Question

calculate the value of the characteristic times for the tiny drop in the millikan Oil Drop experiment when its terminal speed is 6.1*1/100000 m/s .
​

Answer 1

Explanation:

Terminal speed =5.8cm/s

Viscous force =3.9×10  

−10

N  

Radius of the given uncharged drop, r=2.0×10  

−5

m

Density of the uncharged drop, ρ=1.2×10  

−3

kgm  

−3

 

Viscosity of air, η=1.8×10  

−5

Pas

Density of air (ρ  

0

​

) can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g=9.8m/s  

2

 

Terminal velocity (v) is given by the relation:

V=2r  

2

×(ρ−ρ  

0

​

)g/9η

   =2×(2×10  

−5

)  

2

(1.2×10  

3

−0)×9.8/(9×1.8×10  

−5

)

   =5.8×10  

−2

m/s

   =5.8cms  

−1

 

Hence, the terminal speed of the drop is 5.8cms  

−1

.

The viscous force on the drop is given by:

F=6πηrv

∴F=6×3.14×1.8×10  

−5

×2×10  

−5

×5.8×10  

−2

 

       =3.9×10  

−10

N

Hence, the viscous force on the drop is 3.9×10  

−10

N.