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

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(C) The formula for terminal velocity $V_T$ of a sphere falling through a viscous liquid is given by:$V_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$where $

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$0.1$

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(C) The formula for terminal velocity $V_T$ of a sphere falling through a viscous liquid is given by:$V_T = \frac{2r^2(ho - \sigma)g}{9\eta}$where $ho$ is the density of the sphere,$\sigma$ is the density of the liquid,$r$ is the radius,and $\eta$ is the coefficient of viscosity.Since the radius $r$,the liquid density $\sigma$,and the viscosity $\eta$ are the same for both spheres,the terminal velocity is proportional to the difference in densities: $V_T \propto (ho - \sigma)$.For gold: $V_{T1} = 0.2 \, m/s$,$ho_1 = 19.5 	imes 10^3 \, kg/m^3$,$\sigma = 1.5 	imes 10^3 \, kg/m^3$.For silver: $V_{T2} = ?$,$ho_2 = 10.5 	imes 10^3 \, kg/m^3$.Using the ratio: $\frac{V_{T2}}{V_{T1}} = \frac{ho_2 - \sigma}{ho_1 - \sigma}$$\frac{V_{T2}}{0.2} = \frac{10.5 - 1.5}{19.5 - 1.5} = \frac{9}{18} = 0.5$$V_{T2} = 0.2 	imes 0.5 = 0.1 \, m/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.

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