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BITSAT 2018 Slot 2 Solved Paper
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© examsnet.com
Question : 2
Total: 150
A parallel plate capacitor
C
with plates of unit area and separation
d
is filled with a liquid of dielectric constant
K
=
2
, the level of liquid is
d
3
, initially. Suppose, the liquid level decreases at a constant speed
v
, the time constant as a function of time is
6
ε
0
R
5
d
+
3
v
t
(
15
d
+
9
v
t
)
ε
0
R
2
d
3
−
3
d
v
t
−
g
v
2
t
2
6
ε
0
R
5
d
−
3
v
t
(
15
d
−
9
v
t
)
ε
0
R
2
d
3
+
3
d
v
t
−
g
v
2
t
2
Validate
Solution:
We have, the time constant,
τ
=
R
C
′
........(i)
Now,
C
′
=
C
1
C
2
C
1
+
C
2
=
(
A
ε
0
d
−
x
)
(
K
A
ε
0
x
)
A
ε
0
d
−
x
+
K
A
ε
0
x
∴
C
′
=
K
A
ε
0
x
+
K
(
d
−
x
)
Putting the value of C' in Eq. (i), we get
τ
=
R
K
A
ε
0
d
3
−
v
t
+
k
(
d
−
d
3
+
v
t
)
[
∵
x
=
d
3
−
v
t
]
Given, A =1 and K = 2
∴
τ
=
3
×
2
ε
0
×
R
d
−
3
v
t
+
6
d
−
2
d
+
6
v
t
=
6
R
ε
0
5
d
+
3
v
t
© examsnet.com
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