10.28 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.

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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Fluids

10.28 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.

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Vishal Baghel | Contributor-Level 10

5 months ago

Radius of the uncharged drop, r = 2.0 $\times 10^{- 5}$ m

Density of the uncharged drop, $ρ$ = 1.2 $\times 10^{3}$ kg/ $m^{3}$

Viscosity of air, $η$ = 1.8 $\times 10^{- 5}$ Pas

Density of air $ρ_{o}$ , can be taken as zero in order to neglect the buoyancy of air

Acceleration due to gravity, g = 9.8 m/ $s^{2}$

Terminal velocity, v can be written as

v = $\frac{2 \times r^{2} \times (ρ - ρ_{o}) g}{9 \times η}$ = $\frac{2 \times ({2.0 \times 10^{- 5})}^{2} \times (1.2 \times 10^{3} - 0) \times 9.8}{9 \times 1.8 \times 10^{- 5}}$ = 0.05807 m/s = 5.8 cm/s

The viscous force on the drop is given by

F = 6 $π η r v$ = 6 $\times 3.1416 \times$ 1.8 $\times 10^{- 5} \times$ 2.0 $\times 10^{- 5} \times$ 0.05807

= 3.9 $\times 10^{- 10}$ N

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mg = FB + Fv $\Rightarrow \frac{4 π}{3} r^{3} ρ g = \frac{4 π}{3} r^{3} σ g + 6 π η r v$

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