The unit of "impulse per unit area" is same as that of

Options

A. Viscosity
B. Surface tension
C. Bulk modulus
D. Force

Answer

To solve the question regarding the unit of "impulse per unit area," we will go through the following steps:

### Step 1: Understand the concept of impulse
Impulse is defined as the change in momentum of an object when a force is applied over a period of time. Mathematically, impulse (J) can be expressed as:
\[ J = F \cdot t \]
where \( F \) is the force and \( t \) is the time duration.

### Step 2: Determine the dimensions of impulse
The dimension of force \( F \) is given by:
\[ [F] = MLT^{-2} \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

Thus, the dimension of impulse is:
\[ [J] = [F] \cdot [t] = (MLT^{-2}) \cdot (T) = MLT^{-1} \]

### Step 3: Define impulse per unit area
Impulse per unit area can be expressed as:
\[ \text{Impulse per unit area} = \frac{J}{A} \]
where \( A \) is the area.

### Step 4: Determine the dimensions of area
The dimension of area \( A \) is:
\[ [A] = L^2 \]

### Step 5: Calculate the dimensions of impulse per unit area
Now, substituting the dimensions of impulse and area into the equation for impulse per unit area, we get:
\[ \text{Dimensions of impulse per unit area} = \frac{[J]}{[A]} = \frac{MLT^{-1}}{L^2} = ML^{-1}T^{-1} \]

### Step 6: Compare with other physical quantities
Now we will compare the dimensions \( ML^{-1}T^{-1} \) with the dimensions of other quantities:

1. **Viscosity**:
- Viscosity \( \eta \) is defined as:
\[ \eta = \frac{F \cdot r}{A \cdot v} \]
- Dimensions:
\[ [\eta] = \frac{MLT^{-2} \cdot L}{L^2 \cdot LT^{-1}} = \frac{ML}{L^2} \cdot \frac{1}{T} = ML^{-1}T^{-1} \]
- **Match**: Yes, it matches with impulse per unit area.

2. **Surface Tension**:
- Surface tension \( \gamma \) is defined as:
\[ \gamma = \frac{F}{L} \]
- Dimensions:
\[ [\gamma] = \frac{MLT^{-2}}{L} = MT^{-2} \]
- **Match**: No, it does not match.

3. **Bulk Modulus**:
- Bulk modulus \( B \) is defined as:
\[ B = -\frac{dP}{dV} \]
- Dimensions of pressure \( P \):
\[ [P] = \frac{F}{A} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]
- **Match**: No, it does not match.

4. **Force**:
- Force has the dimensions:
\[ [F] = MLT^{-2} \]
- **Match**: No, it does not match.

### Conclusion
The only quantity that has the same dimensions as impulse per unit area is viscosity. Therefore, the answer is viscosity.