The dimensional formula `[ML^(-1)T^(-2)]` is for the quantity

Options

A. Force
B. Acceleration
C. Pressure
D. Work

Answer

To determine the quantity corresponding to the dimensional formula \([ML^{-1}T^{-2}]\), we will analyze each option provided.

### Step 1: Analyze the given dimensional formula
The dimensional formula is \([ML^{-1}T^{-2}]\). This indicates:
- \(M\) represents mass,
- \(L^{-1}\) indicates an inverse relationship with length,
- \(T^{-2}\) indicates an inverse square relationship with time.

### Step 2: Evaluate each option

**Option 1: Force**
- The formula for force is given by Newton's second law:
\[ F = ma \]
where \(m\) is mass and \(a\) is acceleration.
- The SI unit of force is Newton (N), which is equivalent to \(kg \cdot m/s^2\).
- The dimensional formula for force is:
\[ [M^1L^1T^{-2}] \]
- This does not match \([ML^{-1}T^{-2}]\).

**Option 2: Acceleration**
- Acceleration is defined as the change in velocity per unit time:
\[ a = \frac{v}{t} \]
- The SI unit of acceleration is \(m/s^2\).
- The dimensional formula for acceleration is:
\[ [L^1T^{-2}] \]
- This does not match \([ML^{-1}T^{-2}]\).

**Option 3: Pressure**
- Pressure is defined as force per unit area:
\[ P = \frac{F}{A} \]
- The SI unit of pressure is Pascal (Pa), which is equivalent to \(N/m^2\).
- Substituting the dimensional formula for force:
\[ P = \frac{[M^1L^1T^{-2}]}{[L^2]} = [M^1L^{-1}T^{-2}] \]
- This matches \([ML^{-1}T^{-2}]\).

**Option 4: Work**
- Work is defined as force times displacement:
\[ W = F \cdot d \]
- The SI unit of work is Joule (J), which is equivalent to \(N \cdot m\).
- The dimensional formula for work is:
\[ [M^1L^2T^{-2}] \]
- This does not match \([ML^{-1}T^{-2}]\).

### Conclusion
After evaluating all options, the correct answer is:
**Option 3: Pressure**

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