A spherical body of diameter 4cm falls freely through air with the velocity of 0.5m/s. The coefficient of velocity of air is 2*10^-5Nm/m^2.find the viscous force acting on the body.

To find the viscous force acting on the body, we can use Stoke's law, which relates the viscous drag force to the velocity of a spherical object moving through a fluid:

F = 6πηrv

where F is the viscous force, η is the coefficient of viscosity, r is the radius of the spherical body, and v is the velocity of the body.

First, let's convert the diameter of the spherical body from centimeters to meters:

Diameter = 4 cm = 4/100 meters = 0.04 meters

Thus, the radius of the spherical body is half of its diameter:

r = 0.04/2 = 0.02 meters

Next, we are given the velocity of the body, which is 0.5 m/s. Also, the coefficient of viscosity (η) is given as 2 × 10^-5 Nm/m^2.

Now we can substitute the values into the Stoke's law equation to calculate the viscous force:

F = 6π × (2 × 10^-5) × 0.02 × 0.5

Simplifying the equation gives us:

F = 6 × 3.1415 × 2 × 10^-5 × 0.02 × 0.5

F = 0.0000188494 N

Therefore, the viscous force acting on the spherical body is approximately 0.0000188494 Newtons.