Question

The sphere of diameter 0.02 m is falls in the uid of kinematic viscosity 10 stokes with the terminal velocity of 0.02 m/s. What is the value of coefcient of drag on the falling sphere?

Answer 1

Answer:

The drag force$=3.76\times10^{-2}\ \rm N$

Explanation:

Given:

Diameter of the sphere=d=0.02 m

Radius of the sphere=0.01 m

kinematic viscosity $\eta$=10 stokes

Terminal velocity v=0.02 m/s

We know that the the drag force is force is given by

$F_s=6\pi R \eta v$

where

• $F_s$ is the drag force
• R is the radius of the sphere

$F_s=6\times 3.14\times 0.01 \times 10\times 0.02\\=3.76\times10^{-2}\ \rm N$

Hence the force is calculated.