If the dimensions 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`

Answer

To determine which physical quantity corresponds to the dimensions \( M^a L^b T^c \), we will analyze the dimensions of various physical quantities step by step.

### Step 1: Understand the Dimensions of Common Physical Quantities
We need to analyze the dimensions of several physical quantities such as force, pressure, velocity, and acceleration.

### Step 2: Analyze Force
- The formula for force is given by \( F = m \cdot a \), where \( m \) is mass and \( a \) is acceleration.
- The dimension of mass \( m \) is \( M \).
- The dimension of acceleration \( a \) is \( L T^{-2} \).
- Therefore, the dimension of force \( F \) is:
\[ [F] = M \cdot (L T^{-2}) = M^1 L^1 T^{-2} \]
- This gives us \( a = 1, b = 1, c = -2 \).

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

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

### Step 5: Analyze Acceleration
- Acceleration is defined as the change in velocity per unit time: \( a = \frac{v}{t} \).
- We already know the dimension of velocity \( v \) is \( L T^{-1} \).
- Therefore, the dimension of acceleration \( a \) is:
\[ [a] = \frac{L T^{-1}}{T} = L^1 T^{-2} \]
- This gives us \( a = 0, b = 1, c = -2 \).

### Step 6: Compare Dimensions
Now, we compare the dimensions obtained for each physical quantity with the given dimensions \( M^a L^b T^c \):

1. **Force**: \( M^1 L^1 T^{-2} \) → \( a = 1, b = 1, c = -2 \)
2. **Pressure**: \( M^1 L^{-1} T^{-2} \) → \( a = 1, b = -1, c = -2 \)
3. **Velocity**: \( L^1 T^{-1} \) → \( a = 0, b = 1, c = -1 \)
4. **Acceleration**: \( L^1 T^{-2} \) → \( a = 0, b = 1, c = -2 \)

### Conclusion
From the analysis, we find that the physical quantity corresponding to the dimensions \( M^a L^b T^c \) can be identified as **Pressure**, which has the dimensions \( M^1 L^{-1} T^{-2} \).