A) \[\frac{11}{8}\]
B) \[\sqrt{\frac{11}{8}}\]
C) \[\frac{3}{2}\]
D) \[\sqrt{\frac{3}{2}}\]
Correct Answer: D
Solution :
Terminal velocity \[{{\operatorname{V}}_{{{T}_{1}}}}=\frac{2}{9}\frac{\left( \sigma -\ell \right)}{\eta }{{\operatorname{r}}_{1}}^{2}\] \[{{\operatorname{V}}_{{{T}_{2}}}}=\frac{2}{9}\frac{\left( {{\sigma }_{2}}-\ell \right)}{\eta }{{\operatorname{r}}_{2}}^{2}\] \[{{\operatorname{V}}_{{{T}_{2}}}}={{V}_{{{T}_{2}}}}\] \[\frac{2}{9}\frac{\left( 8\times {{10}^{3}}-2\times {{10}^{3}} \right)}{\eta }{{\operatorname{r}}_{1}}^{2}=\frac{2}{9}\frac{\left( 11\times {{10}^{3}}-2\times {{10}^{3}} \right)}{\eta }{{\operatorname{r}}_{2}}^{2}\]\[6\times 1{{0}^{3}}{{r}_{1}}^{2}=9\times 1{{0}^{3}}{{r}_{2}}^{2}\] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{3}{2}}\]
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