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A metal ball \[{{B}_{1}}\] (density 3.2\[gc{{m}^{-3}}\]) is dropped in water while another metal ball \[{{B}_{2}}\](density 6.0\[gc{{m}^{-3}}\]) is dropped in a liquid of density 1.6\[gc{{m}^{-3}}\]. 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.

Last updated date: 23rd Jun 2024
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Hint: We are given two distinct situations of dropping metallic balls of different densities in two different liquids. We need to find the relation between their densities, the terminal velocities and the viscosities of the liquids to solve this problem.

Complete answer:
We know that any solid or denser material passing through a fluid experiences a resistance force due to the viscosity of the medium. This is given by the Stokes’ law. According to the law the weight of the moving substance in a fluid medium due to the change in their densities experience an upthrust which is proportional to the radius of the moving particles, the viscosity of the medium and the terminal velocity attained by the substance in the medium. It is given by –
\[F=6\pi \eta r{{v}_{t}}\]
We can find the terminal velocity of the substances moving by substituting the force with the volume and density.
  & F=mg \\
 & \Rightarrow mg=\dfrac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g \\
 & \Rightarrow \dfrac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g=6\pi \eta r{{v}_{t}} \\
 & \therefore {{v}_{t}}=\dfrac{2}{9\eta }{{r}^{2}}(\rho -\sigma )g \\
Now, let us consider the two situations given two us. We know that the terminal velocities and the diameter of both the balls are the same. So, we can easily equate the terminal velocities to find the required ratio between the viscosities of the water and the liquid.
(i)When metal ball \[{{B}_{1}}\]of density 3.2\[gc{{m}^{-3}}\] is dropped in water of density 1.0\[gc{{m}^{-3}}\], the terminal velocity attained will be –
  & {{v}_{t}}=\dfrac{2}{9\eta }{{r}^{2}}(\rho -\sigma )g \\
 & \Leftrightarrow {{v}_{t}}=\dfrac{2}{9{{\eta }_{w}}}{{r}^{2}}(3.2-1.0)g \\
 & \therefore {{v}_{t}}=\dfrac{2}{9{{\eta }_{w}}}{{r}^{2}}2.2g\text{ --(1)} \\
(ii)When the ball \[{{B}_{2}}\]of density 6.0\[gc{{m}^{-3}}\] is dropped in a liquid of density 1.6\[gc{{m}^{-3}}\], the terminal velocity obtained is given as –
  & {{v}_{t}}=\dfrac{2}{9\eta }{{r}^{2}}(\rho -\sigma )g \\
 & \Leftrightarrow {{v}_{t}}=\dfrac{2}{9{{\eta }_{l}}}{{r}^{2}}(6.0-1.6)g \\
 & \therefore {{v}_{t}}=\dfrac{2}{9{{\eta }_{l}}}{{r}^{2}}4.4g\text{ --(2)} \\
Now, we know that both the terminal velocities are equal. So, we can equate (1) and (2) to get the ratio as –
  & \Rightarrow \dfrac{2}{9{{\eta }_{w}}}{{r}^{2}}2.2g\text{ }=\dfrac{2}{9{{\eta }_{l}}}{{r}^{2}}4.4g\text{ } \\
 & \Rightarrow \dfrac{2.2}{{{\eta }_{w}}}=\dfrac{4.4}{{{\eta }_{l}}} \\
 & \therefore \dfrac{{{\eta }_{w}}}{{{\eta }_{l}}}=0.5 \\
The ratio between the viscosity of water to the viscosity of the unknown liquid is 0.5.

The correct answer is option B.

The Stokes’ law which we applied here is an immediate consequence of the Archimedes’ Principle. The density change in the dropped material and the medium is the reason for the downward movement of the material, which otherwise will be floating.