The viscous drag acting on a metal sphere of diameter \(1 \text{mm}\), falling through a fluid of viscosity \(0.8 \text{Pa} \text{s}\) with a velocity of \(2 \text{m} \text{s}^{-1}\) is equal to
\(1.5 \times 10^{-3} \text{N}\)
\(20 \times 10^{-3} \text{N}\)
\(15 \times 10^{-3} \text{N}\)
\(30 \times 10^{-3} \text{N}\)
Solution:
Using Stokes' Law: \(F = 6\pi\eta r v\). Here, \(r = 0.5 \times 10^{-3} \text{m}\), \(\eta = 0.8 \text{Pa} \text{s}\), and \(v = 2 \text{m/s}\). \(F = 6 \times \pi \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3} \text{N}\).
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