Question 173: 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: