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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 6
Total: 54
A charged shell of radius R carries a total charge Q. Given ϕ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following options is/are correct?
{
∈
0
is the permittivity of free space]
If
h
<
8
R
/
5
and
r
=
3
R
/
5
then
ϕ
=
0
If
h
>
2
R
and
r
>
R
then
ϕ
=
Q
/
∈
0
If
h
>
2
R
and
r
=
3
R
/
5
then
ϕ
=
Q
/
5
∈
0
If
h
>
2
R
and
r
=
4
R
/
5
then
ϕ
=
Q
/
5
∈
0
Validate
Solution:
Option-1:If
h
=
8
R
5
and
r
=
3
R
5
,the cylindrical surface just fits into the sphere. If
h
<
8
R
5
, the cylindrical surface will be well inside the sphere.
ϕ
=
q
i
n
∈
0
=
0
∈
0
=
0
∴ (1) is correct
Option -2: If h>2R and r>R the sphere will be well inside the cylinder
∴
ϕ
=
Q
E
0
∴ (2) is correct
Option 3:
q
in
=
(
Q
AB
)
2
=
Q
2
(
1
−
cos
θ
)
×
2
=
Q
[
1
−
4
5
]
=
Q
5
∴
ϕ
=
q
i
n
∈
0
=
Q
5
∈
0
∴ (3) is correct
Option 4:
q
i
n
=
Q
2
(
1
−
cos
53
∘
)
×
2
=
2
Q
/
5
Q
=
2
Q
/
5
∈
0
∴(4) is wrong
© examsnet.com
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