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

Options

A. pressure if `a = 1, b =- 1, c =- 2`
B. velocity is `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 represented by the dimensions \( M^a L^b T^c \), we can analyze various physical quantities and compare their dimensions with the given values of \( a \), \( b \), and \( c \). Here’s a step-by-step solution:

### Step 1: Understand the Dimensions of Common Physical Quantities
We will start by recalling the dimensions of some common physical quantities:

1. **Velocity**:
- Unit: \( \text{m/s} \)
- Dimension: \( L^1 T^{-1} \)
- This means \( a = 0, b = 1, c = -1 \).

2. **Acceleration**:
- Unit: \( \text{m/s}^2 \)
- Dimension: \( L^1 T^{-2} \)
- This means \( a = 0, b = 1, c = -2 \).

3. **Force**:
- Unit: \( \text{N} = \text{kg} \cdot \text{m/s}^2 \)
- Dimension: \( M^1 L^1 T^{-2} \)
- This means \( a = 1, b = 1, c = -2 \).

4. **Pressure**:
- Unit: \( \text{N/m}^2 \)
- Dimension: \( \text{N} = M^1 L^1 T^{-2} \) divided by \( L^2 \)
- Thus, \( \text{Pressure} = \frac{M^1 L^1 T^{-2}}{L^2} = M^1 L^{-1} T^{-2} \)
- This means \( a = 1, b = -1, c = -2 \).

### Step 2: Compare with Given Dimensions
Now, we compare the dimensions \( M^a L^b T^c \) with the dimensions we calculated for each physical quantity.

- For **Velocity**: \( a = 0, b = 1, c = -1 \) (not matching)
- For **Acceleration**: \( a = 0, b = 1, c = -2 \) (not matching)
- For **Force**: \( a = 1, b = 1, c = -2 \) (not matching)
- For **Pressure**: \( a = 1, b = -1, c = -2 \) (matches)

### Conclusion
The physical quantity represented by the dimensions \( M^a L^b T^c \) is **Pressure**, with \( a = 1, b = -1, c = -2 \).

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