Question

Write dimensional formula of coefficient of viscosity (a) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{-1} \mathrm{~L}^{1} \mathrm{~T}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\)

Short answer

The short answer is: The dimensional formula for the coefficient of viscosity is (d) \([M^1 L^1 T^{-1}]\).

Step-by-step solution

Find formula for viscosity

The first thing we need to do is to find the formula for viscosity. The coefficient of viscosity (η) can be found using the formula for Newton's law of viscosity, which relates the shear stress (τ) to the shear rate (du/dy): \(τ = η \frac{du}{dy}\) where τ is shear stress (force per unit area, N/m^2), η is the coefficient of viscosity (Pa·s), du is the change in velocity (m/s), and dy is the change in height (m).

Express all variables in terms of their dimensions

Now that we have the formula relating the physical quantities, we can express each of the variables in terms of their dimensions. We have: Shear stress (τ) - Force per unit area: [M L T^-2] Shear rate (du/dy) - Change in velocity per change in height: [L T^-1] / [L] = [T^-1] Coefficient of viscosity (η) - We need to find this

Write the relationship in terms of dimensions and solve for the coefficient of viscosity

Now we can write the relationship in terms of the dimensions: [M L T^-2] = [η] x [T^-1] Next, we need to solve for [η]: [η] = [M L T^-2] / [T^-1] [η] = [M L T^-2] x [T^1] [η] = [M L T^-1]

Compare with given options and choose the correct one

The dimension formula we calculated for the coefficient of viscosity is [M L T^-1]. Now let's compare this with the given options in the exercise: (a) [M^1 L^2 T^-1] - Incorrect, as the dimensions don't match. (b) [M^-1 L^1 T^1] - Incorrect, as the dimensions don't match. (c) [M^1 L^-1 T^-1] - Incorrect, as the dimensions don't match. (d) [M^1 L^1 T^-1] - Correct, as the dimensions match our calculated formula. So the correct answer is (d) [M^1 L^1 T^-1].

Key concepts

coefficient of viscosity

The coefficient of viscosity is a fundamental property of fluids that indicates their resistance to flow. In simpler terms, it tells us how "thick" a fluid is. A higher coefficient means the fluid is thicker and requires more force to make it flow, while a lower one means the fluid flows more easily, like water.

In terms of measurements, the coefficient of viscosity, usually denoted by the Greek letter \( \eta \) (eta), is expressed in units of Pascal-seconds (Pa·s) in the SI system. This property plays a crucial role in understanding fluid behavior in various applications, such as when designing engines or piping systems.

By examining the dimensional formula of the coefficient of viscosity, as derived from the exercise, \([M L T^{-1}]\), we observe the interplay between mass, length, and time in fluid dynamics. Here, \(M\) represents mass, \(L\) length, and \(T\) time. This dimensional expression helps in comparing the viscosities of different fluids without directly relying on numeric values.

Newton's law of viscosity

Newton's law of viscosity is a critical concept in fluid mechanics that states that the shear stress between adjacent fluid layers is proportional to the velocity gradient between the layers. In mathematical terms, this is expressed as: \[ \tau = \eta \frac{du}{dy} \]
Here, \( \tau \) is the shear stress, \( \eta \) is the coefficient of viscosity, \( du \) is the change in velocity, and \( dy \) is the change in distance perpendicular to the flow direction.

This law is the functional expression that describes how fluids resist internal flow. Consider honey and water flowing, for instance; honey has a much higher viscosity than water. Consequently, the rate at which layers of honey slide past one another is slower than that of water. This relationship helps engineers and scientists design systems and processes by effectively predicting how a fluid will behave under different conditions.

dimensional analysis

Dimensional analysis is a powerful tool used in physics and engineering to simplify complex problems by examining the dimensions of physical quantities. Instead of dealing with actual numbers, it involves analyzing the type of physical units involved in an equation. This technique helps ensure the accuracy and consistency of equations and conversions.

When approaching problems, dimensional analysis allows us to confirm that both sides of an equation are dimensionally consistent. For instance, in the exercise, we derived the dimensional formula of the coefficient of viscosity as \([M L T^{-1}]\). The process involves determining the basic dimensions of all relevant physical properties (mass \([M]\), length \([L]\), time \([T]\)), which enables us to derive correct relationships between them.

Practically, dimensional analysis simplifies the derivation of equations, allows the checking of derived formulas, and even aids in the estimation of unknown quantities.

shear stress and shear rate

Shear stress and shear rate are two interconnected concepts crucial in the study of fluid dynamics and viscosity.

• Shear stress (\(\tau\)): This represents the force exerted by a fluid per unit area. In the context of viscosity, it's the tangential force that a fluid layer exerts on an adjacent one. Its dimensional formula is \([M L^{-1} T^{-2}]\), indicating how force interacts with area.
• Shear rate (\(\frac{du}{dy}\)): The shear rate describes the change in velocity (\(du\)) per unit change in height or distance perpendicular to the flow (\(dy\)). Its dimensional formula can be simplified to \([T^{-1}]\). This rate is critical in describing how fast the inspired fluid flows.

Understanding these concepts helps in the creation and understanding of models in which fluids behave under stress and flow conditions, which are essential for many industrial and research applications.