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 fw=1000 k gm -3 and Density of air fa=1.2 kgm -3, g=10 m / s 2 Coefficient of viscosity of air =1.8 × 10-5 Nsm -2 ]

Question

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 fw=1000 k gm -3 and Density of air fa=1.2 kgm -3, g=10 m / s 2 Coefficient of viscosity of air =1.8 × 10-5 Nsm -2 ] (A) 250.6 ms -1 (B) 43.56 ms -1 (C) 4.94 ms -1 (D) 14.4 ms -1

Tardigrade Admin · Accepted Answer

At terminal speed a=0 F textnet=0 m g=Fv=6 π η R v v=(m g/6 π η R v) v=(ρw (4 π/3) R3 g/6 π η R) =(2 ρw R2 g/9 η)=(400/81) m / s =4.94 m / s

$250.6 m s^{- 1}$

$43.56 m s^{- 1}$

$4.94 m s^{- 1}$

$14.4 m s^{- 1}$

At terminal speed
$a = 0$
$F_{net} = 0$
$m g = F_{v} = 6 π η R v$
$v = \frac{m g}{6 π η R v}$
$v = \frac{ρ _{w} \frac{4 π}{3} R ^{3} g}{6 π η R}$
$= \frac{2 ρ _{w} R ^{2} g}{9 η} = \frac{400}{81} m / s$
$= 4.94 m / s$

250.6ms−1

43.56ms−1

4.94ms−1

14.4ms−1

At terminal speed a=0 Fnet​=0 mg=Fv​=6πηRv v=6πηRvmg​ v=6πηRρw​34π​R3g​ =9η2ρw​R2g​=81400​m/s =4.94m/s

$250.6 m s^{- 1}$

$43.56 m s^{- 1}$

$4.94 m s^{- 1}$

$14.4 m s^{- 1}$

At terminal speed
$a = 0$
$F_{net} = 0$
$m g = F_{v} = 6 π η R v$
$v = \frac{m g}{6 π η R v}$
$v = \frac{ρ _{w} \frac{4 π}{3} R ^{3} g}{6 π η R}$
$= \frac{2 ρ _{w} R ^{2} g}{9 η} = \frac{400}{81} m / s$
$= 4.94 m / s$