Question

Sol. [Key Concept: Derive the dimensions of R, L and C; the combination $\frac{R}{\sqrt{LC}}$ matches ${ML}^{2}T^{- 4}A^{- 2}$.]

$${R = \frac{V}{I} = \frac{w}{qI} = \frac{{ML}^{2}{\text{ }T}^{- 2}}{I^{2}\text{ }T} = {ML}^{2}{\text{ }T}^{- 3}I^{- 2} }{U = \frac{1}{2}{LI}^{2} \rightarrow {ML}^{2}{\text{ }T}^{- 2} = {LA}^{2} }{L = {ML}^{2}{\text{ }T}^{- 2}{\text{ }A}^{- 2} }{Q = CV }{C = \frac{IT \times q}{{ML}^{2}{\text{ }T}^{- 2}} = \frac{A^{2}{\text{ }T}^{2}}{{ML}^{2}{\text{ }T}^{- 2}} }{C = M^{- 1}{\text{ }L}^{- 2}{\text{ }A}^{- 2}{\text{ }T}^{4} }{LC = {ML}^{2}{\text{ }T}^{- 2}{\text{ }A}^{- 2} \times M^{- 1}{\text{ }L}^{- 1}{\text{ }A}^{2}{\text{ }T}^{4} }{LC = T^{2} }{\frac{R}{\sqrt{LC}} = \frac{{ML}^{2}{\text{ }T}^{- 3}{\text{ }A}^{- 2}}{\text{ }T} = \left( {ML}^{2}{\text{ }T}^{- 4}{\text{ }A}^{- 2} \right) }$$