Dimensions of pressure are same as that of

Options

A. Energy
B. Energy per unit volume
C. Force
D. Force per unit volume

Explanation

To solve the question regarding the dimensions of pressure, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Pressure**:
Pressure (P) is defined as the force (F) applied per unit area (A). Mathematically, this is expressed as:
\[ P = \frac{F}{A} \]

2. **Dimensions of Force**:
The dimension of force can be derived from Newton's second law, which states that force is mass (m) times acceleration (a). The dimension of mass is [M], and acceleration has dimensions of length per time squared ([L][T]^-2). Therefore, the dimension of force is:
\[ [F] = [M][L][T]^{-2} = MLT^{-2} \]

3. **Dimensions of Area**:
Area is defined as length squared. Therefore, the dimension of area is:
\[ [A] = [L]^2 = L^2 \]

4. **Calculating Dimensions of Pressure**:
Now substituting the dimensions of force and area into the pressure formula:
\[ [P] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

5. **Comparing with Energy**:
The dimension of energy (E) is defined as work done, which is force times distance. Therefore, the dimension of energy is:
\[ [E] = [F][L] = (MLT^{-2})(L) = ML^2T^{-2} \]
Now, we need to find the dimension of energy per unit volume. The volume (V) has dimensions:
\[ [V] = [L]^3 = L^3 \]
Thus, the dimension of energy per unit volume is:
\[ \text{Energy per unit volume} = \frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2} \]

6. **Conclusion**:
We find that the dimensions of pressure \( (ML^{-1}T^{-2}) \) are the same as the dimensions of energy per unit volume \( (ML^{-1}T^{-2}) \). Therefore, the correct answer is that the dimensions of pressure are the same as that of energy per unit volume.

### Final Answer:
The dimensions of pressure are the same as that of energy per unit volume.