A) \[1\text{ }cm\text{ }{{s}^{-1}}\]
B) \[2\text{ }cm\text{ }{{s}^{-1}}\]
C) \[4\text{ }cm\text{ }{{s}^{-1}}\]
D) \[8\text{ }cm\text{ }{{s}^{-1}}\]
Correct Answer: D
Solution :
If eight drops of same radius r coalesce, then radius of new drop is given by R. \[\frac{4}{3}\pi {{R}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}\] \[R=2r\] If drop of radius r is falling in viscous medium, then it acquires a critical velocity v and\[v\propto {{r}^{2}}\]. \[\Rightarrow \] \[\frac{{{v}_{2}}}{{{v}_{1}}}={{\left[ \frac{R}{r} \right]}^{2}}={{\left[ \frac{2r}{r} \right]}^{2}}=4\] \[{{v}^{2}}=4{{v}_{1}}\] Or \[{{v}_{2}}=8\,cm{{s}^{-1}}\]
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