Dimension of resistivity is

Options

A. `ML^(2)T^(-2)I^(-1)`
B. `ML^(3)T^(-3)I^(-2)`
C. `ML^(3)T^(-2)I^(-1)`
D. `ML^(2)T^(-2)I^(-2)`

Explanation

To find the dimension of resistivity (ρ), we start with the relationship between resistance (R), resistivity (ρ), length (L), and area (A):

1. **Starting Equation**:
\[ R = \frac{\rho L}{A} \]

2. **Rearranging for Resistivity**:
\[ \rho = \frac{R \cdot A}{L} \]

3. **Substituting Resistance**:
We know that resistance (R) can also be expressed in terms of voltage (V) and current (I):
\[ R = \frac{V}{I} \]
Therefore, substituting this into the equation for resistivity gives:
\[ \rho = \frac{(V/I) \cdot A}{L} \]

4. **Expressing Voltage**:
Voltage (V) can be expressed in terms of work (W) and charge (Q):
\[ V = \frac{W}{Q} \]
Substituting this into the equation for resistivity:
\[ \rho = \frac{(W/Q) \cdot A}{L \cdot I} \]

5. **Substituting Work Done**:
The work done (W) is given by the formula:
\[ W = F \cdot d \]
where force (F) is mass (m) times acceleration (a). The dimension of work done is:
\[ [W] = [F] \cdot [d] = [m \cdot a] \cdot [L] = [m \cdot L^2 \cdot T^{-2}] \]
Thus, we can substitute this into our equation:
\[ \rho = \frac{(m \cdot L^2 \cdot T^{-2}) \cdot A}{L \cdot I} \]

6. **Substituting Area**:
The area (A) has the dimension:
\[ [A] = [L^2] \]
Therefore, substituting this into the equation:
\[ \rho = \frac{(m \cdot L^2 \cdot T^{-2}) \cdot L^2}{L \cdot I} \]

7. **Simplifying the Expression**:
Now we simplify the expression:
\[ \rho = \frac{m \cdot L^4 \cdot T^{-2}}{L \cdot I} = \frac{m \cdot L^3 \cdot T^{-2}}{I} \]

8. **Final Dimension of Resistivity**:
Since current (I) has the dimension:
\[ [I] = [Q/T] \]
Therefore, the dimension of resistivity can be expressed as:
\[ \rho = m \cdot L^3 \cdot T^{-3} \cdot I^{-2} \]

Thus, the final dimension of resistivity (ρ) is:
\[ \text{Dimension of resistivity} = [M^1 L^3 T^{-3} I^{-2}] \]