Which one of the following represents the correct dimensions of the coefficient of viscosity?
(a) [M

Q: What exactly is a 'velocity gradient'?

A velocity gradient measures how fast the velocity of the fluid changes as you move across the layers. If you move a small distance $\Delta z$ perpendicular to the flow, and the velocity changes by $\Delta v$, the gradient is $\frac{\Delta v}{\Delta z}$. Its dimension is simply $[T^{-1}]$ because the length dimensions cancel out.

Q: Why is the dimension of force $[MLT^{-2}]$?

Force is defined as mass times acceleration ($F = ma$). Mass has the dimension $[M]$, and acceleration (which is velocity per unit time) has the dimension $[LT^{-2}]$. Multiplying them gives $[MLT^{-2}]$.

Q: Sometimes I see a negative sign in Newton's law of viscosity. Does that change the dimensions?

Great observation! The negative sign indicates that the viscous force acts in the opposite direction to the fluid's motion (it's a resistive force). However, dimensions only care about the physical nature of the quantities, not their direction or sign. So, the negative sign has absolutely no effect on the dimensional formula.

Q: Can I find the dimensions of $\eta$ using Stokes' Law instead?

Yes, absolutely! Stokes' Law states that the viscous drag force on a small sphere is $F = 6\pi\eta r v$. If you rearrange this to $\eta = \frac{F}{6\pi r v}$ and substitute the dimensions ($[F] = [MLT^{-2}]$, $[r] = [L]$, $[v] = [LT^{-1}]$), you will get the exact same result: $[ML^{-1}T^{-1}]$. The dimensions of a physical constant are always the same, regardless of the formula used.

Q: What are the SI units of the coefficient of viscosity based on this dimensional formula?

Looking at the dimensional formula $[ML^{-1}T^{-1}]$, we can directly translate this to SI units. Mass $[M]$ is in kilograms (kg), Length $[L^{-1}]$ is in meters inverse ($m^{-1}$), and Time $[T^{-1}]$ is in seconds inverse ($s^{-1}$). So, the SI unit is $kg\cdot m^{-1}\cdot s^{-1}$, which is also known as the Poiseuille (Pl) or Pascal-second ($Pa\cdot s$).

Question

Which one of the following represents the correct dimensions of the coefficient of viscosity?

Options:

(A) $[ML^{-1}T^{-2}]$

(B) $[MLT^{-1}]$

(C) $[ML^{-1}T^{-1}]$

(D) $[ML^{-2}T^{-2}]$

Sigma Percentile · Accepted Answer

### The Physics of Flow: Understanding Viscosity

Have you ever wondered why honey flows so much slower than water? The secret lies in a property called **viscosity**, which is essentially the internal friction of a fluid. Imagine a fluid flowing through a pipe; it doesn't move as a single solid block. Instead, it moves in microscopic layers. The layer touching the pipe's wall is practically stationary, while the layer in the exact center moves the fastest.

Because adjacent layers are moving at different speeds, they rub against each other. The slower layer tries to hold back the faster layer, and the faster layer tries to drag the slower layer along. To maintain this flow, a constant force must be applied. This brings us to **Newton's Law of Viscous Force**.

### The Master Equation

Sir Isaac Newton observed that the viscous force $F$ required to maintain a constant velocity difference between two parallel layers of a fluid is directly proportional to the area $A$ of the layers and the **velocity gradient** $\frac{\Delta v}{\Delta z}$ between them.

The velocity gradient simply tells us how rapidly the velocity changes as we move perpendicular to the direction of flow. Mathematically, this is expressed as:

$$F = \eta A \frac{\Delta v}{\Delta z}$$

Here, the proportionality constant $\eta$ (eta) is the star of our show: the **coefficient of viscosity**. It is a measure of how much resistance the fluid offers to flow.

### Dimensional Deconstruction

To find the dimensional formula of $\eta$, we first need to isolate it in our master equation:

$$\eta = \frac{F}{A \left(\frac{\Delta v}{\Delta z}\right)}$$

Now, let's break down the dimensions of each component piece by piece.

First, we have **Force** ($F$). From Newton's second law ($F = ma$), force is mass times acceleration. Therefore, its dimensional formula is:

$$[F] = [MLT^{-2}]$$

Next is **Area** ($A$), which is simply length squared:

$$[A] = [L^2]$$

Then, we tackle the **velocity gradient** $\frac{\Delta v}{\Delta z}$. Velocity has dimensions of $[LT^{-1}]$, and the distance $\Delta z$ has dimensions of $[L]$. Dividing them gives:

$$\left[\frac{\Delta v}{\Delta z}\right] = \frac{[LT^{-1}]}{[L]} = [T^{-1}]$$

### The Final Calculation

Now, we substitute all these individual dimensional formulas back into our isolated equation for $\eta$:

$$[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]}$$

Let's carefully simplify this. The mass $[M]$ remains as it is. For length, we have $[L]$ in the numerator and $[L^2]$ in the denominator, which simplifies to $[L^{1-2}] = [L^{-1}]$. For time, we have $[T^{-2}]$ in the numerator and $[T^{-1}]$ in the denominator, which simplifies to $[T^{-2 - (-1)}] = [T^{-1}]$.

Putting it all together, we arrive at the final dimensional formula for the coefficient of viscosity:

$$[\eta] = [ML^{-1}T^{-1}]$$

This elegant result perfectly matches option (c). Dimensional analysis is a powerful tool that not only helps us verify equations but also deepens our understanding of the physical nature of constants like $\eta$.

Animated Solution for Physics - Units and Measurements: Which one of the following represents the correct dimensions of the coefficient of viscosity?

Select Answer:

Visualized Solution

Newton's Law of Viscosity

Isolating $η$

Dimensions of Force and Area

Dimension of Velocity Gradient

Final Calculation

Conclusion

The Sigma Insight: Dimensional Analysis

The Physics of Flow

The Master Equation

Dimensional Deconstruction

The Final Calculation

Similar Questions

A quantity $x$ is given by $x = \frac{I F v ^{2}}{W L ^{4}}$ where, $I$ is moment of inertia, $F$ is force, $v$ is velocity, $W$ is work and $L$ is length. The dimensional formula for $x$ is same as that of

Dimensions of $\frac{1}{μ _{0} ε _{0}}$ , where symbols have their usual meaning, are

In SI units, the dimensions of $\frac{ε _{0}}{μ _{0}}$ is

The dimension of $\frac{B ^{2}}{2 μ _{0}}$ , where $B$ is magnetic field and $μ_{0}$ is the magnetic permeability of vacuum, is

The pairs of physical quantities that have the same dimensions is (are )

In the relation, $p = \frac{α}{β} e^{- \frac{α Z}{k θ}}$ $p$ is pressure, $Z$ is distance, $k$ is Boltzmann's constant and $θ$ is the temperature. The dimensional formula of $β$ will be

The dimensions of magnetic field in $M, L, T$ and $C$ (coulomb) is given as

If speed ( $v$ ), acceleration ( $A$ ) and force ( $F$ ) are considered as fundamental units, the dimension of Young's modulus will be

Write the dimensions of the following in terms of mass, time, length and charge. (a) Magnetic flux (b) Rigidity modulus

If speed $V$ , area $A$ and force $F$ are chosen as fundamental units, then the dimensional formula of Young's modulus will be