Ncert Exercises & Solutions

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.

The radius of the uncharged drop, r = 2.0 × 10^–5 m

The density of the uncharged drop, ρ = 1.2 × 10³ kg m^–3

The viscosity of air, $η = 1.8 \times 10^{- 5} Pa s$

The density of air $(ρ_{0})$ can be taken as zero in order to neglect buoyancy of air.

Terminal velocity (v) is:

$v = \frac{2 r^{2} \times (ρ - ρ_{0}) g}{9 η} = \frac{2 \times {(2.0 \times 10^{- 5})}^{2} \times (1.2 \times 10^{3} - 0) 9.8}{9 \times 1.8 \times 10^{- 5}} = 5.807 \times 10^{- 2} {ms}^{- 1} = 5.8 cm s^{- 1}$
$The viscous force on the drop = 6 πηrv = 6 \times 3.14 \times 1.8 \times 10^{- 5} \times 2.0 \times 10^{- 5} \times 5.8 \times 10^{- 2} = 3.9 \times 10^{- 10} N$

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