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 if `a=1`, `b=0`, `c=-1`
C. acceleration if `a=0`, `b=-1`, `c=-2`
D. force of `a=0`, `b=-1`, `c=-2`

Answer

To determine what physical quantity corresponds to the dimensions \( M^a L^b T^c \), we need to analyze the dimensions of various physical quantities and compare them to the given dimensions.

### Step-by-Step Solution:

1. **Understand the Given Dimensions**:
The dimensions of the physical quantity are given as \( M^a L^b T^c \). Here, \( M \) represents mass, \( L \) represents length, and \( T \) represents time.

2. **Identify Common Physical Quantities**:
We will check the dimensions of common physical quantities such as pressure, velocity, acceleration, and force.

3. **Calculate Dimensions of Pressure**:
- Pressure is defined as force per unit area.
- The dimension of force is \( [F] = M L T^{-2} \).
- The dimension of area is \( [A] = L^2 \).
- Therefore, the dimension of pressure is:
\[ [P] = \frac{[F]}{[A]} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]
- This gives us \( a = 1, b = -1, c = -2 \).

4. **Calculate Dimensions of Velocity**:
- Velocity is defined as displacement per unit time.
- The dimension of displacement is \( [L] \) and time is \( [T] \).
- Therefore, the dimension of velocity is:
\[ [v] = \frac{[L]}{[T]} = L T^{-1} \]
- This gives us \( a = 0, b = 1, c = -1 \).

5. **Calculate Dimensions of Acceleration**:
- Acceleration is defined as change in velocity per unit time.
- The dimension of velocity is \( L T^{-1} \).
- Therefore, the dimension of acceleration is:
\[ [a] = \frac{[v]}{[T]} = \frac{L T^{-1}}{T} = L T^{-2} \]
- This gives us \( a = 0, b = 1, c = -2 \).

6. **Calculate Dimensions of Force**:
- Force is defined as mass times acceleration.
- The dimension of mass is \( [M] \) and acceleration is \( L T^{-2} \).
- Therefore, the dimension of force is:
\[ [F] = [M] \cdot [a] = M \cdot (L T^{-2}) = M L T^{-2} \]
- This gives us \( a = 1, b = 1, c = -2 \).

7. **Compare Dimensions**:
- Now we compare the calculated dimensions with the given dimensions \( M^a L^b T^c \).
- From our calculations:
- Pressure: \( a = 1, b = -1, c = -2 \) (matches)
- Velocity: \( a = 0, b = 1, c = -1 \) (does not match)
- Acceleration: \( a = 0, b = 1, c = -2 \) (does not match)
- Force: \( a = 1, b = 1, c = -2 \) (does not match)

8. **Conclusion**:
- The only physical quantity that matches the given dimensions \( M^a L^b T^c \) is pressure.

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