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

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

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$0.1$

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(C) The terminal velocity $V_T$ of a sphere of radius $a$ falling through a viscous liquid is given by the formula:$V_T = \frac{2 a^2}{9 \eta} (\rho - \sigma) g$Where $\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,$\eta$ is the coefficient of viscosity,and $g$ is the acceleration due to gravity.Since the size of the spheres and the liquid are the same,$a$,$\eta$,and $g$ are constant.Therefore,$V_T \propto (\rho - \sigma)$.For gold: $V_{T1} = 0.2 \,m/s$,$\rho_1 = 19.5 \,kg/m^3$,$\sigma = 1.5 \,kg/m^3$.For silver: $V_{T2} = V$,$\rho_2 = 10.5 \,kg/m^3$,$\sigma = 1.5 \,kg/m^3$.Using the ratio: $\frac{V_{T1}}{V_{T2}} = \frac{\rho_1 - \sigma}{\rho_2 - \sigma}$$\frac{0.2}{V} = \frac{19.5 - 1.5}{10.5 - 1.5} = \frac{18}{9} = 2$$V = \frac{0.2}{2} = 0.1 \,m/s$.

If the terminal speed of a sphere of gold (density $= 19.5 \,kg/m^3$) is $0.2 \,m/s$ in a viscous liquid (density $= 1.5 \,kg/m^3$),find the terminal speed of a sphere of silver (density $= 10.5 \,kg/m^3$) of the same size in the same liquid.

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