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Test Index
JEE Advanced 2023 Paper 1 Solutions
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Section:
Physics
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© examsnet.com
Question : 5 of 51
Marks:
+1
,
-0
A container has a base of
50
cm
×
5
cm
50\,\text{cm} \times 5\,\text{cm}
50
cm
×
5
cm
and height
50
cm
50\,\text{cm}
50
cm
, as shown in the figure. It has two parallel electrically conducting walls each of area
50
cm
×
50
cm
50\,\text{cm} \times 50\,\text{cm}
50
cm
×
50
cm
. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of
250
cm
3
s
−
1
250\,\text{cm}^3\,\text{s}^{-1}
250
cm
3
s
−
1
. What is the value of the capacitance of the container after 10 seconds?
[Given: Permittivity of free space
ϵ
0
=
9
×
1
0
−
12
C
2
N
−
1
m
−
2
\epsilon_0 = 9 \times 10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}
ϵ
0
=
9
×
1
0
−
12
C
2
N
−
1
m
−
2
, the effects of the non-conducting walls on the capacitance are negligible]
[JEE Adv 2023 P1]
27
pF
27\,\text{pF}
27
pF
63
pF
63\,\text{pF}
63
pF
81
pF
81\,\text{pF}
81
pF
135
pF
135\,\text{pF}
135
pF
Validate
Solution:
h
=
250
×
10
50
×
5
=
10
cm
h = \frac{250 \times 10}{50 \times 5} = 10\,\text{cm}
h
=
50
×
5
250
×
10
=
10
cm
C
1
=
(
0.40
×
0.50
)
×
9
×
1
0
−
12
5
×
1
0
−
2
C_1 = \frac{(0.40 \times 0.50) \times 9 \times 10^{-12}}{5 \times 10^{-2}}
C
1
=
5
×
1
0
−
2
(
0.40
×
0.50
)
×
9
×
1
0
−
12
=
0.36
×
1
0
−
10
F
= 0.36 \times 10^{-10}\,\text{F}
=
0.36
×
1
0
−
10
F
C
2
=
3
×
0.10
×
0.5
×
9
×
1
0
−
12
5
×
1
0
−
2
C_2 = \frac{3 \times 0.10 \times 0.5 \times 9 \times 10^{-12}}{5 \times 10^{-2}}
C
2
=
5
×
1
0
−
2
3
×
0.10
×
0.5
×
9
×
1
0
−
12
C
2
=
0.27
×
1
0
−
10
F
C_2 = 0.27 \times 10^{-10}\,\text{F}
C
2
=
0.27
×
1
0
−
10
F
C
=
C
1
+
C
2
C = C_1 + C_2
C
=
C
1
+
C
2
=
63
pF
= 63\,\text{pF}
=
63
pF
© examsnet.com
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