`'rho'` the electrical resistivity has dimensions

Options

A. `M^(1)L^(2)T^(-2)A^(-2)`
B. `M^(1)L^(3)T^(-2)A^(-2)`
C. `M^(1)L^(-2)T^(-3)A^(-2)`
D. `M^(1)L^(2)T^(-2)A^(-3)`

Answer

To find the dimensions of electrical resistivity (ρ), we start with the relationship between resistance (R), area (A), and length (L). The formula for resistivity is given by:

\[ \rho = \frac{R \cdot A}{L} \]

### Step 1: Find the dimensions of resistance (R)
Resistance (R) can be expressed using Ohm's law, which states:

\[ R = \frac{V}{I} \]

Where:
- \( V \) is voltage (electric potential),
- \( I \) is current.

### Step 2: Find the dimensions of voltage (V)
Voltage can be defined in terms of energy (E) and charge (Q):

\[ V = \frac{E}{Q} \]

The dimension of energy (E) is given by:

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

Thus, the dimension of voltage (V) becomes:

\[ V = \frac{M \cdot L^2 \cdot T^{-2}}{Q} \]

### Step 3: Find the dimensions of current (I)
Current (I) is defined as charge per unit time:

\[ I = \frac{Q}{T} \]

### Step 4: Substitute the dimensions of V and I into the expression for R
Now substituting the dimensions of V and I into the expression for R:

\[ R = \frac{V}{I} = \frac{\left[M \cdot L^2 \cdot T^{-2}\right]/Q}{Q/T} = \frac{M \cdot L^2 \cdot T^{-2}}{Q} \cdot \frac{T}{Q} = \frac{M \cdot L^2 \cdot T^{-1}}{Q^2} \]

### Step 5: Substitute R into the resistivity formula
Now we substitute R back into the formula for resistivity:

\[ \rho = \frac{R \cdot A}{L} \]

The area (A) has dimensions of \( L^2 \), and length (L) has dimensions of \( L \):

\[ \rho = \frac{\left(\frac{M \cdot L^2 \cdot T^{-1}}{Q^2}\right) \cdot L^2}{L} = \frac{M \cdot L^4 \cdot T^{-1}}{Q^2 \cdot L} = \frac{M \cdot L^3 \cdot T^{-1}}{Q^2} \]

### Final Result
Thus, the dimensions of electrical resistivity (ρ) are:

\[ \rho = M \cdot L^3 \cdot T^{-1} \cdot A^{-2} \]