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GATE 2022 ME (ME1) – Question 43

Fluid Mechanics · Viscous flow, boundary layer and elementary turbulent flow · 2 marks · Multiple choice

A solid spherical bead of lead (uniform density = 11000 kg/m$^3$) of diameter $d = 0.1$ mm sinks with a constant velocity $V$ in a large stagnant pool of a liquid (dynamic viscosity = $1.1 \times 10^{-3}$ kg m$^{-1}$ s$^{-1}$). 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 $\rho$ is the density of the liquid. Neglect the buoyancy force. Using $g = 10$ m/s$^2$, the velocity $V$ is __________ m/s.

  1. $\frac{1}{24}$
  2. $\frac{1}{6}$
  3. $\frac{1}{18}$
  4. $\frac{1}{12}$

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Correct answer: (C) $\frac{1}{18}$

Explanation

With $c_D = \frac{24\mu}{\rho Vd}$ the drag is $D = 3\pi\mu dV$ (Stokes' law). Setting it equal to the weight of the bead (the buoyancy neglected), $mg = 3\pi\mu dV$, where $m = \rho_b\frac{\pi d^3}{6}$: $V = \frac{\rho_bgd^2}{18\mu} = \frac{11000 \times 10 \times 10^{-8}}{18 \times 1.1 \times 10^{-3}} = \frac{1.1 \times 10^{-3}}{1.98 \times 10^{-2}} = \frac{1}{18}$ m/s.