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The terminal velocity of a copper ball of radius 5 mm falling through a tank of oil at room temperature is 10 cm s$^{-1}$. If the viscosity of oil at room temperature is 0.9 kg m$^{-1}$s$^{-1}$, the viscous drag force is:
Detailed Solution
$r=5\times10^{-3}$ m, $v_t=10\times10^{-2}$ m/s. $F=6\pi\eta r v_t=6\times\pi\times0.9\times5\times10^{-3}\times10\times10^{-2}=84.78\times10^{-4}\approx8.48\times10^{-3}$ N.