JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
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
Answer and solution
Answer (b) 3.9 × 10⁻¹⁰ N
Solution
Physics sir · 4 steps
The drop falls at terminal speed, so the viscous drag exactly balances its weight. Just find mg.
- Balance of forces. In Millikan's experiment an uncharged drop quickly reaches terminal velocity. Buoyancy is neglected, so only gravity and viscous drag act, and at constant speed they must be equal.
Fᵥ = mg = 4/3π r³ ρ g
- Put in numbers. r³ = (2.0 × 10⁻⁵)³ = 8 × 10⁻¹⁵ m³, so the volume is 4/3π × 8 × 10⁻¹⁵ = 3.35 × 10⁻¹⁴ m³. The mass is 3.35 × 10⁻¹⁴ × 1.2 × 10³ = 4.02 × 10⁻¹¹ kg.
- Multiply by g. Fᵥ = 4.02 × 10⁻¹¹ × 9.8 = 3.94 × 10⁻¹⁰ N. The viscosity is not needed. Stokes' law 6πη r v with the terminal speed v = 5.8 × 10⁻² m s⁻¹ gives the same force.
- So. The viscous force is about 3.9 × 10⁻¹⁰ N, which is option (b).
The idea to keep. At terminal velocity the net force is zero, so the viscous drag equals weight minus buoyancy. You do not need η or v to find the drag force.
Answer from the NTA final answer key. Solution by PrepWiser.