Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm falling through air of viscosity 1.8 × 10-5 N s m -2 will be

Question

Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3 mm falling through air of viscosity 1.8 × 10-5 N s m -2 will be (A) 10.9 m s -1 (B) 7.48 m s -1 (C) 3.7 m s -1 (D) 12.8 m s -1

Tardigrade Admin · Accepted Answer

The terminal velocity of the spherical raindrop of radius

r

is given by

v_{t} = \frac{2 r ^{2} ρ g}{9 η}

where

ρ

is the density of water and

η

the viscosity of air. Substituting

r = 0.3 mm

= 0.3 \times 1 0^{- 3} m, ρ = 1 0^{3} k g / m^{3}, g = 9.8 m / s^{2}

and

η = 1.8 \times 1 0^{- 5} N s / m^{2}

, we get

v_{t} = \frac{2 \times ( 0.3 ) ^{2} \times 1 0 ^{- 3} \times 9.8}{9 \times 1.8 \times 1 0 ^{- 5}}

= 10.9 m s^{- 1}

Q. Neglecting the density of air, the terminal velocity obtained by a raindrop of radius $0.3 mm$ falling through air of viscosity $1.8 \times 1 0^{- 5} N s m^{- 2}$ will be

$10.9 m s^{- 1}$

$7.48 m s^{- 1}$

$3.7 m s^{- 1}$

$12.8 m s^{- 1}$

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

10.9ms−1

7.48ms−1

3.7ms−1

12.8ms−1

The terminal velocity of the spherical raindrop of radius r is given by vt​=9η2r2ρg​ where ρ is the density of water and η the viscosity of air. Substituting r=0.3mm =0.3×10−3m,ρ=103kg/m3,g=9.8m/s2 and η=1.8×10−5Ns/m2, we get vt​=9×1.8×10−52×(0.3)2×10−3×9.8​ =10.9ms−1

Q. Neglecting the density of air, the terminal velocity obtained by a raindrop of radius $0.3 mm$ falling through air of viscosity $1.8 \times 1 0^{- 5} N s m^{- 2}$ will be

$10.9 m s^{- 1}$

$7.48 m s^{- 1}$

$3.7 m s^{- 1}$

$12.8 m s^{- 1}$