\]
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