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Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be
$\frac{79}{72}$
$\frac{19}{36}$
$\frac{39}{72}$
$\frac{79}{36}$
Solution
$v_{T}=\frac{2 r^{2}(\sigma-\rho) g}{9 \eta}$
$\frac{v_{1}}{v_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \frac{\left(\sigma_{1}-\rho\right)}{\left(\sigma_{2}-\rho\right)}$$=\left(\frac{1}{2}\right)^{2}\left(\frac{8 \rho_{2}-0.1 \rho_{2}}{\rho_{2}-0.1 \rho_{2}}\right)$$=\frac{79}{36}$
