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JEE Advanced 2020 Full Test 4 Paper 2
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© examsnet.com
Question : 17
Total: 54
A uniform electric field
E
is present horizontally along the paper throughout region but uniform magnetic field
B
0
is present horizontally (perpendicular to plane of paper in inward direction) right to the line
A
B
as shown. A charge particle having charge
q
and mass
m
is projected vertically upward and crosses the line
A
B
after time
t
0
. The speed of projection if particle moves after
t
0
with constant velocity. (given
q
E
=
m
g
)
is
n
t
0
.
Find the value of 'n'.
(
g
=
9.8
m
∕
s
2
)
Your Answer:
Validate
Solution:
When crosses
A
B
q
v
B
0
cos
θ
=
m
g
q
v
B
0
sin
θ
=
q
E
tan
θ
=
q
E
m
g
=
1
θ
=
π
4
along horizontal
v
cos
θ
=
q
E
m
t
0
u
−
g
t
0
=
v
sin
θ
u
=
(
g
+
q
E
m
)
t
0
=
2
g
t
0
© examsnet.com
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