In the dimension of a physical quantities are given by `M^(0)L^(1)T^(0)` , then the physical quantity will be

Options

A. pressure if `a=1, b= -1, c= -2`
B. velocity if `a=1, b=0, c= -1`
C. acceleration if `a=1, b=1, c= -2`
D. force if `a=0, b= -1, c= -2`

Explanation

To determine the physical quantity corresponding to the given dimensions \( M^{0}L^{1}T^{0} \), we will analyze the dimensions step by step.

### Step-by-Step Solution:

1. **Understanding the Dimensions**:
- The given dimensions are \( M^{0}L^{1}T^{0} \).
- This means that the physical quantity does not depend on mass (M), depends linearly on length (L), and does not depend on time (T).

2. **Identifying the Physical Quantity**:
- Since the dimension is \( L^{1} \), we are looking for a quantity that is measured in units of length.
- Common physical quantities that depend only on length include distance, displacement, and certain types of linear measurements.

3. **Analyzing the Options**:
- Let's consider the options provided in the question:
- **Pressure**: The dimension of pressure is \( M^{1}L^{-1}T^{-2} \) (not matching).
- **Velocity**: The dimension of velocity is \( M^{0}L^{1}T^{-1} \) (not matching).
- **Acceleration**: The dimension of acceleration is \( M^{0}L^{1}T^{-2} \) (not matching).
- **Force**: The dimension of force is \( M^{1}L^{1}T^{-2} \) (not matching).

4. **Conclusion**:
- The only physical quantity that fits the dimension \( M^{0}L^{1}T^{0} \) is a quantity that is purely length-based, such as "length" itself or "displacement".
- Therefore, the physical quantity corresponding to the dimension \( M^{0}L^{1}T^{0} \) is **length**.