Question

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

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

Answer 1

The correct option is B $3:1$

We know that terminal velocity is given by
$V_T=\frac{2r^2}{9\eta }({\rho }_s-{\rho }_l)g$
where, $r\to$ radius of sphere
${\rho }_s\to$ density of sphere
${\rho }_l\to$ density of liquid
$\Rightarrow V_T\propto \frac{r^2}{\eta }({\rho }_s-{\rho }_l)$

So, $\frac{(V_T{)}_P}{(V_T{)}_Q}=\frac{r_P^2}{r_Q^2}\times \frac{{\eta }_Q}{{\eta }_P}\times \frac{({\rho }_s-{\rho }_{l,P})}{({\rho }_s-{\rho }_{l,Q})}$
$={\left(\frac{0.5}{0.25}\right)}^2\times \left(\frac{2}{3}\right)\times \left(\frac{8-0.8}{8-1.6}\right)$
$=\frac{3}{1}=3:1$
Hence, option $\left(b\right)$ is the correct answer.