Let [{varepsilon _0}] denotes dimensional formula of permittivity of vacuum and [{μ _0}] that of pe

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1.Units, Dimensions and Measurement

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Let $[{\varepsilon _0}]$ denotes the dimensional formula of the permittivity of the vacuum and $[{\mu _0}]$ that of the permeability of the vacuum. If $M = {\rm{mass}}$, $L = {\rm{length}}$, $T = {\rm{Time}}$ and $I = {\rm{electric current}}$, then

A

$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^2}I$

B

$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$

C

$[{\mu _0}] = M{L^2}{T^{ - 1}}I$

D

None of these

(IIT-1998)

Solution

Dimension formula of $\varepsilon_0$

$\varepsilon_0=\frac{1}{4 \pi F } \frac{ v _1 v _2}{ r ^2}\left( F = MLT ^{-2}\right)$

$\varepsilon_0=\frac{1}{ MLT ^{-2}} \frac{ LAT \times AT }{1^2}$

$= M ^{-1} L ^{-3} t ^4 A ^2$

Dimension of $\mu_0$

$N / A ^2$ or $WbA A ^{-1} m ^{-1}$

$=[ M ][ L ][ T ]^{-2}[ A ]^{-2}$

Standard 11

Physics

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