An oil drop is of radius 2.0 x 10-5m and density 1.2 x 103 kg/m3. The viscosity of air at the temperature of experiment is 1.8 x 10-5 Ns/m2. Then

V=29r2sgηF=6πηrϑ=4πr33×ρ×g=4π2×10−533×1.2×103×9.8=3.9×10−10N

An oil drop is of radius 2.0 x 10-5m and density 1.2 x 103 kg/m3. The viscosity of air at the temperature of experiment is 1.8 x 10-5 Ns/m2. Then

V=29r2sgηF=6πηrϑ=4πr33×ρ×g=4π2×10−533×1.2×103×9.8=3.9×10−10N

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An oil drop is of radius 2.0 x 10^-5m and density 1.2 x 10³kg/m³. The viscosity of air at the temperature of experiment is 1.8 x 10^-5 Ns/m². Then

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The terminal velocity 5.8 x 10^-2m/ sec

The terminal velocity 3.8 x 10^-2m/ sec

The viscous force is 3.9 x 10^-10N

The viscous force is 5.8 x 10^-2N

answer is A, C.

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$\begin{array}{l} V = \frac{2}{9} \frac{r^{2} sg}{η} \\ F = 6 πηrϑ \end{array}$

$\begin{array}{l} = \frac{4 π r^{3}}{3} \times ρ \times g \\ = \frac{4 π {(2 \times 10^{- 5})}^{3}}{3} \times 1.2 \times 10^{3} \times 9.8 \\ = 3.9 \times 10^{- 10} N \end{array}$

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An oil drop is of radius 2.0 x 10-5m and density 1.2 x 103 kg/m3. The viscosity of air at the temperature of experiment is 1.8 x 10-5 Ns/m2. Then

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An oil drop is of radius 2.0 x 10^-5m and density 1.2 x 10³kg/m³. The viscosity of air at the temperature of experiment is 1.8 x 10^-5 Ns/m². Then