Answer:
Option C
Explanation:
Terminal velocity is given by
$v_{T}=\frac{2}{9}\frac{r^{2}}{\eta}(d-\rho)g$
$v_{T}=\frac{2}{9}\frac{r^{2}}{\eta}(d-\rho)g$
$\left(\frac{1}{0.5}\right)^{2}\times \left(\frac{2}{3}\right)\times \frac{(8-.08)}{(8-1.6)}=4\times \frac{2}{3}\times \frac{7.2}{6.4}=3$