Question

Consider two solid spheres $P$ and $Q$ each of density $8\, g \,cm ^{-3}$ and diameters $1 \,cm$ and $0.5 \,cm$, respectively. Sphere $P$ is dropped into a liquid of density $0.8 \,g\, cm ^{-3}$ and viscosity $\eta=3$ poiseuille. Sphere $Q$ is dropped into a liquid of density $1.6\, g \,cm ^{-3}$ and viscosity $\eta=2$ poiseiulle. The ratio of the terminal velocities of $P$ and $Q$ is

• A. $3: 1$
• B. $9: 1$
• C. $2: 4$
• D. $4: 2$

Answer 1

Answer: (A)

Terminal velocity is given by
$v_{T}=\frac{2}{9} \frac{r^{2}}{\eta}(d-\rho) g$
$\frac{v_{P}}{v_{Q}}=\frac{r_{P}^{2}}{r_{Q}^{2}} \times \frac{\eta_{Q}}{\eta_{P}} \times \frac{\left(d-\rho_{P}\right)}{\left(d-\rho_{Q}\right)}$
Given, $d=8\, g\,cm ^{-3}$
$r_{P}=(1 / 2) \,cm ,$
$ r_{Q}=\frac{0.5}{2} \,cm$
$\rho_{P}=0.8\, g\,cm ^{-3}$
$\rho_{Q}=1.6\, g\,cm ^{-3}$
$\eta_{P}=3$ poise
$\eta_{Q}=2$ poise
$\Rightarrow \frac{v_{P}}{v_{Q}}=\left(\frac{1}{0.5}\right)^{2} \times\left(\frac{2}{3}\right) \times \frac{(8-0.8)}{(8-1.6)}$
$=4 \times \frac{2}{3} \times \frac{7.2}{6.4}=3$
$\Rightarrow v_{P}: v_{Q}=3: 1$