The viscous force on an oil drop is measured to be 6.0 10-13 N when the drop is falling through air with a speed of 9.0 10-4 m/s. If the radius of the drop is 2.5 10-6 m, what is the viscosity of air?

Answer in N x s/m^2

To find the viscosity of air, we can use Stoke's law, which relates the viscous force on an object moving through a viscous medium to its velocity and the properties of the medium. The formula for Stoke's law is:

F = 6πηrv

Where:
F is the viscous force on the object,
η is the viscosity of the medium,
r is the radius of the object, and
v is the velocity of the object.

In this case, the viscous force (F) is given as 6.0 × 10^−13 N, the radius (r) is 2.5 × 10^−6 m, and the velocity (v) is 9.0 × 10^−4 m/s.

Let's rearrange Stoke's law to solve for the viscosity (η):

η = F / (6πrv)

Substituting the given values:

η = (6.0 × 10^−13 N) / (6π × 2.5 × 10^−6 m × 9.0 × 10^−4 m/s)

To simplify the calculation, we can perform the multiplication and division separately:

η = (6.0 × 10^−13 N) / (6π) × (2.5 × 10^−6 m) × (9.0 × 10^−4 m/s)

Now, let's evaluate the numerator:

(6.0 × 10^−13 N) / (6π) ≈ 3.18 × 10^−15 N

Substituting this result back into the equation:

η ≈ (3.18 × 10^−15 N) / [(2.5 × 10^−6 m) × (9.0 × 10^−4 m/s)]

η ≈ 1.41 × 10^−5 N·s/m²

Therefore, the viscosity of air is approximately 1.41 × 10^−5 N·s/m².