Question 201: Units , Dimensions and Error Analysis

Question

In SI units, the dimensions of $\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $ is :

Accepted Answer

$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $ = ${{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}$ = c $ 	imes $ ${{ \in _0}}$
$ 	herefore $ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } ight]$ = $\left[ {L{T^{ - 1}}} ight] 	imes \left[ {{ \in _0}} ight]$
We know, F = ${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$
$ 	herefore $ ${ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}$
$ \Rightarrow $ $\left[ {{ \in _0}} ight] = {{{{\left[ {AT} ight]}^2}} \over {\left[ {ML{T^{ - 2}}} ight] 	imes \left[ {{L^2}} ight]}}$ = $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} ight]$
$ 	herefore $ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } ight]$ = $\left[ {L{T^{ - 1}}} ight]$ $ 	imes $ $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} ight]$
= [A
2
T
3
M
Ã¢ÂÂ1
L
Ã¢ÂÂ2
]

Answer

A
–1
TML
3

Answer

A
2
T
3
M
–1
L
–2

Answer

AT
–3
ML
3/2

Answer

AT
2
M
–1
L
–1

Units , Dimensions and Error Analysis

In SI units, the dimensions of \sqrt {{{{ \in _0}} \over {{\mu _0}}}} is :

Choose the correct answer: