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(B) The terminal velocity $v$ of a sphere of radius $r$ falling through a viscous medium is given by Stokes' Law: $F_v = 6 \pi \eta r v$.At terminal v

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$4$

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(B) The terminal velocity $v$ of a sphere of radius $r$ falling through a viscous medium is given by Stokes' Law: $F_v = 6 \pi \eta r v$.At terminal velocity,the gravitational force equals the viscous drag force (neglecting buoyancy): $mg = 6 \pi \eta r v$.Since $m = ho \cdot \frac{4}{3} \pi r^3$,we have $ho \cdot \frac{4}{3} \pi r^3 g = 6 \pi \eta r v$.This simplifies to $v \propto r^2$.Given $v_r = 1 \, cm \, s^{-1}$ for radius $r$,we find $v_{2r}$ for radius $2r$ as:$\frac{v_{2r}}{v_r} = \frac{(2r)^2}{r^2} = 4$.Therefore,$v_{2r} = 4 	imes v_r = 4 	imes 1 = 4 \, cm \, s^{-1}$.

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