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YesterdaysDreams
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I get the first part (i) and I got 8.57× 10^-4
Could someone answer the second part (ii) with steps.

(b) An oil drop is travelling at terminal velocity.
(i) The oil drop takes 11.9 s to fall a distance of 10.2 mm.
Show that the terminal velocity of the oil drop is about 0.001 m s􀃭􀀔.

(ii) Assuming that the upthrust is negligible, show that the radius of the oil drop is
about 3 μm.
density of oil = 920 kg m􀃭􀀖
viscosity of air = 1.82 × 10^-5􀀘 Pa s

Thanks,
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pineneedles
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(Original post by YesterdaysDreams)
I get the first part (i) and I got 8.57× 10^-4
Could someone answer the second part (ii) with steps.

(b) An oil drop is travelling at terminal velocity.
(i) The oil drop takes 11.9 s to fall a distance of 10.2 mm.
Show that the terminal velocity of the oil drop is about 0.001 m s􀃭􀀔.

(ii) Assuming that the upthrust is negligible, show that the radius of the oil drop is
about 3 μm.
density of oil = 920 kg m􀃭􀀖
viscosity of air = 1.82 × 10^-5􀀘 Pa s

Thanks,
Weight = Upthrust + Viscous drag

As the question tells you upthrust is negligible, we can assume that:

Weight = Viscous drag

Using this, you can write an equation for the weight of the oil drop using volume, density, and g, and equate it to Stoke's law for viscous drag..
I've worked it out in the spoiler below so you can compare.
Spoiler:
Show
I've used p to represent viscosity, and d to represent density.
Density * Volume * g = 6(pi)rpv
(4/3)(pi)(r3)gd = 6(pi)rpv
(4/3)r2gd = 6pv
r2 = (9/2)(6pv/gd)
r2 = (9/2)((1.82 x 10-5 * 8.57 x 10-4)/(9.81 * 920))
r= 2.78 x 10-6m
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