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The viscous drag force on a sphere of radius $r$ moving with a velocity $v$ in a fluid of viscosity $\eta$ is given by Stokes' law \begin{align} F=6\pi\eta r v. \end{align} Stoke's law is valid for laminar flow. The drag on a sphere is proportional to the flow velocity, fluid's viscosity, and sphere's radius. This can be derived using dimensional analysis or by rigorous mathematics.

Consider a sphere of radius $r$ and density $\rho$ moving with a velocity $v$ ina fluid of density $\sigma$ and viscosity $\eta$. The forces acting on the sphere are viscous drag force $F_D=6\pi\eta r v$, gravitational force $F_G=\frac{4}{3}\pi r^3 \rho g$ and buoyant force $F_B=\frac{4}{3}\pi r^3 \sigma g$. The sphere accelerates until the drag force adjusts itself to make net force on the sphere zero. Henceforth, the sphere moves with a constant velocity called terminal velocity. Solve $F_D+F_B=F_G$ to get terminal velocity formula \begin{align} v_T=\frac{2r^2(\rho-\sigma)g}{9\eta}.\nonumber \end{align} The sphere moves downwards if it is denser than fluid otherwise it moves upwards. The terminal velocity is directly proportional to $r^2$ and inversely proportional to the viscosity. If a sphere is released from the rest then its velocity varies with time as shown in the figure.

Stokes' law and terminal velocity find applications in viscometer (viscosity meter), Millikan's oil drop experiment, falling raindrop, sedimentation of particles etc.

Consider two solid spheres P and Q each of density 8 g/cm^{3} and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 g/cm^{3} and viscosity $\eta=3$ poiseulles. Sphere Q is dropped into a liquid of density 1.6 g/cm^{3} and viscosity $\eta=2$ poiseulles. The ratio of the terminal velocities of P and Q is_________

**Solution: **
The terminal velocity of a sphere of radius $r$ and density $\rho$, immersed in a liquid of density $\sigma$ and viscosity $\eta$, is given by
\begin{align}
%\label{yzb:eqn:1}
v=\frac{2}{9}\frac{(\rho-\sigma)r^2 g}{\eta}.\nonumber
\end{align}
Substitute the values of given parameters to get
\begin{align}
\frac{v_{P}}{v_Q}&=\frac{(\rho_P-\sigma_P)\, r_{P}^2\, \eta_Q}{(\rho_Q-\sigma_Q)\, r_{Q}^2\, \eta_P}\\
&=3. \nonumber
\end{align}

- $\frac{|\vec{v}_\text{P}|}{|\vec{v}_\text{Q}|}=\frac{\eta_1}{\eta_2}$
- $\frac{|\vec{v}_\text{P}|}{|\vec{v}_\text{Q}|}=\frac{\eta_2}{\eta_1}$
- $\vec{v}_\text{P}\cdot\vec{v}_\text{Q}>0$
- $\vec{v}_\text{P}\cdot\vec{v}_\text{Q}<0$

**Solution: **
Let $V=\frac{4}{3}\pi r^3$ be the volume of the spheres P and Q of equal radii $r$. The forces acting on the sphere P are its weight $\rho_1Vg$, tension from the string $T$, and the buoyancy force $\sigma_1 Vg$ (see figure). Similarly, forces on the sphere Q are $\rho_2Vg$, $T$, and $\sigma_1 Vg$. In equilibrium, the net force on the spheres P and Q are separately zero i.e.,
\begin{align}
\label{ezb:eqn:1}
&T+\rho_1 V g=\sigma_1 Vg,\\
\label{ezb:eqn:2}
&T+\sigma_2 V g=\rho_2 Vg.
\end{align}

The tension $T>0$ because the string is taut. Thus, first equation gives $\rho_1<\sigma_1$ and second equation gives $\rho_2>\sigma_2$. Eliminate $T$ from first and second equation to get \begin{align} \sigma_1-\rho_1=\rho_2-\sigma_2. \end{align}

Now, consider the situation when the sphere P moves in liquid $L_2$ and the sphere Q moves in liquid $L_1$. These spheres will attain the terminal velocities $\vec{v}_\text{P}$ and $\vec{v}_\text{Q}$ after some time. The direction of the velocity (upwards or downwards) will depend on the density of the sphere in comparison to the density of the liquid. Let us consider the case when $\rho_1>\sigma_2$. In this case, the velocity of the sphere P is downwards. From above equation, if $\rho_1>\sigma_2$ then $\rho_2<\sigma_1$. If the density of a sphere is less than the density of the liquid in which it is immersed, it will move up. Thus, the velocity of the sphere Q is upwards i.e., the directions of $\vec{v}_\text{P}$ and $\vec{v}_\text{Q}$ are opposite (see figure). Hence, $\vec{v}_\text{P}\cdot\vec{v}_\text{Q}<0$.

The forces on the sphere P are its weight $\rho_1Vg$, buoyancy force $\sigma_2 Vg$, and viscous drag $6\pi \eta_2 r v_\text{P}$ (see figure). Similarly, the forces on the sphere Q are $\rho_2Vg$, $\sigma_1 Vg$ and $6\pi \eta_1 r v_\text{Q}$. Net forces on the spheres are zero when they move with terminal velocities i.e., \begin{align} \label{ezb:eqn:4} &6\pi\eta_2 r v_\text{P}+\sigma_2 V g=\rho_1 V g,\\ \label{ezb:eqn:5} &6\pi\eta_1 r v_\text{Q}+\rho_2 V g=\sigma_1 V g. \end{align} Simplify above equations to get, \begin{align} \label{ezb:eqn:6} &v_\text{P}=\frac{2 r^2 (\rho_1-\sigma_2)g}{9\eta_2},\\ \label{ezb:eqn:7} &v_\text{Q}=\frac{2 r^2 (\sigma_1-\rho_2)g}{9\eta_1}. \end{align} Divide to get, \begin{align} \frac{|\vec{v}_\text{P}|}{|\vec{v}_\text{Q}|}=\frac{\eta_1}{\eta_2}\;\frac{\rho_1-\sigma_2}{\sigma_1-\rho_2}=\frac{\eta_1}{\eta_2}. \end{align} The readers are encouraged to show the results for the case $\rho_1<\sigma_2$.

A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.

**Solution: **
The forces acting on the sphere are its weight $\frac{4}{3}\pi r^3 \rho g$ downwards, buoyancy force $\frac{4}{3}\pi r^3 \sigma g$ upwards, and viscous force $6\pi\eta r v$ upwards. The sphere attains the terminal velocity $v_t$ when the resultant force on it is zero i.e.,
\begin{align}
\label{ilb:eqn:1}
\tfrac{4}{3}\pi r^3 \rho g=\tfrac{4}{3}\pi r^3 \sigma g+6\pi\eta r v_t.
\end{align}
Solve above equation to get the terminal velocity,
\begin{align}
v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}.
\end{align}
The rate of heat generation is equal to the rate of work done by the viscous force which, in turn, is equal to its power. Thus,
\begin{align}
\frac{\mathrm{d}Q}{\mathrm{d}t} & =(6\pi\eta rv_t) v_t \\
&=\frac{8\pi(\rho-\sigma)^2 g^2 r^5}{27\eta}.\nonumber
\end{align}

**Question 1:** Two identical spherical drops of water are falling through air with a steady velocity of 20 cm/s. If the drops combine to form a single drop, what would be the terminal velocity of the single drop?

**Question 2:** The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil is 6.5 cm/s. Compute the viscosity of the oil if density of oil is $1.5\times10^3$ kg/m^{3} and density of copper is $8.9 \times 10^3$ kg/m^{3}.