Give the quantities for which the following are the dimensions:
`M^(0)L^(0)T^(-2)`

Answer

To solve the question, we need to identify the physical quantity that has the dimensions \( M^0 L^0 T^{-2} \).

### Step-by-Step Solution:

1. **Understanding the Dimensions**:
- The dimensions \( M^0 L^0 T^{-2} \) indicate that the quantity does not depend on mass (M) or length (L) and has a time dimension of \( T^{-2} \). This means it is related to a quantity that involves time but is independent of mass and length.

2. **Identifying the Quantity**:
- A known physical quantity that has the dimension \( T^{-2} \) is angular acceleration (denoted as \( \alpha \)). Angular acceleration is defined as the rate of change of angular velocity.

3. **Angular Velocity and Its Dimensions**:
- Angular velocity (denoted as \( \omega \)) is defined as the rate of change of angular displacement (denoted as \( \theta \)). The angular displacement \( \theta \) is dimensionless (it is measured in radians, which are dimensionless).
- Therefore, the dimension of angular velocity \( \omega \) is \( T^{-1} \) (since it is the change in angle per unit time).

4. **Calculating Angular Acceleration**:
- Angular acceleration \( \alpha \) is given by:
\[ \alpha = \frac{d\omega}{dt} \]
- Since \( \omega \) has dimensions \( T^{-1} \), the dimensions of \( \alpha \) can be calculated as:
\[ \text{Dimensions of } \alpha = \frac{T^{-1}}{T} = T^{-2} \]

5. **Final Conclusion**:
- Since \( \alpha \) has dimensions \( M^0 L^0 T^{-2} \), we conclude that the quantity corresponding to the dimensions \( M^0 L^0 T^{-2} \) is **angular acceleration**.

### Final Answer:
The quantity with dimensions \( M^0 L^0 T^{-2} \) is **angular acceleration** (α).

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