If `eta` represents the coefficient of viscosity and T the surface tension. Then the dimensions of `T//eta` are same as that of

Options

A. length
B. mass
C. time
D. speed

Answer

To solve the problem, we need to find the dimensions of the expression \( \frac{T}{\eta} \), where \( T \) is the surface tension and \( \eta \) is the coefficient of viscosity.

### Step-by-Step Solution:

1. **Identify the Dimensions of Surface Tension (T)**:
Surface tension is defined as force per unit length. The dimension of force is given by:
\[ [F] = [M][L][T^{-2}] \]
Therefore, the dimensions of surface tension \( T \) can be expressed as:
\[ [T] = \frac{[F]}{[L]} = \frac{[M][L][T^{-2}]}{[L]} = [M][T^{-2}] \]

2. **Identify the Dimensions of Coefficient of Viscosity (η)**:
The coefficient of viscosity \( \eta \) is defined as the ratio of shear stress to shear rate. Shear stress has dimensions of force per unit area, and shear rate has dimensions of time inverse. Thus:
\[ [\text{Shear Stress}] = \frac{[F]}{[A]} = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}] \]
The shear rate is defined as the change in velocity per unit distance, which has dimensions:
\[ [\text{Shear Rate}] = \frac{[L][T^{-1}]}{[L]} = [T^{-1}] \]
Therefore, the dimensions of the coefficient of viscosity \( \eta \) are:
\[ [\eta] = \frac{[\text{Shear Stress}]}{[\text{Shear Rate}]} = \frac{[M][L^{-1}][T^{-2}]}{[T^{-1}]} = [M][L^{-1}][T^{-1}] \]

3. **Calculate the Dimensions of \( \frac{T}{\eta} \)**:
Now, we can find the dimensions of the expression \( \frac{T}{\eta} \):
\[ \frac{T}{\eta} = \frac{[M][T^{-2}]}{[M][L^{-1}][T^{-1}]} = \frac{[M][T^{-2}]}{[M][L^{-1}][T^{-1}]} = [L][T^{-1}] \]

4. **Interpret the Result**:
The dimensions \( [L][T^{-1}] \) represent velocity (or speed).

Thus, the dimensions of \( \frac{T}{\eta} \) are the same as that of speed.

### Final Answer:
The dimensions of \( \frac{T}{\eta} \) are the same as that of speed.