Question

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

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

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}$
$=0.1 \,m \,s ^{-1}$