Give the quantities for which the following are the dimensions:
`M^(0)L^(0)T^(0)`

Answer

To solve the question of identifying quantities that have the dimensions \( M^0 L^0 T^0 \), we need to understand what this notation means. The notation \( M^0 L^0 T^0 \) indicates that the quantity is dimensionless, meaning it does not depend on mass (M), length (L), or time (T).

### Step-by-Step Solution:

1. **Understanding Dimensionless Quantities**:
- A dimensionless quantity is one that has no physical dimensions associated with it. This means that it is a pure number and does not change with the system of units used.

2. **Identifying Examples**:
- **Strain**:
- Strain is defined as the change in length divided by the original length.
- Mathematically, it is expressed as:
\[ \text{Strain} = \frac{\Delta L}{L} \]
- Both \(\Delta L\) and \(L\) have the dimension of length (L), so:
\[ \text{Dimensions of Strain} = \frac{L}{L} = L^0 \]
- Therefore, strain is dimensionless: \( M^0 L^0 T^0 \).

3. **Another Example - Angle**:
- An angle is defined as the ratio of the length of an arc to the radius of the circle.
- Mathematically, it is expressed as:
\[ \text{Angle} = \frac{\text{Arc Length}}{\text{Radius}} = \frac{L}{L} \]
- Thus, the dimensions of angle are also:
\[ \text{Dimensions of Angle} = \frac{L}{L} = L^0 \]
- Hence, angle is dimensionless: \( M^0 L^0 T^0 \).

4. **Specific Gravity**:
- Specific gravity is defined as the ratio of the density of a substance to the density of a reference substance (usually water).
- Mathematically, it is expressed as:
\[ \text{Specific Gravity} = \frac{\text{Density of Substance}}{\text{Density of Reference}} = \frac{ML^{-3}}{ML^{-3}} \]
- Therefore, the dimensions of specific gravity are:
\[ \text{Dimensions of Specific Gravity} = \frac{ML^{-3}}{ML^{-3}} = M^0 L^0 T^0 \]
- Thus, specific gravity is also dimensionless: \( M^0 L^0 T^0 \).

5. **Conclusion**:
- The quantities that have the dimensions \( M^0 L^0 T^0 \) include:
- Strain
- Angle
- Specific Gravity