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(B) The density of the ball $ho$ is given by $ho = \frac{m}{V} = \frac{\pi/2}{\frac{4}{3}\pi(0.5)^3} = \frac{3}{8 	imes 0.125} = 3 \ gm/cc$.The t

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$9.8$

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(B) The density of the ball $ho$ is given by $ho = \frac{m}{V} = \frac{\pi/2}{\frac{4}{3}\pi(0.5)^3} = \frac{3}{8 	imes 0.125} = 3 \ gm/cc$.The terminal velocity $V_T$ is $V_T = \frac{	ext{distance}}{	ext{time}} = \frac{50 \ cm}{5 \ s} = 10 \ cm/s$.According to Stoke's Law, the terminal velocity is $V_T = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$, where $\sigma = 1.2 \ gm/cc$ is the density of the liquid and $g = 980 \ cm/s^2$.Rearranging for viscosity $\eta$: $\eta = \frac{2}{9} \frac{r^2(ho - \sigma)g}{V_T}$.Substituting the values: $\eta = \frac{2}{9} 	imes \frac{(0.5)^2 	imes (3 - 1.2) 	imes 980}{10}$.$\eta = \frac{2}{9} 	imes \frac{0.25 	imes 1.8 	imes 980}{10} = \frac{2}{9} 	imes \frac{0.45 	imes 980}{10} = \frac{2}{9} 	imes 44.1 = 9.8 \ 	ext{poise}$.

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