Q.29. A spherical metal ball of radius ‘r’ falls through viscous liquid with velocity ‘V’. Another metal ball of same material but of radius ( \frac{r}{3}) falls through same liquid, then its terminal velocity will be
A. \frac{V}{3}
B. \frac{V}{4}
C. \frac{V}{6}
D. \frac{V}{9}
Answer :- D. \frac{V}{9}
Explanation :-
The terminal velocity of a sphere falling through a viscous fluid is governed by Stokes’ law, which states that:
V=\frac{2}{9}\frac{r^{2}(\rho -\sigma )}{\eta }where:
V is the terminal velocity,
r is the radius of the sphere,
ρ is the density of the sphere,
σ is the density of the fluid,
g is the acceleration due to gravity,
η is the dynamic viscosity of the fluid.
Given that the original ball has a radius r and terminal velocity V, for the smaller ball with radius \frac{r}{3}, the terminal velocity V′ can be calculated by substituting the new radius into the equation:
V^{'}=\frac{2}{9}\frac{(\frac{r}{3})^{2}(\rho -\sigma )g}{\eta }Simplifying, we get:
V^{'}=\frac{2}{9}\frac{(\frac{r^{2}}{3})(\rho -\sigma )g}{\eta }This can be further reduced to:
V^{'}=\frac{1}{9}\left ( \frac{2}{9}\frac{(r^{2}(\rho -\sigma )g}{\eta }\right )Recognizing that the term in the parentheses is the expression for the original terminal velocity V, we find:
V′= \frac{1}{9} V
Therefore, the terminal velocity of the smaller ball is:
Option D: D. \frac{V}{9}