Match list (Physical Quantity ) with List II (Dimensions) and select the correct answer using the codes givn below the list
`{:("List I",,"List II"),("(A) Relative density",,(1) ML^(2)T^(-3)),("(B) Potential energy",,(2) MLT^(-1)),("(C) Viscosity",,(3) M^(@)L^(@)T^(@)),("(D) Linear Momentum",,(4) ML^(-1)T^(-1)),(,,(5)ML^(2)T^(2)):}`

Options

A. `{:(A,,B,,C,,D),(3,,5,,2,,1):}`
B. `{:(A,,B,,C,,D),(5,,3,,1,,2):}`
C. `{:(A,,B,,C,,D),(3,,5,,4,,2):}`
D. `{:(A,,B,,C,,D),(5,,3,,4,,1):}`

Answer

To solve the problem of matching physical quantities with their corresponding dimensions, we will analyze each item in List I and determine its dimensional formula.

### Step-by-Step Solution:

1. **Relative Density (A)**:
- Relative density is defined as the ratio of the density of a substance to the density of water.
- Since both densities have the same units (mass per unit volume), the units cancel out.
- Therefore, relative density is a dimensionless quantity.
- **Dimensional Formula**: \( M^0 L^0 T^0 \) (which corresponds to option 3).

2. **Potential Energy (B)**:
- Potential energy (PE) is given by the formula \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.
- The dimensions of mass \( m \) are \( M \), height \( h \) has dimensions \( L \), and \( g \) can be expressed as \( L T^{-2} \).
- Therefore, the dimensional formula for potential energy is:
\[ [PE] = M \cdot (L T^{-2}) \cdot L = M L^2 T^{-2} \]
- **Dimensional Formula**: \( ML^2 T^{-2} \) (which corresponds to option 5).

3. **Viscosity (C)**:
- Viscosity is defined as the resistance of a fluid to flow and is given by the formula:
\[ \text{Viscosity} = \frac{\text{Force}}{\text{Area} \cdot \text{Velocity}} = \frac{F}{A \cdot v} \]
- The dimensions of force \( F \) are \( M L T^{-2} \), area \( A \) is \( L^2 \), and velocity \( v \) is \( L T^{-1} \).
- Therefore, the dimensional formula for viscosity is:
\[ [\text{Viscosity}] = \frac{M L T^{-2}}{L^2 \cdot (L T^{-1})} = \frac{M L T^{-2}}{L^3 T^{-1}} = M L^{-1} T^{-1} \]
- **Dimensional Formula**: \( M L^{-1} T^{-1} \) (which corresponds to option 4).

4. **Linear Momentum (D)**:
- Linear momentum \( p \) is defined as the product of mass and velocity:
\[ p = mv \]
- The dimensions of mass \( m \) are \( M \) and velocity \( v \) is \( L T^{-1} \).
- Therefore, the dimensional formula for linear momentum is:
\[ [p] = M \cdot (L T^{-1}) = M L T^{-1} \]
- **Dimensional Formula**: \( M L T^{-1} \) (which corresponds to option 2).

### Summary of Matches:
- (A) Relative Density → (3) \( M^0 L^0 T^0 \)
- (B) Potential Energy → (5) \( M L^2 T^{-2} \)
- (C) Viscosity → (4) \( M L^{-1} T^{-1} \)
- (D) Linear Momentum → (2) \( M L T^{-1} \)

### Final Answer:
The correct matching is:
- A - 3
- B - 5
- C - 4
- D - 2