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(B) According to Ohm's Law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.Therefore,$R = \frac{V}{I}$.We

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$[ML^{2}T^{-3}A^{-2}]$

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(B) According to Ohm's Law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.Therefore,$R = \frac{V}{I}$.We know that potential difference $V = \frac{W}{q}$,where $W$ is work done and $q$ is charge.The dimensional formula for work $W$ is $[ML^{2}T^{-2}]$ and for charge $q$ is $[AT]$.Thus,the dimensional formula for potential $V$ is $[V] = \frac{[ML^{2}T^{-2}]}{[AT]} = [ML^{2}T^{-3}A^{-1}]$.Substituting this into the expression for resistance:$[R] = \frac{[ML^{2}T^{-3}A^{-1}]}{[A]} = [ML^{2}T^{-3}A^{-2}]$.Hence,the correct option is $B$.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(p)$ $M L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(q)$ $M L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(r)$ $M L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(s)$ None of these

What is the dimensional formula of electrical resistance?

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