Match the following two columns.
`{:(,"Column I",,"Column II"),("A.","Electrical resistance",1.,["ML"^(3)"T"^(-3)"A"^(-2)]),("B.","Electrical potential",2.,["ML"^(2)"T"^(-3)"A"^(-2)]),("C.","Specific resistance",3.,["ML"^(2)"T"^(-3)"A"^(-1)]),("D.","Specific conductance",4.,"None"):}`

Options

A. A-2, B-3, C-1, D-4
B. A-2, B-4, C-3, D-1
C. A-1, B-2, C-4, D-3
D. A-1, B-3, C-2, D-4

Explanation

To solve the problem of matching the two columns based on the dimensions of the physical quantities, we will follow these steps:

### Step 1: Calculate the dimensions of Electrical Resistance (R)
Electrical resistance (R) is given by the formula:
\[ R = \frac{V}{I} \]
where \( V \) is the electrical potential (voltage) and \( I \) is the current.

### Step 2: Find the dimensions of Electrical Potential (V)
Electrical potential (V) can be expressed in terms of work (W) and charge (Q):
\[ V = \frac{W}{Q} \]
The dimensions of work (W) are:
\[ [W] = [ML^2T^{-2}] \]
The dimensions of charge (Q) are:
\[ [Q] = [AT] \]
Thus, the dimensions of electrical potential (V) are:
\[ [V] = \frac{[W]}{[Q]} = \frac{ML^2T^{-2}}{AT} = [ML^2T^{-3}A^{-1}] \]

### Step 3: Substitute dimensions of V into the formula for R
Now substituting the dimensions of V into the formula for R:
\[ [R] = \frac{[V]}{[I]} = \frac{[ML^2T^{-3}A^{-1}]}{[A]} = [ML^2T^{-3}A^{-2}] \]

### Step 4: Calculate the dimensions of Specific Resistance (ρ)
Specific resistance (ρ) is related to resistance (R) by the formula:
\[ R = \frac{\rho L}{A} \]
Rearranging gives:
\[ \rho = \frac{R \cdot A}{L} \]
We already found the dimensions of R:
\[ [R] = [ML^2T^{-3}A^{-2}] \]
The dimensions of area (A) are:
\[ [A] = [L^2] \]
And the dimensions of length (L) are:
\[ [L] = [L] \]
Thus, the dimensions of specific resistance (ρ) are:
\[ [\rho] = \frac{[R] \cdot [A]}{[L]} = \frac{[ML^2T^{-3}A^{-2}] \cdot [L^2]}{[L]} = [ML^3T^{-3}A^{-2}] \]

### Step 5: Calculate the dimensions of Specific Conductance (σ)
Specific conductance (σ) is the reciprocal of specific resistance:
\[ \sigma = \frac{1}{\rho} \]
Thus, the dimensions of specific conductance are:
\[ [\sigma] = \frac{1}{[ML^3T^{-3}A^{-2}]} = [M^{-1}L^{-3}T^{3}A^{2}] \]
However, for the purpose of this matching, we can note that specific conductance is dimensionally related to specific resistance.

### Step 6: Match the dimensions with Column II
Now we can match the dimensions calculated with the options in Column II:
- **Electrical Resistance**: \( [ML^2T^{-3}A^{-2}] \) matches with **B**.
- **Electrical Potential**: \( [ML^2T^{-3}A^{-1}] \) matches with **A**.
- **Specific Resistance**: \( [ML^3T^{-3}A^{-2}] \) matches with **C**.
- **Specific Conductance**: The remaining option is **D** which is "None".

### Final Matching
- A. Electrical resistance → 2. [ML^2T^{-3}A^{-2}]
- B. Electrical potential → 1. [ML^2T^{-3}A^{-1}]
- C. Specific resistance → 3. [ML^3T^{-3}A^{-2}]
- D. Specific conductance → 4. None