If X represents force, Y represents area and Z represents time. The quantities having same dimensions as coefficient of viscosity in are

Options

A. `(XY)/(Z)`
B. `(XZ)/(Y)`
C. `(YZ)/(X)`
D. `(X)/(YZ)`

Explanation

To find the quantities having the same dimensions as the coefficient of viscosity, we can start by recalling the definition and formula for the coefficient of viscosity (η).

### Step-by-Step Solution:

1. **Understanding the Coefficient of Viscosity**:
The coefficient of viscosity (η) is defined in the context of Newton's law of viscosity, which states that the shear stress (τ) is proportional to the shear rate (du/dy). The formula can be expressed as:
\[ \tau = \eta \frac{du}{dy} \]

2. **Relating Shear Stress and Shear Rate**:
Shear stress (τ) is defined as force (F) per unit area (A):
\[ \tau = \frac{F}{A} \]
Therefore, we can rewrite the equation for shear stress in terms of force and area:
\[ \frac{F}{A} = \eta \frac{du}{dy} \]

3. **Rearranging for Coefficient of Viscosity**:
Rearranging the equation gives us:
\[ \eta = \frac{F}{A} \cdot \frac{dy}{du} \]

4. **Substituting Dimensions**:
- Force (F) has dimensions of \( [M L T^{-2}] \)
- Area (A) has dimensions of \( [L^2] \)
- The shear rate \( \frac{du}{dy} \) has dimensions of \( [T^{-1}] \) since it is a velocity gradient (change in velocity over change in distance).

5. **Calculating Dimensions of η**:
Plugging in the dimensions into the equation:
\[ \eta = \frac{[M L T^{-2}]}{[L^2]} \cdot [T^{-1}] = [M L^{-1} T^{-1}] \]

6. **Conclusion**:
The dimensions of the coefficient of viscosity (η) are \( [M L^{-1} T^{-1}] \). Thus, any quantity that has the same dimensions as \( [M L^{-1} T^{-1}] \) will be equivalent to the coefficient of viscosity.