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JEE Advanced 2020 Full Test 7 Paper 2
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© examsnet.com
Question : 8
Total: 54
A capacitor with air as dielectric has capacity
C
as shown in figure is completely filled with dielectric slab of mass
M
and is connected to a block of mass
M
. The system is released form rest at time
t
=
0
when the capacitor is completely filled with dielectric. Then:
(Dielectric constant of slab
=
K
)
Charge on the plate of the capacitor when it is half filled with air is
(
K
+
1
)
C
V
2
The kinetic energy of the block when the capacitor is half filled with dielectric is
m
g
ℓ
4
−
C
V
2
(
K
−
1
)
8
The potential difference between the plates when the capacitor is half filled with air is
V
The energy stored in the capacitor when it is half filled with dielectric is
K
2
C
V
2
.
Validate
Solution:
V remain same as capacitor is connected with battery charge on
C
=
C
′
V
C
e
q
=
C
2
+
K
C
2
=
(
K
+
1
)
C
2
Charge on
C
=
(
K
+
1
)
2
C
V
energy stored in
C
when slab is
half filled
=
1
2
C
′
V
2
=
1
2
(
K
+
1
)
C
2
V
2
Total work done
=
K
E
−
0
W
by battery
+
T
×
ℓ
2
+
W
by
F
electric
=
K
E
W
by battery
+
T
×
ℓ
2
−
[
1
2
C
′
V
2
−
1
2
C
V
2
]
=
K
E
W
by battery
+
T
×
ℓ
2
−
1
2
V
2
[
(
K
+
1
)
2
C
−
K
C
]
W
by battery
+
T
×
ℓ
2
−
1
2
V
2
[
1
−
K
]
C
2
=
K
E
W
by battery
=
(
Q
t
−
Q
i
)
×
V
=
[
(
K
+
1
)
C
V
2
−
C
V
]
V
=
[
K
−
1
]
2
C
V
2
M
g
−
T
=
M
a
T
=
M
a
T
=
m
g
2
[
1
−
K
]
C
V
2
2
−
1
4
C
V
2
[
K
−
1
]
+
m
g
ℓ
4
=
K
E
© examsnet.com
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