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

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(C) The formula for terminal velocity $v$ is given by $v = \frac{2}{9 \eta} r^2 g (\rho - \sigma)$,where $r$ is the radius of the sphere,$\rho$ is the

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$0.1$

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(C) The formula for terminal velocity $v$ is given by $v = \frac{2}{9 \eta} r^2 g (ho - \sigma)$,where $r$ is the radius of the sphere,$ho$ is the density of the sphere,$\sigma$ is the density of the liquid,and $\eta$ is the coefficient of viscosity.Since $r$,$g$,and $\eta$ are constant for both spheres in the same liquid,we have $v \propto (ho - \sigma)$.For the gold sphere: $v_g = k (ho_g - \sigma)$,where $k = \frac{2 r^2 g}{9 \eta}$.$0.2 = k (19.5 	imes 10^3 - 1.5 	imes 10^3) = k (18 	imes 10^3)$.For the silver sphere: $v_s = k (ho_s - \sigma) = k (10.5 	imes 10^3 - 1.5 	imes 10^3) = k (9 	imes 10^3)$.Taking the ratio: $\frac{v_s}{v_g} = \frac{9 	imes 10^3}{18 	imes 10^3} = \frac{1}{2}$.Therefore,$v_s = \frac{v_g}{2} = \frac{0.2}{2} = 0.1\, m/s$.

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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