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Standard XII

Physics

Terminal Velocity

Neglecting th...

Question

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius $0.3 m m$ falling through air of viscosity $1.8 \times 10^{- 5} N s m^{- 2}$ will be:

10.9 m s^{- 1}

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7.48 m s^{- 1}

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3.7 m s^{- 1}

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12.8 m s^{- 1}

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Solution

The correct option is A $10.9 m s^{- 1}$
Neglecting the density of air, and hence buoyant force due to air, at equilibrium,
$\frac{4}{3} ρ π r^{3} g = 6 π η r v$
$⟹ v = \frac{2 r^{2}}{9 η} ρ g$
$= \frac{2 \times {0.0003}^{2}}{9 \times 1.8 \times 10^{- 5}} \times 1000 \times 9.8 m / s$
$= 10.9 m / s$

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Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm falling through air of viscosity 1.8×10−5Nsm−2 will be:

The correct option is A 10.9 ms−1Neglecting the density of air, and hence buoyant force due to air, at equilibrium,43ρπr3g=6πηrv⟹v=2r29ηρg=2×0.000329×1.8×10−5×1000×9.8m/s=10.9m/s

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius $0.3 m m$ falling through air of viscosity $1.8 \times 10^{- 5} N s m^{- 2}$ will be: