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JEE Advanced 2020 Full Test 6 Paper 1
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© examsnet.com
Question : 2
Total: 54
Consider a bob of mass
m
and having charge
q
attached with a light string of length I and pivoted at point O. It is released at rest at
60
∘
with vertical. There are two regions- Region-I (left of line
P
Q
) has a uniform and constant magnetic field B directed inside plane of paper. Region-II (right of line
P
Q
) has a constant and uniform electric field E directed vertically up as shown. Consider no effect of gravity in both the regions.
Time taken by particle to cross region-I for
1
st
time is
π
√
2
m
l
5
q
E
Time taken by particle to cross region-I for
2
nd
time is
π
√
2
m
l
13
q
E
Angular speed of
1
st
revolution in magnetic field is
√
5
q
E
2
m
l
Angular speed of
2
nd
revolution in magnetic field is
√
13
q
E
m
l
Validate
Solution:
Using work-energy theorem till string become taut Work done by
q
E
=
Change in K.E.
⇒
q
E
l
=
1
2
m
v
2
−
0
⇒
v
=
√
2
q
E
l
2
m
When string become taut an impulse is received and speed of the bob becomes
v
cos
30
∘
=
√
3
q
E
l
2
m
Further using work energy theorem speed of the bob becomes
v
′
=
√
5
q
E
l
2
m
∴
time period
=
2
π
l
v
′
=
2
π
√
2
m
l
5
q
E
After 1 st half revolution in magnetic field work done by
q
E
=
Change in K.E., from lowermost to uppermost point.
q
E
.2
l
=
m
v
1
2
2
−
m
2
5
q
E
l
2
m
v
1
=
√
13
q
E
l
2
m
∴
Time period of
2
nd
revolution
=
2
π
l
v
1
=
π
√
2
m
l
13
q
E
∴
Angular speed of
1
st
revolution
=
√
5
q
E
2
m
l
∴
Angular speed of
2
nd
revolution
=
√
13
q
E
2
m
l
∴
A
B
C
© examsnet.com
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