Question

A metal ball $ B_{1} $ (density $ 3.2 \,g \,cm^{-3} $ ) is dropped in water while another metal ball $ B_{2} $ (density $ 6.0 \,g \,cm^{-3} $ ) is dropped in a liquid of density $ 1.6\, g \,cm^{-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 \ldots\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 $ =1\,g\,cm^{-3} $ )
$ =\frac{2.2}{4.4}=0.5 $