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

Asked by Abhisek | 1 year ago |  188

1 Answer

Solution :-

Acceleration due to gravity, g = 9.8 m/s2

Radius of the uncharged drop, r = 2.0 × 10-5 m

Density of the uncharged drop, ρ = 1.2 × 103 kg m-3

Viscosity of air, η =  1.8 × 10-5 Pa s

We consider the density of air to be zero in order to neglect the buoyancy of air.

Therefore terminal velocity (v) is :

v = \( \dfrac{ 2 r^{ 2 } g \rho }{ 9 \eta }\)

\( \dfrac{ 2 \left ( 2.0 \times 10^{ -5 } \right )^{ 2 } \times 9.8 \times 1.2 \times 10^{ 3 } }{ 9 \times 1.8 \times 10^{ -5 } }\)

And the viscous force on the drop is :

F = 6πηrv

= 6 x 3.14 x 1.8 × 10-5x2 x 10-5 x  5.8 10-2

= 3.91 x 10-10 N

Answered by Abhisek | 1 year ago

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