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: (a) 4.23 × 10 - 6 N (b) 8.48 × 10 - 3 N (c) 8.48 × 10 - 5 N (d) 4.23 × 10 - 3 N

Answer: (b). Stokes' law: F = 6πη r v, with r = 5×10⁻³ m, v = 0.1 m/s. F = 6π(0.9)(5×10⁻³)(0.1) = 2.7π×10⁻³ F ≈ 8.48×10⁻³ N So the answer is (b) 8.48×10⁻³ N.

The terminal velocity of a copper ball of radius 5 mm falling through… | NEET 2022 (Phase 2)

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Viscosity & Stokes' Law

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:

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Answer: (b)

Stokes' law: $F = 6\pi\eta r v$ , with $r = 5\times10^{-3}$ m, $v = 0.1$ m/s.
$F = 6\pi(0.9)(5\times10^{-3})(0.1) = 2.7\pi\times10^{-3}$
$F \approx 8.48\times10^{-3}$ N
So the answer is (b) $8.48\times10^{-3}$ N.

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