A) \[1.5\text{ }cm-{{s}^{-1}}\]
B) \[6\text{ }c{{m}^{-1}}\]
C) \[24\text{ }cm-{{s}^{-1}}\]
D) \[32\text{ }cm-{{s}^{-1}}\]
Correct Answer: C
Solution :
Let now radius of big drop is R. Then, \[\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\times \pi {{r}^{3}}.8\] \[R=2r\] where r is radius of small drops. Now, terminal velocity of drop in liquid. \[{{v}_{e}}=\frac{2}{9}\times \frac{{{r}^{2}}}{\eta }(\rho -\sigma )g\] where \[\eta \] is coefficient of viscosity and p is density of drop a is density of liquid. Terminal speed drop is \[6\text{ }cm\text{ }{{s}^{-1}}\] \[\therefore \] \[6=\frac{2}{9}\times \frac{{{r}^{2}}}{\eta }(\rho -\sigma )g\] ...(i) Let terminal velocity becomes v after coalesce, then \[v=\frac{2}{9}\frac{{{R}^{2}}}{\eta }(\rho -\sigma )g\] ?..(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{6}{v}=\frac{\frac{2}{9}\frac{{{r}^{2}}}{\eta }(\rho -\sigma )g}{\frac{2}{9}\frac{{{R}^{2}}}{\eta }(\rho -\sigma )g}\] or \[\frac{6}{v}=\frac{{{r}^{2}}}{{{(2r)}^{2}}}\] or \[v=24cm\,{{s}^{-1}}\]
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