Give the quantities for which the following are the dimensions:
`M^(1)T^(-2)`

Answer

To find the physical quantities that have the dimensions \( M^1 T^{-2} \), we can analyze the dimensions of various physical concepts.

### Step-by-Step Solution:

1. **Understanding the Dimensions**:
The given dimensions are \( M^1 T^{-2} \). This indicates that the quantity has a mass dimension of 1 and a time dimension of -2.

2. **Identifying Relevant Physical Quantities**:
We can start by recalling some fundamental physical quantities that might fit these dimensions. Two common quantities that often come to mind are force and surface tension.

3. **Calculating the Dimensions of Force**:
- The dimension of force is given by Newton's second law, \( F = ma \) (Force = mass × acceleration).
- Here, mass \( m \) has dimensions \( M^1 \) and acceleration \( a \) has dimensions \( L T^{-2} \).
- Therefore, the dimensions of force can be calculated as:
\[ [F] = [m][a] = M^1 (L T^{-2}) = M^1 L^1 T^{-2} \]
- However, we need to find a quantity that has dimensions of \( M^1 T^{-2} \), so we will consider surface tension next.

4. **Calculating the Dimensions of Surface Tension**:
- Surface tension \( \gamma \) is defined as force per unit length.
- Thus, the dimensions of surface tension can be expressed as:
\[ [\gamma] = \frac{[F]}{[L]} = \frac{M^1 L^1 T^{-2}}{L^1} = M^1 T^{-2} \]
- This matches the required dimensions.

5. **Calculating the Dimensions of Spring Constant**:
- The spring constant \( k \) is defined as the force per unit displacement.
- Thus, the dimensions of spring constant can be expressed as:
\[ [k] = \frac{[F]}{[L]} = \frac{M^1 L^1 T^{-2}}{L^1} = M^1 T^{-2} \]
- This also matches the required dimensions.

### Conclusion:
The quantities that have the dimensions \( M^1 T^{-2} \) are:
- Surface Tension
- Spring Constant