Question

A metal ball $B_1$ (density $3.2gc{m}^{-3}$ ) is dropped in water while another metal ball $B_2$ (density $6.0gc{m}^{-3}$ ) is dropped in a liquid of density $1.6gc{m}^{-3}$ . If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is

• A. $2.0$
• B. $0.5$
• C. $4.0$
• D. indeterminate due to insufficient data

Answer 1

Answer: (B)

The terminal velocity of the body of radius $r$ , density $\rho$ falling through a medium of density $\sigma$ is given by
$v=\frac{2}{9}\frac{r^2\left(\rho -\sigma \right)g}{\eta }$
where $\eta$ is the coefficient of viscosity of the medium
$\therefore v_{B_1}=\frac{2}{9}\frac{r^2{B}_1}{{\eta }_{water}}\left({\rho }_{B_1}-{\sigma }_{water}\right)g\dots \left(i\right)$
and $v_{B_2}=\frac{2}{9}\frac{r_{B_2}^2}{{\eta }_{liquid}}\left(\rho B_2-{\sigma }_{liquid}\right)g...\left(ii\right)$
where the subscripts $B_1$ and $B_2$ represent metal ball $B_1$ and metal ball $B_2$ respectively.
$\because r_{B_1}=r_{B_2}$ and $v_{B_1}=v_{B_2}$ (Given)
Substituting these values in $\left(i\right)$ and $\left(ii\right)$ , we get
$\frac{{\eta }_{water}}{{\eta }_{liquid}}=\frac{\left({\rho }_{B_1}-{\sigma }_{water}\right)}{\left({\rho }_{B_2}-{\sigma }_{liquid}\right)}$
Substituting the given values, we get
$\frac{{\eta }_{water}}{{\eta }_{liguid}}=\frac{\left(3.2-1\right)}{\left(6.0-1.6\right)}$ ( $\because$ Density of water $=1gc{m}^{-3}$ )
$=\frac{2.2}{4.4}=0.5$