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1.Units, Dimensions and Measurement
hard
In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu } {\varepsilon }$ of the permeability $\mu $ and permittivity $\varepsilon $ is
A$\left[ {R{T^{ - 2}}} \right]$
B$\left[ {{R^2}{T^{ - 1}}} \right]$
C$\left[ {{R^2}} \right]$
D$\left[ {{R^2}{T^2}} \right]$
(JEE MAIN-2014)
Solution
$\begin{array}{l}
{\rm{Dimensions of}}\,\mu = \left[ {ML{T^{ – 2}}{A^{ – 2}}} \right]\\
{\rm{Dimensions of}}\, \in \, = \,\left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\\
{\rm{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ – 3}}{A^{ – 2}}} \right]\\
\therefore \frac{{{\rm{Dimensions of}}\,\mu }}{{{\rm{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ – 2}}{A^{ – 2}}}}{{{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}}}} \right]\\
= \left[ {{M^2}{L^4}{T^{-6}}{A^{ – 4}}} \right] = \left[ {{R^2}} \right]
\end{array}$
{\rm{Dimensions of}}\,\mu = \left[ {ML{T^{ – 2}}{A^{ – 2}}} \right]\\
{\rm{Dimensions of}}\, \in \, = \,\left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\\
{\rm{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ – 3}}{A^{ – 2}}} \right]\\
\therefore \frac{{{\rm{Dimensions of}}\,\mu }}{{{\rm{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ – 2}}{A^{ – 2}}}}{{{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}}}} \right]\\
= \left[ {{M^2}{L^4}{T^{-6}}{A^{ – 4}}} \right] = \left[ {{R^2}} \right]
\end{array}$
Standard 11
Physics
Similar Questions
Match List $I$ with List $II$
| LIST$-I$ | LIST$-II$ |
| $(A)$ Torque | $(I)$ $ML ^{-2} T ^{-2}$ |
| $(B)$ Stress | $(II)$ $ML ^2 T ^{-2}$ |
| $(C)$ Pressure of gradient | $(III)$ $ML ^{-1} T ^{-1}$ |
| $(D)$ Coefficient of viscosity | $(IV)$ $ML ^{-1} T ^{-2}$ |
