Question

The dimensional formula for specific resistance (resistivity) is:

• A. $[M L^{-3} T^{-2} A^{-2}]$
• B. $[M L^3 T^{-3} A^{-2}]$
• C. $[M L^3 T^{-3} A^2]$
• D. $[M L^3 T^3 A^2]$

Hint:

Instead of deriving everything from scratch, remember the resistance dimension $[M L^2 T^{-3} A^{-2}]$. Since resistivity multiplies resistance by a unit of length ($[L]$), simply increase the exponent of $L$ by $1$ to get $[M L^3 T^{-3} A^{-2}]$.

Answer 1

The Correct Option is B

Concept: Specific resistance or resistivity ($\rho$) is related to electrical resistance ($R$) by the expression $\rho = R \frac{A}{l}$. Resistance can be derived from Ohm's Law ($R = \frac{V}{I}$), where potential difference $V$ is defined as work done per unit charge ($V = \frac{W}{q}$).

Step 1: Break down individual component dimensions.

• Work, $[W] = [M L^2 T^{-2}]$
• Charge, $[q] = [A T]$
• Potential, $[V] = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$
• Resistance, $[R] = \frac{[V]}{[I]} = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}]$

Step 2: Evaluate the final dimensional formula for resistivity.
Using the geometry tracking multiplier $\frac{\text{Area}}{\text{Length}} \rightarrow \frac{[L^2]}{[L]} = [L]$: \[ [\rho] = [R] \times [L] = [M L^2 T^{-3} A^{-2}] \times [L] = [M L^3 T^{-3} A^{-2}] \]