Discussion
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·s). The coefficient of drag is given by \(c_{D} = \frac{24}{Re}\), 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 \(D = (c_{D})(0.5\rho V^{2})\left(\frac{\pi d^{2}}{4}\right)\), where ρ is the density of the liquid. Neglect the buoyancy force. Using \(g\) = 10 m/s2, the velocity \(V\) is __________ m/s.
Answer: C
The settling velocity is \(\frac{1}{18}\) m/s.
With \(c_{D} = 24/Re = 24\mu/(\rho V d)\), the drag reduces to Stokes' law, \(D = 3\pi\mu V d\); the liquid density cancels. At constant velocity the drag balances the weight (buoyancy is neglected):
\[ \begin{aligned} 3\pi\mu V d &= \rho_{s}\,g\,\frac{\pi d^{3}}{6} \\ V &= \frac{\rho_{s} g d^{2}}{18\mu} = \frac{11000 \times 10 \times 10^{-8}}{18 \times 1.1 \times 10^{-3}} \\ &= \frac{1}{18}\ \text{m/s} \end{aligned} \]
The other fractions do not satisfy the force balance; 1/24 is the trap of using the 24 of the drag law in place of the 18 of Stokes' settling velocity.
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