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Standard XII
Physics
Dimensional Analysis
Question
Dimensions of electrical resistance are :
[
M
L
2
T
−
3
A
−
2
]
[
M
L
2
T
−
3
A
−
1
]
[
M
L
3
T
−
3
A
−
2
]
[
M
L
2
T
3
A
2
]
A
[
M
L
2
T
−
3
A
−
2
]
B
[
M
L
3
T
−
3
A
−
2
]
C
[
M
L
2
T
−
3
A
−
1
]
D
[
M
L
2
T
3
A
2
]
Open in App
Solution
Verified by Toppr
Since according to Ohm's Law :
V
=
I
R
Hence,
R
=
V
I
Where
V
=
W
q
=
w
o
r
k
d
o
n
e
c
h
a
r
g
e
Hence; Potential
=
[
M
1
L
2
T
−
2
]
[
A
T
]
∴
[
V
]
=
[
M
1
L
2
T
−
3
A
−
1
]
From above equations,
∴
Resistance
=
[
R
]
=
[
M
1
L
2
T
−
3
A
−
1
]
[
A
]
=
[
M
1
L
2
T
−
3
A
−
2
]
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21
Similar Questions
Q1
Match the following two columns:
Column I
Column II
A.
Electrical resistance
1.
[
M
L
3
T
−
3
A
−
2
]
B.
Electrical potential
2.
[
M
L
2
T
−
3
A
−
2
]
C.
Specific resistance
3.
[
M
L
2
T
−
3
A
−
1
]
D.
Specific conductance
4.
None of these
View Solution
Q2
Match the Column I with Column II.
Column I
(Physical quantity)
Column II
(Dimensional formula)
(A)
Premittivity of free space
(p)
[
M
0
L
0
T
−
1
]
(B)
Radiant flux
(q)
[
M
L
3
T
−
3
A
−
2
]
(C)
Resistivity
(r)
[
M
L
2
T
−
3
]
(D)
Hubble constant
(s)
[
M
−
1
L
−
3
T
4
A
2
]
View Solution
Q3
[
M
L
2
T
−
2
]
are dimensions of
View Solution
Q4
A physical quantity
X
is expressed as velocity
=
√
(
X
density
)
. The dimensions of
X
are
View Solution
Q5
If
2
tan
3
A
cos
3
A
−
tan
3
A
+
1
=
2
cos
3
A
, smallest magnitude of
A
is
View Solution