Q. If a ball of steel $density\ \rho = 7.8\ g\ {cm}^{- 3})$ attains a terminal velocity of $10{cms}^{- 1}$ when falling in a tank of water (coefficient of viscosity $ɳ_{water} = 8.5 \times 10^{- 4}\ Pa - s)$ then its terminal velocity in glycerine $(\rho = 12\ g\ {cm}^{- 3},\ ɳ = 13.2\ Pa - s)$ would be nearly
A) $1.06 \times 10^{- 5}\ {cms}^{- 1}$
B) $6.25 \times 10^{- 4}\ {cms}^{- 1}$
C) $6.45 \times 10^{- 4}\ {cms}^{- 1}$
D) $1.5 \times 10^{- 5}\ {cms}^{- 1}$
Explanation
(b) $v \propto \frac{\rho - \rho_{0}}{ɳ}$
$\frac{v_{2}}{v_{1}} = \frac{\rho - \rho_{02}}{\rho - \rho_{01}} \times \frac{ɳ_{1}}{ɳ_{2}}$
$= \frac{7.8 - 1.2}{7.8 - 1} \times \frac{8.5 \times 10^{- 4} \times 10}{13.2}$
$= 6.25 \times 10^{- 4}\ cm/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.