Question

The terminal speed of a sphere of gold (density $=19.5kg{m}^{-3}$ ) is $0.2m{s}^{-1}$ in a viscous liquid (density $=1.5kg{m}^{-3}$ ). Then the terminal speed of a sphere of silver (density $=10.5kg{m}^{-3}$ ) of the same size in the same liquid is

• A. $0.1m{s}^{-1}$
• B. $0.4m{s}^{-1}$
• C. $0.2m{s}^{-1}$
• D. $0.3m{s}^{-1}$

Answer 1

Answer: (A)

The terminal speed of a spherical body of radius $r$, density $\rho$ falling through a medium of density $\sigma$ is given by
$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$
where $\eta$ is the coefficient of viscosity of the medium.
$\therefore v_g=\frac{2}{9}\frac{r_g^2\left({\rho }_g-{\sigma }_{\text{liquid }}\right)g}{{\eta }_{\text{liquid }}}$
and $v_s=\frac{2}{9}\frac{r_s^2\left({\rho }_s-{\sigma }_{\text{liquid }}\right)g}{{\eta }_{\text{liquid }}}$
where the subscripts $g$ and $s$ represent gold and silver spheres respectively.
Since both the spheres are of same size and falling in the same liquid,
$\therefore \frac{v_g}{v_s}=\frac{{\rho }_g-{\sigma }_{\text{liquid }}}{{\rho }_s-{\sigma }_{\text{liquid }}}$
$=\frac{(19.5-1.5)kg{m}^{-3}}{(10.5-1.5)kg{m}^{-3}}$
$=\frac{18}{9}=2$
or $v_s=\frac{v_g}{2}=\frac{0.2}{2}m{s}^{-1}$
$=0.1m{s}^{-1}$