According to Newton, the viscous force acting between liquid layers of area `A` and velocity gradient `Deltav//Deltaz` is given by `F=-etaA(Deltav)/(Deltaz)` where `eta` is constant called coefficient of viscosity. The dimension of `eta` is :

Options

A. `[ML^(2)T^(-2)]`
B. `[ML^(-1)T^(-1)]`
C. `[ML^(-2)T^(-2)]`
D. `[M^(0)L^(0)T^(0)]`

Answer

To find the dimension of the coefficient of viscosity (η), we start with the given formula for the viscous force:

\[ F = -\eta A \frac{\Delta v}{\Delta z} \]

Where:
- \( F \) is the viscous force,
- \( A \) is the area,
- \( \Delta v \) is the change in velocity,
- \( \Delta z \) is the change in distance.

### Step 1: Rearranging the equation
We can rearrange the equation to solve for η:

\[ \eta = -\frac{F}{A \frac{\Delta v}{\Delta z}} \]

### Step 2: Identifying the dimensions
Now, we need to identify the dimensions of each term in the equation.

1. **Dimension of Force (F)**:
The dimension of force is given by:
\[ [F] = MLT^{-2} \]

2. **Dimension of Area (A)**:
The dimension of area is:
\[ [A] = L^2 \]

3. **Dimension of Velocity Gradient (\(\frac{\Delta v}{\Delta z}\))**:
The velocity gradient is the change in velocity per unit distance. The dimension of velocity (\(v\)) is:
\[ [v] = LT^{-1} \]
Therefore, the dimension of velocity gradient is:
\[ \left[\frac{\Delta v}{\Delta z}\right] = \frac{LT^{-1}}{L} = T^{-1} \]

### Step 3: Substituting dimensions into η
Now, substituting the dimensions into the equation for η:

\[ [\eta] = \frac{[F]}{[A] \cdot \left[\frac{\Delta v}{\Delta z}\right]} = \frac{MLT^{-2}}{L^2 \cdot T^{-1}} \]

### Step 4: Simplifying the expression
Now we simplify the expression:

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

### Conclusion
Thus, the dimension of the coefficient of viscosity (η) is:

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

### Final Answer
The dimension of η is \( ML^{-1}T^{-1} \).

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