Question
Question: A square plate \[0.1m\] side moves parallel to a second plate with a velocity of \[0.1m{s^{ - 1}}\]....
A square plate 0.1m side moves parallel to a second plate with a velocity of 0.1ms−1. Both plates are immersed in water. If the viscous force is 0.002N and the coefficient of viscosity 0.001poise, distance between the plates is:
(A) 0.1m
(B) 0.05m
(C ) 0.005m
(D) 0.0005m
Solution
We have the values of all required quantities in the question. We would calculate the area of the square plate at first as the dimension of one of the sides is given.
Using that value of area, and the formula for viscous force we can obtain the required value of distance between the plates.
Complete step-by-step answer:
As per the question, the following values of given to us:
Side of the plate =0.1m=a
Velocity of the plate =dv=0.1ms−1
Viscous force =F=0.002N
Coefficient of viscosity =0.001poise
From here, we can calculate the area of the plate is =a2=(0.1×0.1)m2
Thus, area =0.01m2
Coefficient of Viscosity =η=0.001poise
But, we need to take the value in decaPoise, therefore we divide it by 10
Therefore, η=100.001decapoise=0.0001decapoise
We have the velocity of the plate given,
Now, using the expression for viscous force, we get:
F=ηAdxdv
Where:
F= Viscous force
η= Coefficient of viscosity
A= Area of the plate
dv= Velocity with which the plate moves
dx= Distance between the plates.
For the above expression, we need to find the value of dx.
Putting the values, in the expression we get:
0.002=dx0.0001×0.01×0.1
Rearranging the equation we get:
dx=0.0020.0001×0.01×0.1
Thus, we obtain:
dx=0.00005m
This is the required solution.
Note: The unit poise is in the MKS unit, therefore, we need to convert it to deca poise, and otherwise we may get erroneous results. Viscosity is defined as a resistance experienced by a fluid while it flows. This is a property of the fluid. Viscous force is defined as the force between a body and the fluid, while it flows.