Question

A metal ball $B_1$ (density 3.2 g $c{m}^{-3}$) is dropped in water while another metal ball $B_2$ (density 6.0 g $c{m}^{-3}$) is dropped in a liquid of density 1.6 g $c{m}^{-3}$. 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
• B. 0.5
• C. 4
• D. indeterminate due to insufficient data

Answer 1

Answer: (B)

0.5

The terminal velocity of the body of radius r, density $\rho$ falling through a medium of density $\sigma$ is given by
$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

where $\eta$ is the coefficient of viscosity of the medium
$\therefore v_{B_1}=\frac{2}{9}\frac{r_{B_1}^2}{{\eta }_{water}}({\rho }_{B_1}-{\sigma }_{water})g$ ....(i)

and $v_{B_2}=\frac{2}{9}\frac{r_{B_2}^2}{{\eta }_{liquid}}({\rho }_{B_2}-{\sigma }_{liquid})g$ ....(ii)

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 $\because v_{B_1}=v_{B_2}$ (Given)

Substituting these values in (i) and (ii), we get

$\frac{{\eta }_{water}}{{\eta }_{liquid}}=\frac{({\rho }_{B_1}-{\sigma }_{water})}{({\rho }_{B_2}-{\sigma }_{liquid})}$

Substituting the given values, we get

$\frac{{\eta }_{water}}{{\eta }_{liquid}}=\frac{(3.2-1)}{(6.0-1.6)}$ ($\because$ Density of water $=1gc{m}^{-3}$)

$=\frac{2.2}{4.4}=0.5$