A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 2.0×10−5kgm−1s−1, the terminal velocity of the drop will be:(The density of water = 103kgm−3 and g=10ms−2)

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A drop of water of radius 0.0015 mm is falling in air. If the

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A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is $2.0 \times 1 0^{- 5} k g m^{- 1} s^{- 1}$ , the terminal velocity of the drop will be:
(The density of water = $1 0^{3} k g m^{- 3}$ and $g = 10 m s^{- 2}$ )

1.0 \times 1 0^{- 4} m s^{- 1}

2.0 \times 1 0^{- 4} m s^{- 1}

2.5 \times 1 0^{- 4} m s^{- 1}

5.0 \times 1 0^{- 4} m s^{- 1}

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Michael Discussed

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is

2.0 \times 1 0^{- 5} k g m^{- 1} s^{- 1}

, the terminal velocity of the drop will be:
(The density of water =

1 0^{3} k g m^{- 3}

and

g = 10 m s^{- 2}

)

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10 mins ago

Here,

r = 0.0015 mm = 0.0015 \times 1 0^{- 3} m

η = 2.0 \times 1 0^{- 5} k g m^{- 1} s^{- 1}

ρ = 1.0 \times 1 0^{3} k g m^{- 3}

g = 10 m s^{- 2}

Neglecting the density of air, the terminal velocity of the water drop is

v_{T} = \frac{2}{9} \frac{r ^{2} ρ g}{η} = \frac{2 \times ( 0.0015 \times 1 0 ^{- 3} ) ^{2} \times 1.0 \times 1 0 ^{3} \times 10}{9 \times 2.0 \times 1 0 ^{- 5}}

v_{T} = 2.5 \times 1 0^{- 4} m s^{- 1}

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Question Text	A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is $2.0 \times 1 0^{- 5} k g m^{- 1} s^{- 1}$ , the terminal velocity of the drop will be: (The density of water = $1 0^{3} k g m^{- 3}$ and $g = 10 m s^{- 2}$ )
Answer Type	Text solution:1
Upvotes	151