Question

A container filled with viscous liquid is moving vertically downwards with constant speed \[3{v_0}\] . At the instant shown, a sphere of radius \[r\] is moving vertically downwards (in liquid) has speed \[{v_0}\] . The coefficient of viscosity is \[\eta \] . There is no relative motion between the liquid and the container. Then at the shown instant, the magnitude of viscous force acting on sphere is:

Answer 1

Hint: First of all, we will find out the speed of the sphere with respect to the fluid. After that we will use the formula of viscous force and substitute the required values, manipulate accordingly and obtain the result.


Complete step by step answer:
In the given question, we are supplied the following data:
The liquid inside the contained is moving vertically downward with a constant speed of \[3{v_0}\] .
At that moment, a sphere is dropped inside the liquid which is moving down at a speed of \[{v_0}\] .
The radius of the sphere is given as \[r\] .
The coefficient of viscosity is given as \[\eta \] .
We are asked to find out the magnitude of viscous force which is acting on the sphere.
To find, we proceed as follows:
Whenever a body or a sphere moves through a liquid or fluid, then it must overcome the frictional force of the liquid to continue the motion inside it. The viscous force which is the drag force is proportional to the radius of the object or size of the object and the velocity at which the sphere of the body is moving inside the fluid.
The relative velocity of the spherical ball with respect to the fluid is given by subtraction of the velocities of the fluid and the ball, which is given by:
\[
v' = 3{v_0} - {v_0} \\
v' = 2{v_0} \\
\]
We apply the formula, which gives the viscous force, which is written in mathematical form as:
\[F = 6\pi r\eta v'\] …… (1)
Where,
\[F\] indicates viscous force.
\[r\] indicates radius of the sphere.
\[\eta \] indicates viscosity of the fluid.
\[v'\] indicates the relative velocity of the sphere with respect to the fluid.
Now, by substituting the required values in the equation (1), we get:
\[
F = 6\pi r\eta v \\
F = 6\pi r\eta \times 2{v_0} \\
F = 12\pi r\eta {v_0} \\
\]
Hence, the magnitude of viscous force which is acting on the sphere is \[12\pi r\eta {v_0}\] .

Note: While solving this problem, many students tend to make mistakes by taking account directly the speed of the fluid while finding the viscous force, which is wrong. We need to find out the relative speed of the ball with respect to the fluid. \[\eta \] is a constant which is the property of a material and it does not depend on the geometry of the body.