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 $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is give

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(C) The terminal velocity $V_T$ of a sphere of radius $r$ and density $ho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by: $V_T = \frac{2r^2}{9\eta}(ho - \sigma)g$. Since $r$,$\eta$,and $\sigma$ are the same for both spheres,$V_T \propto (ho - \sigma)$. Let $V_{Au}$ and $V_{Ag}$ be the terminal velocities of gold and silver spheres respectively. $\frac{V_{Ag}}{V_{Au}} = \frac{ho_{Ag} - \sigma}{ho_{Au} - \sigma}$. Given: $ho_{Au} = 19.5 	imes 10^3 \, kg/m^3$,$ho_{Ag} = 10.5 	imes 10^3 \, kg/m^3$,$\sigma = 1.5 	imes 10^3 \, kg/m^3$,and $V_{Au} = 0.2 \, m/s$. $V_{Ag} = 0.2 	imes \frac{10.5 - 1.5}{19.5 - 1.5} = 0.2 	imes \frac{9}{18} = 0.2 	imes 0.5 = 0.1 \, m/s$.

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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