A copper ball of radius 3.0 mm falls in an oil tank of viscosity 2 kg/m-s. Then, the terminal velocity of the copper ball will be 

(Density of oil $=1.5\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3$ and  Density of Copper $=9\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3 \mathrm{and} \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2$)

Given,

$\mathrm{radius} \mathrm{of} \mathrm{copper} \mathrm{ball}, \mathrm{r}=3\times {10}^{-3}\mathrm{m},$

$\mathrm{viscosity} \mathrm{of} \mathrm{oil}, \mathrm{\eta }=1\mathrm{kg}/\mathrm{ms},$

$\mathrm{density} \mathrm{of} \mathrm{oil}, \mathrm{\rho }=1.5\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$

$\mathrm{density} \mathrm{of} \mathrm{copper}, \mathrm{\sigma }=9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$

we know that the terminal velocity,

${\mathrm{v}}_{\mathrm{T}}=\frac{2}{9}\frac{{\mathrm{r}}^2(\mathrm{\sigma }-\mathrm{\rho })\mathrm{g}}{\mathrm{n}}$

${\mathrm{v}}_{\mathrm{T}}=\frac{2}{9}\times \frac{{(3\times {10}^{-3})}^2(9\times {10}^3-1.5\times {10}^3)\times 10}{1}$

${\mathrm{v}}_{\mathrm{T}}=\frac{2}{9}\times 9\times {10}^{-6}\times 7.5\times {10}^3\times 10=15\times {10}^{-2}\mathrm{m}/\mathrm{s}$ 

Hence the correct answer is $15\times {10}^{-2}\mathrm{m}/\mathrm{s}.$