Question

Can someone help solve and explain this question?
The viscous drag F between two liquid layers of surface area of contact A in the region of velocity gradient dv/dx is given by F=nA dv/dx where n is the coefficient of viscosity of the liquid.Find the dimension of n and write down its unit in SI base units.

Answer 1

You can apply the pi theorem. Or basically, knowing the dimensions of the other variables, you can know the unknown.

The equation,

\[F=\eta A \frac{dv}{dx}\]

The same equation, but with dimensions,
\[\mathrm{N}=\eta \mathrm{\ m^2}\frac{\mathrm{1}}{\mathrm{s}}\]
Solve to obtain,
\[[\eta]=\frac{\mathrm{Ns}}{\mathrm{m}^2}=Pa\cdot s\]

Answer 2

I don't understand...And the book answer is [ML ^{-1}T ^{-1}\]

Answer 3

ok, let's put all the magnitudes in international units,
\[F=Force=newton=Kg⋅m/s^2=[M][L]/[T]^2\\
A=Area=m^2=[L]^2\\
dv/dx=[T]^{-1}\] Now, from the original equation, you can obtain,
\[η=\frac{F}{A⋅dv/dx}=\frac{[M][L][T]^{-2}}{[L]^2[T]^{-1}}=[M][L]^{-1}[T]^{-1}\]
Use the equation I gave you to obtain the correct solution.

Answer 4

ok...Thanks for your help...

Answer 5

And U know what is Young Modulus ?

Answer 6

You can find more information in wikipedia,

http://en.wikipedia.org/wiki/Young's_modulus

But basically, it tells you the pressure you need to do in order to deformate something.

Answer 7

Or the force per unit area and per unit of deformation.

Answer 8

ok...thanks...This is another question . How do you check the dimensional consistency of a formula? Why can't this method give definite confirmation that an equation is correct? The first part I can do but the second part I can't do...Please help me...

Answer 9

The dimensional consistency can be checked if the terms at the right and at the left of a dimensional equation are the same.

This method only gives information about the functional dependency but not about constants or adimensional factors that could be included in the correct equation. As an example, dimensional analysis will let you obtain the relation between time, length and gravity in a pendulum, but will not give you the 2 pi factor that is included in the right formula for the period.

Answer 10

Thanks...^_^

Answer 11

State two ways in which a dimensionally consistent equation may be physically incorrect.