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Physics Question on mechanical properties of fluid

If a ball of steel (density ρ=7.8 g cm−3\rho = 7.8 \,g\, cm ^{-3}) attains a terminal velocity of 10 cm s−110\, cm \,s^{-1} when falling in a water (Coefficient of Viscosity ηwater\eta_{water} =8.5×10−4 Pa.s= 8.5 \times 10^{-4}\, Pa.s) then its terminal velocity in glycerine (ρ=1.2 g cm−3,η=13.2Pa.s.\rho = 1.2 \,g \,cm^{-3}, \eta = 13.2 Pa.s.) would be, nearly :

A

6.25??10−4 cm s−16.25 ??10^{-4}\, cm\, s^{-1}

B

6.45??10−4 cm s−16.45 ??10^{-4}\, cm\, s^{-1}

C

1.5??10−5 cm s−11.5 ??10^{-5}\, cm\, s^{-1}

D

1.6??10−4 cm s−11.6 ??10^{-4}\, cm\, s^{-1}

Answer

6.25??10−4 cm s−16.25 ??10^{-4}\, cm\, s^{-1}

Explanation

Solution

Vρg=6πηrv+vρℓgV\rho g = 6\pi\eta rv + v\rho_{\ell}g Vg(ρ−ρℓ)=6πηrvVg\left(\rho - \rho_{\ell}\right) = 6\pi\eta rv Vg(ρ−ρℓ′)=6πη′rv′Vg\left(\rho-\rho_{\ell}'\right) = 6\pi\eta'rv' V′η′=(ρ−ρℓ′)(ρ−ρℓ)×vηV' \eta' = \frac{\left(\rho -\rho _{\ell} '\right)}{\left(\rho -\rho_{\ell }\right)}\times v\eta V′=(ρ−ρℓ′)(ρ−ρℓ)×vηη′V' = \frac{\left(\rho -\rho _{\ell} '\right)}{\left(\rho -\rho_{\ell }\right)}\times \frac{v\eta}{\eta'} =(7.8−1.2)(7.8−1)×10×8.5×10−413.2= \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.v' = 6.25 ??10^{-4} \,cm/s.