Question

19.A metal sphere of radius 2.2mm and
density 8 x10^3 kg/m^3 falls through a liquid
of density 1.3 x^3 kg/m^3 .find terminal
velocity​

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