A) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{I}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}{{\text{I}}^{-1}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{I}}^{-1}}\text{ }\!\!]\!\!\text{ }\]
Correct Answer: D
Solution :
Resistance \[R=\frac{potential\,\,difference}{current}=\frac{V}{i}=\frac{W}{qi}\] (\[\because \]Potential difference is equal to work done per unit charge) So, Dimensions of\[R\] \[\text{=}\frac{\text{ }\!\![\!\!\text{ Dimensions}\,\,\text{of}\,\,\text{work }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Dimensions}\,\,\text{of}\,\,\text{charge }\!\!]\!\!\text{ }\,\,\text{ }\!\![\!\!\text{ Dimensions}\,\,\text{of}\,\,\text{current }\!\!]\!\!\text{ }}\] \[=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[IT][I]}=[M{{L}^{2}}{{T}^{-3}}{{I}^{-2}}]\]
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