In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kgm -3 ? Take viscosity of liquid =1.8 × 10-5 Nsm -2. (Neglect buoyancy due to air).

Question

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kgm -3 ? Take viscosity of liquid =1.8 × 10-5 Nsm -2. (Neglect buoyancy due to air). (A) 3.8 × 10-11 N (B) 3.9 × 10-10 N (C) 1.8 × 10-10 N (D) 5.8 × 10-10 N

Tardigrade Admin · Accepted Answer

Viscous force = Weight =ρ ×((4/3) π r 3) g =3.9 × 10-10

Q. In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0 \times 1 0^{- 5} m$ and density $1.2 \times 1 0^{3} k g m^{- 3}$ ? Take viscosity of liquid $= 1.8 \times 1 0^{- 5} N s m^{- 2}$ . (Neglect buoyancy due to air).

$3.8 \times 1 0^{- 11} N$

$3.9 \times 1 0^{- 10} N$

$1.8 \times 1 0^{- 10} N$

$5.8 \times 1 0^{- 10} N$

Viscous force $=$ Weight
$= ρ \times (\frac{4}{3} π r^{3}) g$
$= 3.9 \times 1 0^{- 10}$

Q. In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0×10−5m and density 1.2×103kgm−3 ? Take viscosity of liquid =1.8×10−5Nsm−2. (Neglect buoyancy due to air).

3.8×10−11N

3.9×10−10N

1.8×10−10N

5.8×10−10N