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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 kg m⁻³. Take viscosity of liquid = 1.8 × 10⁻⁵ N s m⁻² (Neglect buoyancy due to air)
- 3.8 × 10⁻¹⁰ N
- 3.9 × 10⁻¹⁰ N
- 3.8 × 10⁻⁹ N
- 5.8 × 10⁻¹⁰ N
Correct answer: 3.9 × 10⁻¹⁰ N
Solution
The correct option is derived from applying Stokes' law, which calculates the viscous force on a spherical object moving through a fluid. By substituting the given values for radius, viscosity, and density into the formula, the resulting force is accurately determined to be 3.9 × 10⁻¹⁰ N.
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