If m is the mass of drop of a liquid of radius 'r' then `(mg)/(pi r)` has the same dimensions of :

Options

A. surface tension
B. tension
C. Young's modulus
D. coefficient of viscosity

Answer

To solve the problem, we need to determine the dimensions of the expression \(\frac{mg}{\pi r}\) and compare it with the dimensions of the given options.

### Step 1: Identify the components of the expression
The expression we need to analyze is \(\frac{mg}{\pi r}\), where:
- \(m\) is the mass of the liquid drop,
- \(g\) is the acceleration due to gravity,
- \(r\) is the radius of the drop.

### Step 2: Write down the dimensions of each component
1. **Mass (\(m\))**: The dimension of mass is \([M^1 L^0 T^0]\).
2. **Acceleration due to gravity (\(g\))**: The dimension of acceleration is given by \([L^1 T^{-2}]\).
3. **Radius (\(r\))**: The dimension of length is \([L^1]\).

### Step 3: Calculate the dimensions of \(mg\)
Now, we can find the dimensions of the product \(mg\):
\[ \text{Dimensions of } mg = [M^1][L^1 T^{-2}] = [M^1 L^1 T^{-2}] \]

### Step 4: Calculate the dimensions of \(\frac{mg}{\pi r}\)
Next, we divide \(mg\) by \(r\):
\[ \text{Dimensions of } \frac{mg}{\pi r} = \frac{[M^1 L^1 T^{-2}]}{[L^1]} = [M^1 L^{1-1} T^{-2}] = [M^1 L^0 T^{-2}] = [M^1 T^{-2}] \]

### Step 5: Compare with the dimensions of the options
Now, we need to find which of the given options has the same dimensions as \([M^1 T^{-2}]\):

1. **Surface Tension**: Surface tension is defined as force per unit length.
- Dimensions of force: \([M^1 L^1 T^{-2}]\)
- Dimensions of length: \([L^1]\)
- Therefore, dimensions of surface tension: \(\frac{[M^1 L^1 T^{-2}]}{[L^1]} = [M^1 L^{-2}]\)

2. **Tension**: Tension is a force, so its dimensions are \([M^1 L^1 T^{-2}]\).

3. **Young's Modulus**: Young's modulus is defined as stress (force per unit area) divided by strain (dimensionless).
- Dimensions of stress: \(\frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]\)
- Therefore, dimensions of Young's modulus: \([M^1 L^{-1} T^{-2}]\).

4. **Coefficient of Viscosity**: Coefficient of viscosity is defined as shear stress divided by shear rate.
- Dimensions of shear stress: \(\frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]\)
- Therefore, dimensions of coefficient of viscosity: \([M^1 L^{-1} T^{-2}]\).

### Conclusion
The dimensions of \(\frac{mg}{\pi r}\) are \([M^1 T^{-2}]\), which matches the dimensions of tension. Thus, the correct answer is:

**Tension**.