Question

Consider two solid spheres $P$ and $Q$ each of density $8gc{m}^{-3}$ and diameters $1cm$ and $0.5cm$, respectively. Sphere $P$ is dropped into a liquid of density $0.8gc{m}^{-3}$ and viscosity $\eta =3$ poiseuille. Sphere $Q$ is dropped into a liquid of density $1.6gc{m}^{-3}$ 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, $d=8gc{m}^{-3}$
$r_P=(1/2)cm,$
$r_Q=\frac{0.5}{2}cm$
${\rho }_P=0.8gc{m}^{-3}$
${\rho }_Q=1.6gc{m}^{-3}$
${\eta }_P=3$ poise
${\eta }_Q=2$ poise
$\Rightarrow \frac{v_P}{v_Q}={\left(\frac{1}{0.5}\right)}^2\times \left(\frac{2}{3}\right)\times \frac{(8-0.8)}{(8-1.6)}$
$=4\times \frac{2}{3}\times \frac{7.2}{6.4}=3$
$\Rightarrow v_P:v_Q=3:1$