Question

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8 × 10⁻⁵ kg/m-s), what will be the terminal velocity of the drop? (Density of water = 1.0 × 10³ kg/m³ and g = 9.8 N/kg.) Density of air can be neglected.
​

Answer 1

Answer:

Explanation:By Stoke's law the terminal velocity of a water drop of radius r is given by

v=  

9

2

​  

 

η

r  

2

(ρ−σ)g

​  

 

where ρ is the density of water σ is the density of air and η the coefficient of viscosity of air Here σ is negligible and r = 0.0015 mm = 1.5 x 10  

−3

 = 1.5 x 10  

−6

 m Substituting the values

v=  

9

2

​  

×  

1.8×10  

−5

 

(1.5×10  

−6

)  

2

×(1.0×10  

3

)×9.8

​  

=2.72×10  

−4

m/s

Answer 2

Answer:

$v_T$ = 2.72 × 10^–4 m/s

Explanation:

Given,

A drop of water of radius 0.0015 mm = 1.5*10^-6 m

The coefficient of viscosity of air = 1.8 × 10⁻⁵ kg/m-s

To find: Terminal velocity of drop

Assumption: Density of air can be neglected.

​We know,

If ρ is the density of water, σ is the density of air and η the coefficient of viscosity of air, then the equation of terminal velocity can be given as:

$v_T = \dfrac{2}{9} \dfrac{r^2(\rho - \sigma)g}{\eta}~~~~~~(\sigma = neglibible)$

or, $v_T = \dfrac{2}{9}\dfrac{(1.5\times10^{-6})^2\times 1.0\times10^3 \times9.8}{1.8\times10^{-5}}\\\\\boxed{v_T = 2.72 \times 10^{-4} \ m/s}$