Question

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius \(2.0 \times 10^{-5}\) m and density \(1.2 \times 10^3\) kgm\(^{-3}\)? Take viscosity of liquid \(= 1.8 \times 10^{-5}\) Nsm\(^{-2}\). (Neglect buoyancy due to air).

• A. \(5.8 \times 10^{-10}\) N
• B. \(1.8 \times 10^{-10}\) N
• C. \(3.8 \times 10^{-11}\) N
• D. \(3.9 \times 10^{-10}\) N

Hint:

In problems involving Millikan's experiment, for an uncharged drop falling freely, the key is to equate the viscous force (from Stokes' Law) with the gravitational force at terminal velocity. The viscosity value is sometimes given to distract you from the simpler force balance equation if terminal velocity is assumed.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the viscous force on an uncharged oil drop. In the absence of an electric field, the drop falls under gravity. It will reach a terminal velocity where the upward viscous force balances the downward gravitational force. The question implies we need to find this force at terminal velocity. Buoyancy is to be neglected.
Step 2: Key Formula or Approach:
At terminal velocity, the net force on the oil drop is zero.
Viscous Force (\(F_v\)) = Gravitational Force (\(F_g\))
The gravitational force is given by \(F_g = mg\), where \(m\) is the mass of the drop.
The mass can be calculated using the density (\(\rho\)) and volume (\(V\)): \(m = \rho \times V\).
The volume of a spherical drop is \(V = \frac{4}{3}\pi r^3\).
Step 3: Detailed Explanation:
First, we calculate the mass of the oil drop.
Given: Radius, \(r = 2.0 \times 10^{-5}\) m
Density, \(\rho = 1.2 \times 10^3\) kg/m\(^3\)
Acceleration due to gravity, \(g \approx 9.8\) m/s\(^2\)
Volume of the drop: \[ V = \frac{4}{3}\pi r^3 = \frac{4}{3} \pi (2.0 \times 10^{-5})^3 = \frac{4}{3} \pi (8 \times 10^{-15}) = \frac{32}{3} \pi \times 10^{-15} \text{ m}^3 \] Mass of the drop: \[ m = \rho \times V = (1.2 \times 10^3) \times \left(\frac{32}{3} \pi \times 10^{-15}\right) = 0.4 \times 32 \pi \times 10^{-12} = 12.8 \pi \times 10^{-12} \text{ kg} \] Now, calculate the gravitational force (\(F_g\)): \[ F_g = mg = (12.8 \pi \times 10^{-12}) \times 9.8 \] Using \(\pi \approx 3.14\): \[ F_g \approx (12.8 \times 3.14 \times 10^{-12}) \times 9.8 \approx (40.192 \times 10^{-12}) \times 9.8 \approx 393.88 \times 10^{-12} \text{ N} \] Step 4: Final Answer:
The gravitational force is approximately \(3.9388 \times 10^{-10}\) N.
At terminal velocity, the viscous force equals the gravitational force.
\[ F_v = F_g \approx 3.9 \times 10^{-10} \text{ N} \] This matches option (D).