Find the viscosity of glycerine (having densi | vedclass

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(A) The formula for terminal velocity $v_T$ is given by $v_T = \frac{2}{9} \frac{g(\rho - \sigma)r^2}{\eta}$.Rearranging for viscosity $\eta$,we get $

Vedclass · Accepted Answer

$14.6$

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(A) The formula for terminal velocity $v_T$ is given by $v_T = \frac{2}{9} \frac{g(ho - \sigma)r^2}{\eta}$.Rearranging for viscosity $\eta$,we get $\eta = \frac{2}{9} \frac{g(ho - \sigma)r^2}{v_T}$.Given values: $g = 980 \ cm \ s^{-2}$,$ho = 8 \ g \ cm^{-3}$,$\sigma = 1.3 \ g \ cm^{-3}$,$r = 2 \ mm = 0.2 \ cm$,and $v_T = 4 \ cm \ s^{-1}$.Substituting these values into the formula:$\eta = \frac{2}{9} 	imes \frac{980 	imes (8 - 1.3) 	imes (0.2)^2}{4}$.$\eta = \frac{2}{9} 	imes \frac{980 	imes 6.7 	imes 0.04}{4}$.$\eta = \frac{2}{9} 	imes 980 	imes 6.7 	imes 0.01$.$\eta = \frac{131.32}{9} \approx 14.59 \ 	ext{Poise} \approx 14.6 \ 	ext{Poise}$.

Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

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