A) 32 cm/sec
B) 16 cm/sec
C) 4 cm/sec
D) 8 cm/sec
Correct Answer: C
Solution :
Here: Radius of 1st drop \[{{r}_{1}}=2\,mm\] Terminal velocity of 1st drop \[{{\upsilon }_{1}}=16\,cm/\sec \] Radius of second drop \[{{r}_{2}}=1\,mm\] The terminal velocity of a spherical body \[\upsilon =\frac{2}{9}\times \frac{(\rho -\sigma )g{{r}^{2}}}{m}\] So, \[\upsilon \propto {{r}^{2}}\] Hence, \[\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] or \[\frac{16}{{{v}_{2}}}=\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=\frac{4}{1}\] So, \[{{\upsilon }_{2}}=\frac{16}{4}=4\,cm/\sec \]
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