An oil drop falls through air with a terminal velocity of 5 × 10-4 m/s. Viscosity of oil is 1.8 × 10 -5 Ns/m2 and density of oil is 900 kg/m3. Neglecting density of air as compared to that of the oil

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An oil drop falls through air with a terminal velocity of 5 × 10^-4 m/s. Viscosity of oil is 1.8 × 10 ^-5 Ns/m² and density of oil is 900 kg/m³. Neglecting density of air as compared to that of the oil

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radius of the drop is 6.20 × 10^-2 m

radius of the drop is 2.14 × 10^-6 m

terminal velocity of the drop at half of this radius is 1.25 × 10^-4 m/s

terminal velocity of the drop at half of this radius is 2.5 × 10^-4 m/s

answer is B, C.

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Detailed Solution

$v_{T} = \frac{2 r^{2} g (σ - ρ)}{9 η} \Rightarrow r = \sqrt{\frac{9 ηv}{2 (σ - ρ) g}} = \sqrt{\frac{9 (1.8 \times 10^{- 5}) (5 \times 10^{- 4})}{2 (900 - 0) 10}} = 2.14 \times 10^{- 6} m$
Also v $\propto$ r²
$\frac{v^{1}}{v} = {(\frac{r^{1}}{r})}^{2} \Rightarrow v^{1} = \frac{v}{4} = 1.25 \times 10^{- 4} m / s$