A) \[{{\upsilon }_{O}}=4{{\upsilon }_{H}}\]
B) \[{{\upsilon }_{H}}=4{{\upsilon }_{O}}\]
C) \[{{\upsilon }_{O}}={{\upsilon }_{H}}\]
D) \[{{\upsilon }_{H}}=16{{\upsilon }_{O}}\]
Correct Answer: B
Solution :
\[{{u}_{O}}=\sqrt{\frac{3RT}{{{M}_{O}}}}{{M}_{H}}=2,\,{{M}_{O}}=32\] \[{{\upsilon }_{H}}=\sqrt{\frac{3RT}{{{M}_{H}}}}\] \[\frac{{{\upsilon }_{O}}}{{{\upsilon }_{H}}}=\sqrt{\frac{{{M}_{H}}}{{{M}_{O}}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}\] Volume of small drop (64) \[=64.\frac{4}{3}\pi {{r}^{3}}\] volume of big drop \[=\frac{4}{3}\pi r{{R}^{3}}\] \[\therefore \] \[64\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[R=4r\] \[{{q}_{1}}=\]charge on small drop. \[(q)\] \[{{q}_{2}}=\]charge on big drop formed by combining 64 drops \[=64\,q\] \[\therefore \]\[{{T}_{1}}=\] surface charge density for small \[{{T}_{2}}=\]surface charge density for big \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{q}_{1}}/4\pi {{r}^{2}}}{{{q}_{2}}/4\pi r{{R}^{2}}}=\frac{q/{{r}^{2}}}{64q/{{R}^{2}}}\] \[=\frac{q/{{r}^{2}}}{64q/{{(4r)}^{2}}}=\frac{1}{4}\]
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