An oil drop is of radius 2.0 x 10-5m and density 1.2 x 103 kg/m3. The viscosity of air at the temperature of experiment is 1.8 x 10-5 Ns/m2. Then

Question

An oil drop is of radius `2.0 x 10^-5 m` and density `1.2 x 10^3 kg/m^3`. The viscosity of air at the temperature of experiment is `1.8 x 10^-5 Ns/m^2`. Then

🧐 Not the exact question you are looking for?Go ask a question

Solution

It seems like you're asking for the terminal velocity of the oil drop, but you didn't specify it in the question. The terminal velocity can be calculated using Stokes' Law, which is given by:

v = (2/9) * (r^2) * (ρ - ρ') * g / η

where:

v is the terminal velocity,
r is the radius of the oil drop,
ρ is the density of the oil drop,
ρ' is the density of the air (which is approximately 1.2 kg/m^3),
g is the acceleration due to gravity (approximately 9.81 m/s^2), and
η is the dynamic viscosity of the air.

However, since the density of the oil drop and the air are the same in this case, the oil drop will not fall, meaning the terminal velocity is zero.

This problem has been solved

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.

An oil drop is of radius 2.0 x 10-5m and density 1.2 x 103 kg/m3. The viscosity of air at the temperature of experiment is 1.8 x 10-5 Ns/m2. Then

Question