If the dimension of a physical quantity are given by `M^a L^b T^c,` then the physical quantity will be

Options

A. Force if a = 0, b = –1, c = –2
B. Pressure if a = 1, b = –1, c = –2
C. Velocity if a = 1, b = 0, c = –1
D. Acceleration if a = 1, b = 1, c = –2

Explanation

To determine the physical quantity based on its dimensions given by \( M^a L^b T^c \), we need to analyze the dimensions of various physical quantities and compare them with the provided dimensions.

### Step-by-Step Solution:

1. **Identify the Dimensions of Common Physical Quantities:**
- **Velocity (v)**:
\[ v = \frac{\text{displacement}}{\text{time}} = \frac{L}{T} \implies [v] = M^0 L^1 T^{-1} \]
Here, \( a = 0, b = 1, c = -1 \).

2. **Check if the Dimensions Match:**
- Given \( M^a L^b T^c \), if \( a = 1 \), \( b = 1 \), and \( c = -1 \), then velocity does not match because \( a \neq 1 \).

3. **Next, Analyze Acceleration (a):**
- **Acceleration**:
\[ a = \frac{\text{change in velocity}}{\text{time}} = \frac{L/T}{T} = \frac{L}{T^2} \implies [a] = M^0 L^1 T^{-2} \]
Here, \( a = 0, b = 1, c = -2 \).

4. **Check if the Dimensions Match:**
- Again, since \( a \neq 1 \), acceleration does not match the given dimensions.

5. **Now, Analyze Force (F):**
- **Force**:
\[ F = m \cdot a \implies [F] = [M] \cdot [A] = M^1 L^1 T^{-2} \]
Here, \( a = 1, b = 1, c = -2 \).

6. **Check if the Dimensions Match:**
- The dimensions of force match \( M^1 L^1 T^{-2} \), but we need to check if this corresponds to the given dimensions.

7. **Lastly, Analyze Pressure (P):**
- **Pressure**:
\[ P = \frac{F}{A} = \frac{M^1 L^1 T^{-2}}{L^2} = M^1 L^{-1} T^{-2} \]
Here, \( a = 1, b = -1, c = -2 \).

8. **Check if the Dimensions Match:**
- The dimensions of pressure also do not match the given dimensions.

9. **Conclusion:**
- The only physical quantity that corresponds to the dimensions \( M^a L^b T^c \) with \( a = 1, b = -1, c = -2 \) is **Pressure**.

### Final Answer:
The physical quantity corresponding to the dimensions \( M^a L^b T^c \) is **Pressure**.