A rain drop of radius 0.3 mm has a terminal velocity of 1 m / s in air. The viscosity of air is 18 × 10-5 poise. The viscous force on the drop is

Question

A rain drop of radius 0.3 mm has a terminal velocity of 1 m / s in air. The viscosity of air is 18 × 10-5 poise. The viscous force on the drop is (A) 16.95 × 10-7 N (B) 1.695 × 10-8 N (C) 1.017 × 10-11 N (D) 101.73 × 10-9 N

Tardigrade Admin · Accepted Answer

From Stoke's formula F=6 π η r v Given, η=18 × 10-5=18 × 10-6 kg / mass, r=0.3 mm =0.3 × 10-3 m, v=1 m / s ∴ F=6 × 3.14 × 18 × 10-6 × 0.3 × 10-3 × 1 ⇒ F=101.73 × 10-9 N

Q. A rain drop of radius $0.3 mm$ has a terminal velocity of $1 m / s$ in air. The viscosity of air is $18 \times 1 0^{- 5}$ poise. The viscous force on the drop is

$16.95 \times 1 0^{- 7} N$

$1.695 \times 1 0^{- 8} N$

$1.017 \times 1 0^{- 11} N$

$101.73 \times 1 0^{- 9} N$

From Stoke's formula
$F = 6 π ηr v$
Given, $η = 18 \times 1 0^{- 5} = 18 \times 1 0^{- 6} k g$ / mass,
$r = 0.3 mm$
$= 0.3 \times 1 0^{- 3} m$ ,
$v = 1 m / s$
$∴ F = 6 \times 3.14 \times 18 \times 1 0^{- 6} \times 0.3 \times 1 0^{- 3} \times 1$
$\Rightarrow F = 101.73 \times 1 0^{- 9} N$

Q. A rain drop of radius 0.3mm has a terminal velocity of 1m/s in air. The viscosity of air is 18×10−5 poise. The viscous force on the drop is

16.95×10−7N

1.695×10−8N

1.017×10−11N

101.73×10−9N