If the terminal speed of a sphere of gold (de | vedclass

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Question

Terminal speed $\quad \mathrm{v}=\frac{2}{9 \eta} \mathrm{r}^{2}(\rho-\sigma)$

Here $\mathrm{r}=$ Radius of spherical body

$\rho=$ density of bo

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

Terminal speed $\quad \mathrm{v}=\frac{2}{9 \eta} \mathrm{r}^{2}(ho-\sigma)$

Here $\mathrm{r}=$ Radius of spherical body

$ho=$ density of body

$\sigma=$ density of liquid

$\eta=$ coefficient of viscosity of liquid

If body is of gold, $ho_{g}=19.5 	imes 10^{3} \mathrm{kg} / \mathrm{m}^{3}$

$\sigma=1.5 	imes 10^{3} \mathrm{kg} / \mathrm{m}^{3}, \sigma_{\mathrm{silver}=}=10.5 	imes 10^{3} \mathrm{kg} / \mathrm{m}^{3}$

$\mathrm{v}_{\mathrm{gold}}=\frac{2}{9 \eta} \mathrm{r}^{2}(19.5-1.5) 	imes 10^{3}$

$=\frac{2 r^{2}}{9 \eta} 	imes 18 	imes 10^{3}$

$v_{silver}=\frac{2 r^{2}}{9 \eta}(10.5-1.5) 	imes 10^{3}$

$=\frac{2 r^{2}}{9 \eta} 	imes 9 	imes 10^{3}$

$\frac{\mathrm{v}_{\mathrm{gald}}}{\mathrm{v}_{	ext {siver }}}=\frac{2}{1} \Rightarrow \mathrm{v}_{	ext {silver }}=\frac{\mathrm{V}_{	ext {gold }}}{2}=\frac{0.2}{2}$

$\mathrm{v}_{	ext {silver }}=0.1 \mathrm{m} / \mathrm{s}$

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 ........... $m/s$

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