Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = $1.8 \times 10^{- 4}$ poise, g = $9.9 \times 10 m s^{- 2}$ and density of water = $1000 k g m^{- 3}$

Open in App

Solution

r = 0.02 cm = $2 \times 10^{- 4} m$

$η = 1.8 \times 10^{- 4} p o i s e$

= $1.8 \times 10^{- 5}$ decapoise

g= 9.9 $m / s^{2}$

$ρ = 1000 k g / m^{3}$

To find terminal velocity of rain drops, the force acting on the drops are,

(i) The weight $(\frac{4}{3}) π r^{3} ρ g$ downward.

(ii) Force of buoyancy $(\frac{4}{3}) π r^{3} ρ g$ upward.

(iii) Force of viscosity $6 π η r v$ upward.

Because , $σ$ of air is very small, the force of buoyancy may be neglected.

Thus, $6 π η r v = (\frac{4}{3}) π r^{3} ρ g$

or v = $2 r^{2} \frac{π g}{9 η}$

= $2 \times (0.02 \times 10^{- 2})^{2} \times 1000 \times \frac{(9.9)}{9} \times (1.8 \times 10^{- 5})$

= 5m /sec.

Suggest Corrections

Similar questions

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

2.0 \times 10^{- 5} k g m^{- 1} s^{- 1}

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

10^{3} k g m^{- 3}

and

g = 10 m s^{- 2}

)

Q. A raindrop with radius

R = 0.2 mm

falls from a cloud at a height

h = 2000 m

above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water

ρ_{w} = 1000 {kg m}^{- 3}

and density of air

ρ_{a} = 1.2 {kg m}^{- 3}, g = 10 {m/s}^{2}

Coefficient of viscosity of air

= 1.8 \times 10^{- 5} {Nsm}^{- 2}

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

1.8 \times 10^{- 5} k g / m s

, what will be the terminal velocity of the drop? (density of water =

1.0 \times 10^{3} k g / m^{2}

and g = 9.8 N/kg). The density of air can be neglected

Q. 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:

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm falling through the air of viscosity $1.8 \times 10^{- 5} N s / m^{2}$ will be $(ρ_{w a t e r} = 10^{3} k g / m^{3})$

Join BYJU'S Learning Program

Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = 1.8×10−4 poise, g = 9.9×10ms−2 and density of water = 1000kgm−3

Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm, viscosity of air = $1.8 \times 10^{- 4}$ poise, g = $9.9 \times 10 m s^{- 2}$ and density of water = $1000 k g m^{- 3}$