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If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3\, kg/m^3$) is $0.2\, m/s$ in a viscous liquid (density $= 1.5 \times 10^3\, kg/m^3$), find the terminal speed of a sphere of silver (density $= 10.5 \times 10^3\, kg/m^3$) of the same size in the same liquid ........... $m/s$
$0.4$
$0.133$
$0.1$
$0.2$
Solution
Terminal speed $\quad \mathrm{v}=\frac{2}{9 \eta} \mathrm{r}^{2}(\rho-\sigma)$
Here $\mathrm{r}=$ Radius of spherical body
$\rho=$ density of body
$\sigma=$ density of liquid
$\eta=$ coefficient of viscosity of liquid
If body is of gold, $\rho_{g}=19.5 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$
$\sigma=1.5 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}, \sigma_{\mathrm{silver}=}=10.5 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$
$\mathrm{v}_{\mathrm{gold}}=\frac{2}{9 \eta} \mathrm{r}^{2}(19.5-1.5) \times 10^{3}$
$=\frac{2 r^{2}}{9 \eta} \times 18 \times 10^{3}$
$v_{silver}=\frac{2 r^{2}}{9 \eta}(10.5-1.5) \times 10^{3}$
$=\frac{2 r^{2}}{9 \eta} \times 9 \times 10^{3}$
$\frac{\mathrm{v}_{\mathrm{gald}}}{\mathrm{v}_{\text {siver }}}=\frac{2}{1} \Rightarrow \mathrm{v}_{\text {silver }}=\frac{\mathrm{V}_{\text {gold }}}{2}=\frac{0.2}{2}$
$\mathrm{v}_{\text {silver }}=0.1 \mathrm{m} / \mathrm{s}$
