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The dimensions of coefficient of viscosity are _______.
- a)ML-1T-1
- b)ML-1T-2
- c)ML-2T-2
- d)ML-2T-1
Correct answer is option 'A'. Can you explain this answer?
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The dimensions of coefficient of viscosity are _______.a)ML-1T-1b)ML-1...
Coefficient of viscosity, η=Fdx/Adv
force F=[MLT^-2]
Area A=[L^2]
dv/dx=[LT^-1]/L=T^-1
Coefficient of viscosity, η=[MLT^-2]/[L^2][T^-1]
Coefficient of viscosity, η=[ML^-1T^-1]
force F=[MLT^-2]
Area A=[L^2]
dv/dx=[LT^-1]/L=T^-1
Coefficient of viscosity, η=[MLT^-2]/[L^2][T^-1]
Coefficient of viscosity, η=[ML^-1T^-1]
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The dimensions of coefficient of viscosity are _______.a)ML-1T-1b)ML-1...
The dimensions of the coefficient of viscosity are ML-1T-1.
Viscosity is a measure of a fluid's resistance to flow. It is also known as the internal friction of a fluid. The coefficient of viscosity, denoted by the symbol η (eta), represents the proportionality constant between the shear stress and the velocity gradient in a fluid.
To understand the dimensions of the coefficient of viscosity, we need to consider the dimensions of the variables involved in the equation for shear stress and velocity gradient.
Shear stress (τ) is defined as the force per unit area acting tangentially to the surface of a fluid. It can be expressed as:
τ = η(dv/dy)
where η is the coefficient of viscosity and dv/dy is the velocity gradient.
The dimensions of shear stress are:
[τ] = [Force/Area] = [MLT-2/L2] = [ML-1T-2]
The dimensions of velocity gradient are:
[dv/dy] = [Velocity/Distance] = [LT-1/L] = [T-1]
Now, substituting the dimensions of shear stress and velocity gradient into the equation, we get:
[ML-1T-2] = η[T-1]
Simplifying the equation, we find that the dimensions of the coefficient of viscosity are:
[η] = ML-1T-1
Therefore, the correct answer is option 'A' - ML-1T-1.
Viscosity is a measure of a fluid's resistance to flow. It is also known as the internal friction of a fluid. The coefficient of viscosity, denoted by the symbol η (eta), represents the proportionality constant between the shear stress and the velocity gradient in a fluid.
To understand the dimensions of the coefficient of viscosity, we need to consider the dimensions of the variables involved in the equation for shear stress and velocity gradient.
Shear stress (τ) is defined as the force per unit area acting tangentially to the surface of a fluid. It can be expressed as:
τ = η(dv/dy)
where η is the coefficient of viscosity and dv/dy is the velocity gradient.
The dimensions of shear stress are:
[τ] = [Force/Area] = [MLT-2/L2] = [ML-1T-2]
The dimensions of velocity gradient are:
[dv/dy] = [Velocity/Distance] = [LT-1/L] = [T-1]
Now, substituting the dimensions of shear stress and velocity gradient into the equation, we get:
[ML-1T-2] = η[T-1]
Simplifying the equation, we find that the dimensions of the coefficient of viscosity are:
[η] = ML-1T-1
Therefore, the correct answer is option 'A' - ML-1T-1.
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