A) 40 cm/s
B) 20 cm/s
C) 25 cm/s
D) 50 cm/s
Correct Answer: A
Solution :
Let R be radius of coalesced drop, and r the radius of 8 small drops. Also, Mass of coalesced drop = 8 x mass of small drop \[\Rightarrow \] \[\frac{4}{3}\pi \,{{R}^{3}}\times \rho =8\times \frac{4}{3}\pi \,{{r}^{3}}\times \rho \] \[\Rightarrow \] \[R=2\,r\] Also terminal velocity \[{{v}_{T}}=\frac{2}{9}\frac{(\rho -\sigma )\,{{r}^{2}}g}{\eta }\] ?. (i) and of coalesced drop is \[v{{}_{T}}=\frac{2}{9}\frac{(\rho -\sigma )\,{{R}^{2}}g}{\eta }\] ... (ii) From Eqs. (i) and (ii), we get \[\frac{v{{}_{T}}}{{{v}_{T}}}={{\left( \frac{R}{r} \right)}^{2}}={{(2)}^{2}}=4\] \[\therefore \] \[v{{}_{T}}=4\times 10=40\,cm/s\]You need to login to perform this action.
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