Question

An air bubble of radius 1 cm rises with a constant speed of 3.5 mm/sec. through a liquid of density $1.75 \times 10^3 kg m^{-3}$. Neglecting the density of air, the coefficient of viscosity of the liquid (in kg $m^{-1} s^{-1}$)

• A. 54.5
• B. 109
• C. 163.5
• D. 218

Answer 1

Answer: (B)

109

Answer 2

At constant speed (terminal velocity), the upward buoyant force equals the downward viscous drag force. Neglecting the weight of the air bubble: $F_B = F_d$ $\rho_{liquid} V g = 6 \pi \eta r v$ Given $V = \frac{4}{3} \pi r^3$, we have: $\rho_{liquid} (\frac{4}{3} \pi r^3) g = 6 \pi \eta r v$ Solving for $\eta$: $\eta = \frac{2 r^2 \rho_{liquid} g}{9 v}$ Substituting the values: $r = 10^{-2}$ m, $v = 3.5 \times 10^{-3}$ m/s, $\rho_{liquid} = 1.75 \times 10^3$ kg/m$^3$, $g = 9.8$ m/s$^2$. $\eta = \frac{2 \times (10^{-2})^2 \times (1.75 \times 10^3) \times 9.8}{9 \times (3.5 \times 10^{-3})} \approx 108.89$ kg m$^{-1}$ s$^{-1}$. This is closest to 109.