A. Permittivity |
B. Resistance |
C. Magnetic permeability |
D. Stress |
A) A, B, C, D
B) D, C, B, A
C) A, D,
D) C, B, C, A
Correct Answer: C
Solution :
For dimension of permittivity is given by \[\because \] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] or \[{{\varepsilon }_{0}}=\frac{1\times {{q}_{1}}{{q}_{2}}}{F\times 4\pi {{r}^{2}}}\] Dimensions of \[[{{\varepsilon }_{0}}]=\frac{[IT\times IT]}{[ML{{T}^{-2}}][{{L}^{2}}]}\] \[=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}]\] For dimensions of resistance \[\because \] \[R=\frac{V}{I}=\frac{W}{QI}=\frac{W}{ITI}\] \[\left( \begin{align} & \because \,V=\frac{W}{Q} \\ & Q=IT \\ \end{align} \right)\] \[=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{I}^{2}}T]}=[M{{L}^{2}}{{T}^{-3}}{{I}^{-2}}]\] For dimensions of magnetic permeability \[\because \] \[F=\frac{{{\mu }_{0}}}{2\pi }\times \frac{{{I}_{1}}{{I}_{1}}l}{r}\] or \[{{\mu }_{0}}=\frac{F\times 2\pi r}{{{I}_{1}}{{I}_{2}}l}\] Dimensions of \[[{{\mu }_{0}}]=\frac{[ML{{T}^{-2}}][L]}{[{{I}^{2}}][L]}=[ML{{T}^{-1}}{{I}^{-2}}]\] For dimensions of stress \[\because \] \[Stress=\frac{force}{area}\] \[=\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}]}=[M{{L}^{-1}}{{T}^{-2}}]\]You need to login to perform this action.
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