A steel ball of radius 0.3 mm falls at 2 m/s through glycerine ($\eta = 0.833$ N·s/m²). The viscous drag is:
A$9.4 \times 10^{-3}$ N
B$9.4 \times 10^{-5}$ N
C$0.094$ N
D$1.5 \times 10^{-3}$ N
Answer & Solution
Correct answer: A. $9.4 \times 10^{-3}$ N
$F = 6\pi\eta r v = 6 \times 3.142 \times 0.833 \times 0.3 \times 10^{-3} \times 2 \approx 9.42 \times 10^{-3}$ N.
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