Millikan's oil drop experiment-terminal velocity.

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SUMMARY

The terminal velocity of an oil drop with a radius of 1x10^-6 m at 20 degrees Celsius, calculated using Stokes' law, is determined to be 1.04 x 10^-4 m/s. The mass of the droplet is computed as 3.66 x 10^-15 kg, leading to a calculated fall time of 48 seconds over a distance of 5 mm. The results indicate that the terminal velocity is significantly low, suggesting that small particles may be influenced by air currents rather than simply falling under gravity. This highlights the importance of considering environmental factors in practical scenarios.

PREREQUISITES
  • Understanding of Stokes' law for drag force calculations
  • Knowledge of basic physics concepts such as weight and terminal velocity
  • Familiarity with fluid dynamics, specifically viscosity and its effects
  • Ability to perform calculations involving mass, density, and volume
NEXT STEPS
  • Explore advanced applications of Stokes' law in different fluid mediums
  • Investigate the effects of temperature on air viscosity and terminal velocity
  • Learn about the impact of environmental factors on particle motion in fluids
  • Study the principles of drag force in various contexts, such as in aerodynamics
USEFUL FOR

Students studying physics, particularly those focusing on fluid dynamics and terminal velocity, as well as researchers interested in the behavior of small particles in various environments.

Jenkz
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Homework Statement


Find the terminal velocity of an oil drop with radius 1x10^-6 m at 20 degress under zero field conditions. How long would it take to fall 5mm (range of microscope graticule)?

Homework Equations


Terminal velocity; Weight = Drag
Weight = mg

Assuming Stokes law holds; Drag = 6(pi)rηv
v - terminal velocity
r - radius of droplet
η -viscosity of air, 1.832×10–5 Pa s (at 20 degrees)

Mass, m = (4/3) pi (r^3) \rho
\rho = Density of oil, 874 kgm^-3 (at 20 degrees)

time= distance/ velocity

The Attempt at a Solution



Mass = 3.66 x 10^-15 Kg
v = (mg) / (6(pi)rη) = 1.04 x10^-4 ms^-1

time = 48s

The velocity seems too small, and the time taken is definitely too long. Please help?
 
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I'd say the result is theorically true.
Small particles tends to "fly" in air rather than drop down like stones.

In real world the particle will actually fly away following some slow air flow generated eg. from one near hot lamp, or by the body of someone near.
 
Really? I'm still skeptical about my answers... Thanks though.
 

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