A) 10 cm per sec
B) 2.5 cm per sec
C) \[5\times {{(4)}^{1/3}}cm\] per sec
D) \[5\times \sqrt{2}\,cm\] per sec
Correct Answer: C
Solution :
If two drops of same radius r coalesce then radius of new drop is given by R \[\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi {{r}^{3}}+\frac{4}{3}\pi {{r}^{3}}\]Þ \[{{R}^{3}}=2{{r}^{3}}\Rightarrow R={{2}^{1/3}}r\] If drop of radius r is falling in viscous medium then it acquire a critical velocity v and \[v\propto {{r}^{2}}\] \[\frac{{{v}_{2}}}{{{v}_{1}}}={{\left( \frac{R}{r} \right)}^{2}}={{\left( \frac{{{2}^{1/3}}r}{r} \right)}^{2}}\] Þ \[{{v}_{2}}={{2}^{2/3}}\times {{v}_{1}}={{2}^{2/3}}\times (5)=5\times {{(4)}^{1/3}}m/s\]
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