Question

If a ball of steel (density $\rho = 7.8 \,g\, cm ^{-3}$) attains a terminal velocity of $10\, cm \,s^{-1}$ when falling in a water (Coefficient of Viscosity $\eta_{water}$ $= 8.5 \times 10^{-4}\, Pa.s$) then its terminal velocity in glycerine ($\rho = 1.2 \,g \,cm^{-3}, \eta = 13.2 Pa.s.$) would be, nearly :

• A. $6.25 × 10^{-4}\, cm\, s^{-1}$
• B. $6.45 × 10^{-4}\, cm\, s^{-1}$
• C. $1.5 × 10^{-5}\, cm\, s^{-1}$
• D. $1.6 × 10^{-4}\, cm\, s^{-1}$

Answer 1

Answer: (A)

$V\rho g = 6\pi\eta rv + v\rho_{\ell}g$
$Vg\left(\rho - \rho_{\ell}\right) = 6\pi\eta rv$
$Vg\left(\rho-\rho_{\ell}'\right) = 6\pi\eta'rv'$
$V' \eta' = \frac{\left(\rho -\rho _{\ell} '\right)}{\left(\rho -\rho_{\ell }\right)}\times v\eta$
$V' = \frac{\left(\rho -\rho _{\ell} '\right)}{\left(\rho -\rho_{\ell }\right)}\times \frac{v\eta}{\eta'}$
$= \frac{\left(7.8 -1.2\right)}{\left(7.8 - 1\right)}\times \frac{10 \times 8.5 \times10^{-4}}{13.2}$
$v' = 6.25 × 10^{-4} \,cm/s.$