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 $\rho_L$ and viscosity $\eta$ is give

Vedclass · Accepted Answer

$0.1$

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(C) The terminal velocity $v_T$ of a sphere of radius $r$ and density $ho$ falling through a liquid of density $ho_L$ and viscosity $\eta$ is given by the formula:$v_T = \frac{2}{9} \frac{r^2 g}{\eta} (ho - ho_L)$Since the size of the spheres and the liquid are the same,$r$,$g$,and $\eta$ are constant.Therefore,$v_T \propto (ho - ho_L)$.For gold: $v_{Tg} = 0.2 \ m/s$,$ho_g = 19.5 	imes 10^3 \ kg/m^3$,$ho_L = 1.5 	imes 10^3 \ kg/m^3$.$v_{Tg} \propto (19.5 - 1.5) 	imes 10^3 = 18 	imes 10^3$.For silver: $v_{Ts} = ?, ho_s = 10.5 	imes 10^3 \ kg/m^3$.$v_{Ts} \propto (10.5 - 1.5) 	imes 10^3 = 9 	imes 10^3$.Taking the ratio:$\frac{v_{Ts}}{v_{Tg}} = \frac{9 	imes 10^3}{18 	imes 10^3} = \frac{1}{2}$.$v_{Ts} = 0.2 	imes \frac{1}{2} = 0.1 \ m/s$.

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