Which of the following is dimensionally same?

Options

A. Pressure = momentum per unit volume
B. Pressure = momentum per unit volume per unit time
C. Pressure = energy per unit volume
D. Pressure = energy per unit area

Answer

To determine which of the given options is dimensionally the same as pressure, we will first calculate the dimensions of pressure and then compare them with the dimensions of the options provided.

### Step 1: Calculate the dimension of pressure
Pressure is defined as force per unit area.

\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \]

### Step 2: Calculate the dimension of force
Force is defined as mass times acceleration.

\[ \text{Force} = \text{Mass} \times \text{Acceleration} \]

The dimension of mass is \( [M] \) and the dimension of acceleration is \( [L][T^{-2}] \). Therefore, the dimension of force is:

\[ \text{Dimension of Force} = [M][L][T^{-2}] = [M L T^{-2}] \]

### Step 3: Calculate the dimension of area
Area is defined as length squared.

\[ \text{Area} = [L]^2 = [L^2] \]

### Step 4: Substitute the dimensions into the pressure formula
Now we can substitute the dimensions of force and area into the pressure formula:

\[ \text{Dimension of Pressure} = \frac{[M L T^{-2}]}{[L^2]} = [M L T^{-2}] \cdot [L^{-2}] = [M L^{-1} T^{-2}] \]

### Step 5: Analyze the options
Now we will analyze each option to see if their dimensions match with \( [M L^{-1} T^{-2}] \).

#### Option A: Momentum per unit volume
Momentum is defined as mass times velocity.

\[ \text{Momentum} = [M][L][T^{-1}] = [M L T^{-1}] \]

The dimension of volume is \( [L^3] \). Therefore, the dimension of momentum per unit volume is:

\[ \text{Dimension of A} = \frac{[M L T^{-1}]}{[L^3]} = [M L^{-2} T^{-1}] \]

This does not match \( [M L^{-1} T^{-2}] \).

#### Option B: Momentum per unit volume per unit time
Using the dimension from option A:

\[ \text{Dimension of B} = \frac{[M L^{-2} T^{-1}]}{[T]} = [M L^{-2} T^{-2}] \]

This does not match \( [M L^{-1} T^{-2}] \).

#### Option C: Energy per unit volume
Energy is defined as work done, which is force times distance.

\[ \text{Energy} = \text{Force} \times \text{Distance} = [M L T^{-2}] \cdot [L] = [M L^2 T^{-2}] \]

The dimension of volume is \( [L^3] \). Therefore, the dimension of energy per unit volume is:

\[ \text{Dimension of C} = \frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}] \]

This matches \( [M L^{-1} T^{-2}] \).

#### Option D: Energy per unit area
Using the dimension of energy:

\[ \text{Dimension of D} = \frac{[M L^2 T^{-2}]}{[L^2]} = [M T^{-2}] \]

This does not match \( [M L^{-1} T^{-2}] \).

### Conclusion
The only option that has the same dimensions as pressure is option **C**: Energy per unit volume.