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Standard XII
Physics
Dimensional Analysis
Question
Dimensions of
1
μ
0
ε
0
, where symbols have their usual meaning, are
:
[
L
−
1
T
]
[
L
−
2
T
2
]
[
L
T
−
1
]
[
L
2
T
−
2
]
A
[
L
2
T
−
2
]
B
[
L
−
1
T
]
C
[
L
−
2
T
2
]
D
[
L
T
−
1
]
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Solution
Verified by Toppr
Vaccum permittivity:
F
c
=
1
4
π
ε
0
×
q
1
q
2
r
2
Where
ε
0
is permttivity of vaccum.
Dimension will be
[
M
−
1
L
−
3
T
4
I
2
]
vaccum permeability:
μ
0
=
4
π
×
10
−
7
N
/
A
2
Dimensions will be
[
M
L
T
−
2
I
−
2
]
Dimensions for
1
μ
0
ε
0
will be
1
[
M
−
1
L
−
3
T
4
I
2
]
×
[
M
1
L
1
T
−
2
I
−
2
]
=
[
L
2
T
−
2
]
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, where symbols have their usual meaning, are
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