Match List I with List II and select the correct answer using the codes given below the lists.
`{:("List I",,"List II"),("(Physical Quantity)",,"(Dimension)"),("A. Angular momentum",, (1) ML^(-1)T^(-2)),("B. Torque",,(2) MT^(-2)),("C. Surface Tension",,(3) ML^(2)T^(-1)),("D. Coefficient of viscosity",,(4) ML^(2)T^(-2)):}`

Options

A. `{:(A,,B,,C,,D),(4,,3,,2,,1):}`
B. `{:(A,,B,,C,,D),(3,,4,,2,,1):}`
C. `{:(A,,B,,C,,D),(1,,4,,2,,3):}`
D. `{:(A,,B,,C,,D),(3,,1,,4,,2):}`

Answer

To solve the problem of matching the physical quantities in List I with their corresponding dimensions in List II, we will derive the dimensional formulas for each physical quantity step by step.

### Step 1: Angular Momentum
- **Definition**: Angular momentum (L) is given by the formula \( L = M \cdot V \cdot R \), where \( M \) is mass, \( V \) is velocity, and \( R \) is the radius (or distance).
- **Dimensional Formula**:
- Mass (M) has the dimension [M].
- Velocity (V) has the dimension [LT^{-1}].
- Radius (R) has the dimension [L].

Therefore, the dimensional formula for angular momentum is:
\[ [L] = [M] \cdot [LT^{-1}] \cdot [L] = [ML^2T^{-1}] \]

**Conclusion**: Angular momentum corresponds to (3) ML²T⁻¹.

### Step 2: Torque
- **Definition**: Torque (τ) is defined as the product of force and the distance from the pivot point, \( τ = F \cdot R \).
- **Dimensional Formula**:
- Force (F) has the dimension [MLT^{-2}].
- Distance (R) has the dimension [L].

Therefore, the dimensional formula for torque is:
\[ [τ] = [F] \cdot [L] = [MLT^{-2}] \cdot [L] = [ML^2T^{-2}] \]

**Conclusion**: Torque corresponds to (4) ML²T⁻².

### Step 3: Surface Tension
- **Definition**: Surface tension (γ) is defined as force per unit length.
- **Dimensional Formula**:
- Force (F) has the dimension [MLT^{-2}].
- Length (L) has the dimension [L].

Therefore, the dimensional formula for surface tension is:
\[ [γ] = \frac{[F]}{[L]} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}] \]

**Conclusion**: Surface tension corresponds to (2) MT⁻².

### Step 4: Coefficient of Viscosity
- **Definition**: The coefficient of viscosity (η) is defined as the ratio of shear stress to shear rate.
- **Dimensional Formula**:
- Shear stress has the dimension [MLT^{-2}] (force per unit area).
- Shear rate has the dimension [LT^{-1}].

Therefore, the dimensional formula for the coefficient of viscosity is:
\[ [η] = \frac{[F/A]}{[V/L]} = \frac{[MLT^{-2}]}{[L^2]} \cdot [LT^{-1}] = \frac{[MLT^{-2}]}{[L^2]} \cdot [L] = [ML^{-1}T^{-1}] \]

**Conclusion**: Coefficient of viscosity corresponds to (1) ML⁻¹T⁻¹.

### Final Matching:
- A. Angular momentum → (3) ML²T⁻¹
- B. Torque → (4) ML²T⁻²
- C. Surface Tension → (2) MT⁻²
- D. Coefficient of viscosity → (1) ML⁻¹T⁻¹

### Correct Code:
The correct matching code is **3 4 2 1**.