The dimensions of resistivity in terms of M, L, T and Q, where Q stands for the dimensions of charge is 

Options

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

Answer

To find the dimensions of resistivity (ρ) in terms of mass (M), length (L), time (T), and charge (Q), we start with the relationship between resistivity, resistance, length, and area.

1. **Understanding the formula for resistivity**:
The formula for resistance (R) is given by:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the conductor,
- \( A \) is the cross-sectional area.

2. **Rearranging the formula**:
We can rearrange the formula to express resistivity (ρ):
\[ \rho = \frac{R A}{L} \]

3. **Finding the dimensions of resistance (R)**:
The dimensions of resistance (R) can be expressed as:
\[ [R] = \frac{V}{I} \]
where \( V \) is voltage and \( I \) is current. The dimensions of voltage (V) can be expressed as:
\[ [V] = [I][R] = [Q][T^{-1}] \]
Therefore, the dimensions of resistance become:
\[ [R] = [M L^2 T^{-3} Q^{-1}] \]

4. **Finding the dimensions of area (A)**:
The area (A) has dimensions:
\[ [A] = L^2 \]

5. **Finding the dimensions of length (L)**:
The length (L) has dimensions:
\[ [L] = L \]

6. **Substituting into the resistivity formula**:
Now, substituting the dimensions of resistance (R), area (A), and length (L) into the formula for resistivity:
\[ [\rho] = \frac{[R][A]}{[L]} = \frac{[M L^2 T^{-3} Q^{-1}] [L^2]}{[L]} \]

7. **Simplifying the expression**:
Simplifying the dimensions:
\[ [\rho] = \frac{[M L^2 T^{-3} Q^{-1}] [L^2]}{[L]} = [M L^3 T^{-3} Q^{-1}] \]

8. **Final result**:
Therefore, the dimensions of resistivity (ρ) in terms of M, L, T, and Q are:
\[ [\rho] = M L^3 T^{-3} Q^{-1} \]