GATE 2023 CH – Question 53
A large tank is filled with water (density = 1 g cm$^{-3}$) upto a height of 5 m. A 100 µm diameter solid spherical particle (density = 0.8 g cm$^{-3}$) is released at the bottom of the tank. The particle attains its terminal velocity ($v_t$) after traveling to a certain height in the tank. Use acceleration due to gravity as 10 m s$^{-2}$ and water viscosity as $10^{-3}$ Pa s. Neglect wall effects on the particle. If Stokes law is applicable, the absolute value of $v_t$ (in mm s$^{-1}$) is ____________ (rounded off to two decimal places).
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Correct answer: 1.10 to 1.12
Explanation
By Stokes' law, $v_t = \frac{d^2(\rho_p - \rho)g}{18\mu} = \frac{(10^{-4})^2 \times (800 - 1000) \times 10}{18 \times 10^{-3}} = -1.111 \times 10^{-3}$ m/s. The particle is lighter than water, so it rises. The magnitude is 1.11 mm/s.