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If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly
$1.6 \times 10^{-5} \;cms ^{-1}$
$6.25 \times 10^{-4} \;cms ^{-1}$
$6.45 \times 10^{-4}\; cms ^{-1}$
$1.5 \times 10^{-5}\; cms ^{-1}$
Solution
Terminal velocity of any liquid
$v =\frac{2 r ^{2}\left(\rho-\rho_{0}\right) g }{9 \eta}$
$v \propto \frac{\left(\rho-\rho_{0}\right)}{\eta}$
$\frac{ v _{2}}{ v _{1}}=\frac{\rho-\rho_{ g }}{\eta_{ g }} \times \frac{\eta_{ w }}{\rho-\rho_{ w }}$
$\frac{ v _{2}}{10}=\frac{7.8-1.2}{13.2} \times \frac{8.5 \times 10^{-4}}{7.8-1}=0.625 \times 10^{-4}$
$v _{2}=10 \times 0.625 \times 10^{-4}$
$v _{2}=6.25 \times 10^{-4} \;cm / s$
