A solid spherical bead of lead (uniform density = 11000 kg/m3) of diameter d = 0.1 mm sinks with a constant velocity V in a large stagnant pool of a liquid (dynamic viscosity = 1.1 × 10-3 kg∙m-1∙s-1). The coefficient of drag is given by ${\mathrm{C}}_{\mathrm{D}}=\frac{24}{\mathrm{R}\mathrm{e}}$, where the Reynolds number (Re) is defined on the basis of the diameter of the bead. The drag force acting on the bead is expressed as $\mathrm{D}=({\mathrm{C}}_{\mathrm{D}})(0.5\rho {\mathrm{V}}^2)\left(\frac{\pi {\mathrm{d}}^2}{4}\right),$ where ρ is the density of the liquid. Neglect the buoyancy force. Using g = 10 m/s2, the velocity V is __________ m/s.

The correct option is 1/18.

1. Understanding the Forces Acting on the Bead
When the solid spherical bead of lead sinks with a constant velocity $V$ (terminal velocity) in a stagnant pool of liquid, it is in a state of dynamic equilibrium. This means the net force acting on the bead is zero.
Since we are instructed to neglect the buoyancy force, the only two forces acting on the bead are:
1. The downward gravitational force ($F_g$) acting on the bead due to its mass.
2. The upward drag force ($D$) exerted by the viscous liquid opposing its motion.

For constant velocity $V$, these forces must balance each other:
$D=F_g$

2. Expressing the Drag Force
The coefficient of drag is defined as:
$C_D=\frac{24}{Re}$
where the Reynolds number based on the bead's diameter is:
$Re=\frac{\rho Vd}{\mu }$
Here, $\rho$ is the density of the liquid, $V$ is the sinking velocity, $d$ is the bead diameter, and $\mu$ is the dynamic viscosity of the liquid.

The drag force acting on the bead is given by:
$D=C_D\left(0.5\rho V^2\right)\left(\frac{\pi d^2}{4}\right)$

Substituting the expression for $C_D$ into the drag force equation:
$D=\left(\frac{24}{Re}\right)\left(0.5\rho V^2\right)\left(\frac{\pi d^2}{4}\right)$

Substituting the definition of the Reynolds number:
$D=\left(\frac{24\mu }{\rho Vd}\right)\left(0.5\rho V^2\right)\left(\frac{\pi d^2}{4}\right)$

Simplifying the expression:
$D=3\pi \mu Vd$

3. Expressing the Gravitational Force
The gravitational force on the solid spherical lead bead of uniform density ${\rho }_s$ is:
$F_g=m_{s}g={\rho }_s{V}_{sphere}g$
where $V_{sphere}$ is the volume of the spherical bead:
$V_{sphere}=\frac{\pi d^3}{6}$

Thus, the gravitational force is:
$F_g={\rho }_s\left(\frac{\pi d^3}{6}\right)g$

4. Equating the Forces
Equating the drag force to the gravitational force:
$3\pi \mu Vd={\rho }_s\left(\frac{\pi d^3}{6}\right)g$

Solving for $V$:
$V=\frac{{\rho }_s{d}^{2}g}{18\mu }$

5. Substituting the Given Values
Let us convert all parameters to standard SI units:
• Density of lead bead, ${\rho }_s=11000{\text{kg/m}}^3$
• Diameter of bead, $d=0.1\text{mm}={10}^{-4}\text{m}$
• Acceleration due to gravity, $g=10{\text{m/s}}^2$
• Dynamic viscosity of the liquid, $\mu =1.1\times {10}^{-3}{\text{kg⋅m}}^{-1}{\text{⋅s}}^{-1}$

Substitute these values into the velocity equation:
$V=\frac{11000\times {\left({10}^{-4}\right)}^2\times 10}{18\times (1.1\times {10}^{-3})}$

Simplify the numerator:
$11000\times {10}^{-8}\times 10=110000\times {10}^{-8}=1.1\times {10}^{-3}$

Simplify the fraction for velocity:
$V=\frac{1.1\times {10}^{-3}}{18\times 1.1\times {10}^{-3}}=\frac{1}{18}$