A) 5 cm/s
B) 10 cm/s
C) 7.9 cm/s
D) 6 cm/s
Correct Answer: C
Solution :
A body attains terminal velocity when gravitational pull is equal to viscous drag. It is given by \[v=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] where r is radius of drop,\[\eta \]the coefficient of viscosity\[\rho \]and\[\sigma \]are density of liquid and drop respectively. Also, Volume of two drops = volume of one drop \[2\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[R={{2}^{1/3}}.r\] \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] where, \[{{v}_{1}}=5\,cm/s,{{r}_{1}}=r,\,{{r}_{2}}={{2}^{1/3}}.r\] \[\Rightarrow \] \[{{v}_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}{{v}_{1}}\] \[\Rightarrow \] \[{{v}_{2}}=\frac{{{({{2}^{1/3}}.r)}^{2}}}{{{r}^{2}}}\times 5\] \[={{2}^{2/3}}\times 5=7.9\text{ }cm/s\]
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