According to Newton, the viscous force acting between liquid layers of area A and velocity gradient `(Deltav)/(Deltaz)` is given by `F =- eta A (dv)/(dz)`, where `eta` is constant called

Options

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

Explanation

To solve the problem, we need to find the dimensions of the coefficient of viscosity (η) in the equation for viscous force. The equation given is:

\[ F = -\eta A \frac{dv}{dz} \]

where:
- \( F \) is the viscous force,
- \( A \) is the area,
- \( \frac{dv}{dz} \) is the velocity gradient,
- \( \eta \) is the coefficient of viscosity.

### Step 1: Identify the dimensions of each term in the equation.

1. **Viscous Force (F)**:
The dimensional formula for force is:
\[ [F] = [M][L][T^{-2}] \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

2. **Area (A)**:
The dimensional formula for area is:
\[ [A] = [L^2] \]

3. **Velocity Gradient (\( \frac{dv}{dz} \))**:
The velocity \( v \) has dimensions of:
\[ [v] = [L][T^{-1}] \]
Therefore, the velocity gradient \( \frac{dv}{dz} \) has dimensions:
\[ \left[\frac{dv}{dz}\right] = \frac{[L][T^{-1}]}{[L]} = [T^{-1}] \]

### Step 2: Rearranging the equation to find the dimensions of η.

From the equation \( F = -\eta A \frac{dv}{dz} \), we can rearrange it to solve for \( \eta \):
\[ \eta = -\frac{F}{A \frac{dv}{dz}} \]

### Step 3: Substitute the dimensions into the equation.

Substituting the dimensions we found:
\[ [\eta] = \frac{[F]}{[A][\frac{dv}{dz}]} = \frac{[M][L][T^{-2}]}{[L^2][T^{-1}]} \]

### Step 4: Simplify the expression.

Now, simplifying the dimensions:
\[ [\eta] = \frac{[M][L][T^{-2}]}{[L^2][T^{-1}]} = [M][L^{-1}][T^{-1}] \]

### Conclusion

Thus, the dimensions of the coefficient of viscosity \( \eta \) are:
\[ [\eta] = [M][L^{-1}][T^{-1}] \]

### Final Answer
The coefficient of viscosity (η) has the dimensions of \( [M][L^{-1}][T^{-1}] \).

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