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Terminal velocity

An oil drop falls through air with a terminal velocity of 5 1

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Solving time: 3 mins

An oil drop falls through air with a terminal velocity of $5 \times 1 0^{- 4} m / s$ . The radius of the drop will be? Neglect density of air as compared to that of oil.
(Viscosity of air $= \frac{18 \times 1 0 ^{- 5}}{5} N - s / m^{2}, g = 10 m / s^{2}$ , density of oil $= 900 kg / m^{3}$ )

2.5 \times 1 0^{- 6} m

2 \times 1 0^{- 6} m

3 \times 1 0^{- 6} m

4 \times 1 0^{- 6} m

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Samuel Discussed

An oil drop falls through air with a terminal velocity of

5 \times 1 0^{- 4} m / s

. The radius of the drop will be? Neglect density of air as compared to that of oil.
(Viscosity of air

= \frac{18 \times 1 0 ^{- 5}}{5} N - s / m^{2}, g = 10 m / s^{2}

, density of oil

= 900 kg / m^{3}

)

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

Viscosity,

η = \frac{18 \times 1 0 ^{- 5}}{5} N - sm^{- 2}

Terminal velocity

v_{T} = 5 \times 1 0^{- 4} ms^{- 1}

v_{T} = \frac{2 r ^{2} ( p - σ ) g}{9 η}

\Rightarrow 5 \times 1 0^{- 4} = \frac{2 r ^{2} ( 900 ) ( 10 )}{9 ( \frac{18 \times 1 0 ^{- 5}}{5} )}

\Rightarrow r^{2} = 9 \times 1 0^{- 12} m^{2}

\Rightarrow r = 3 \times 1 0^{- 6} m

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Question Text	An oil drop falls through air with a terminal velocity of $5 \times 1 0^{- 4} m / s$ . The radius of the drop will be? Neglect density of air as compared to that of oil. (Viscosity of air $= \frac{18 \times 1 0 ^{- 5}}{5} N - s / m^{2}, g = 10 m / s^{2}$ , density of oil $= 900 kg / m^{3}$ )
Topic	Mechanical Properties of Fluids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	142

An oil drop falls through air with a terminal velocity of $5 \times 1 0^{- 4} m / s$ . The radius of the drop will be? Neglect density of air as compared to that of oil.
(Viscosity of air $= \frac{18 \times 1 0 ^{- 5}}{5} N - s / m^{2}, g = 10 m / s^{2}$ , density of oil $= 900 kg / m^{3}$ )

Text solutionVerified

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Text solutionVerified

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An oil drop falls through air with a terminal velocity of $5 \times 1 0^{- 4} m / s$ . The radius of the drop will be? Neglect density of air as compared to that of oil.
(Viscosity of air $= \frac{18 \times 1 0 ^{- 5}}{5} N - s / m^{2}, g = 10 m / s^{2}$ , density of oil $= 900 kg / m^{3}$ )