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JEE Main 9 Jan 2019 Shift 2 Solved Paper
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© examsnet.com
Question : 11
Total: 90
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants
K
1
,
K
2
,
K
3
,
K
4
arranged as shown in the figure. The effective dielectric constant K will be :
K =
(
K
1
+
K
2
)
(
K
3
+
K
4
)
2
(
K
1
+
K
2
+
K
3
+
K
4
)
K =
(
K
1
+
K
2
)
(
K
3
+
K
4
)
(
K
1
+
K
2
+
K
3
+
K
4
)
K =
(
K
1
+
K
4
)
(
K
2
+
K
3
)
2
(
K
1
+
K
2
+
K
3
+
K
4
)
K =
(
K
1
+
K
3
)
(
K
2
+
K
4
)
(
K
1
+
K
2
+
K
3
+
K
4
)
All of the above
Validate
Solution:
C
12
=
C
1
C
2
C
1
+
C
2
=
k
1
∊
0
L
2
×
L
d
2
k
2
[
∊
0
L
2
×
L
]
d
2
(
k
1
+
k
2
)
[
∊
0
L
2
×
L
d
2
]
C
12
=
k
1
K
2
k
1
+
k
2
∊
0
L
2
d
in the same way we get,
C
34
=
k
3
k
4
k
3
+
k
4
∊
0
L
2
d
... (i)
Now if
k
eq
= k ,
C
eq
=
k
∊
0
L
2
d
... (ii)
on comparing equation (i) to equation (ii), weget
k
eq
=
k
1
k
2
(
k
3
+
k
4
)
+
k
3
k
4
(
k
1
+
k
2
)
(
k
1
+
k
2
)
(
k
3
+
k
4
)
This does not match with any of the optionsso probably they have assumed the wrongcombination
C
13
=
k
1
∊
0
L
L
2
d
2
+
k
3
∊
0
L
.
L
2
d
2
=
(
k
1
+
k
3
)
∊
0
L
2
d
C
24
=
(
k
2
+
k
4
)
∊
0
L
2
d
C
eq
=
C
13
C
24
C
13
C
24
=
(
k
1
+
k
3
)
(
k
2
+
k
4
)
(
k
1
+
k
2
+
k
3
+
k
4
)
∊
0
L
2
d
=
k
∊
0
L
2
d
k = K =
(
K
1
+
K
3
)
(
K
2
+
K
4
)
(
K
1
+
K
2
+
K
3
+
K
4
)
However this is one of the four options.It must be a "Bonus" logically but of the givenoptions probably they might go with (4)
© examsnet.com
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