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(C) The terminal velocity $V_t$ of a spherical body of radius $R$ and density $\rho$,falling through a liquid of density $\sigma$,is given by $V_t = \

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$\sqrt{8/5}$

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(C) The terminal velocity $V_t$ of a spherical body of radius $R$ and density $ho$,falling through a liquid of density $\sigma$,is given by $V_t = \frac{2}{9} \frac{R^2(ho - \sigma)g}{\eta}$,where $\eta$ is the coefficient of viscosity.Since both spheres fall with the same uniform speed,$V_{t1} = V_{t2}$.Thus,$R_1^2(ho_1 - \sigma) = R_2^2(ho_2 - \sigma)$.Rearranging for the ratio of radii: $\frac{R_1^2}{R_2^2} = \frac{ho_2 - \sigma}{ho_1 - \sigma}$.Given $ho_1 = 8 	imes 10^3 \ kg/m^3$,$ho_2 = 11 	imes 10^3 \ kg/m^3$,and $\sigma = 3 	imes 10^3 \ kg/m^3$.Substituting the values: $\frac{R_1^2}{R_2^2} = \frac{11 	imes 10^3 - 3 	imes 10^3}{8 	imes 10^3 - 3 	imes 10^3} = \frac{8 	imes 10^3}{5 	imes 10^3} = \frac{8}{5}$.Therefore,$\frac{R_1}{R_2} = \sqrt{\frac{8}{5}}$.

Two metal spheres are falling through a liquid of density $3 \times 10^3 \ kg/m^3$ with the same uniform speed. The material densities of sphere $1$ and sphere $2$ are $8 \times 10^3 \ kg/m^3$ and $11 \times 10^3 \ kg/m^3$ respectively. The ratio of their radii is

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