# In Millikan’S Oil Drop Experiment, What is the Terminal Speed of an Uncharged Drop of Radius 2.0 × 10–5 M and Density 1.2 × 103 Kg M–3? Take the Viscosity of Air at the Temperature of the Experiment to Be 1.8 × 10–5 Pa S. How Much is the Viscous Force on the Drop at that Speed? Neglect Buoyancy of the Drop Due to Air - Physics

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

#### Solution 1

Terminal speed = 5.8 cm/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 × 103 kg m–3

Viscosity of air, eta = 1.8 xx 10^(-5) Pa s

Density of air (rho_0) can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g = 9.8 m/s2

Terminal velocity (v) is given by the relation:

v = (2r^2 xx (rho - rho_0)g)/(9eta)

= (2xx(2.0xx10^(-5))^2 (1.2xx10^3 - 0)xx 9.8)/(9xx1.8xx10^(-5))

= 5.807 xx10^(-2) ms^(-1)

= 5.8 cm^1

Hence, the terminal speed of the drop is 5.8 cm s–1.

The viscous force on the drop is given by:

F = 6pietarv

:. F = 6xx3.14xx1.8xx10^(-5)xx2.0xx^(-5)xx 5.8xx10^(-2)

= 3.9 xx 10^(-10) N

Hence, the viscous force on the drop is 3.9 × 10–10 N.

#### Solution 2

Here radius of drop, r = 2.0 x 10-5 m, density of drop, p = 1.2 x 103 kg/m3, viscosity of air TI = 1.8 x 10-5Pa-s.

Neglecting upward thrust due to air, we find that terminal speed is

v_r = 2/9 (r62rhog)/eta = (2xx(2.0xx10^(-5))^2xx(1.2xx10^3))xx9.8)/(9xx(1.8xx10^(-5)))

= 5.81 xx 10^(-2) ms^(-1) " or " 5.81 cm s^(-1)

Viscous force at this speed

F = 6pietarv = 6 xx 3.14 xx (1.8xx10^(-5))xx (2.0xx10^(-5))xx (5.81xx10^(-2))

= 3.94 xx 10^(10) N

Concept: Viscosity
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#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 10 Mechanical Properties of Fluids
Q 28 | Page 271