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JEE Advanced 2020 Paper 1
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© examsnet.com
Question : 10
Total: 54
A uniform electric field,
→
E
=
−
400
√
3
^
y
N
C
−
1
is applied in a region. A charged particle of mass
m
carrying positive charge
q
is projected in this region with an initial speed of
2
√
10
×
10
6
m
s
−
1
.
This particle is aimed to hit a target
T
,
which is
5
m
away from its entry point into the field as shownschematically in the figure. Take
q
m
=
10
10
C
k
g
−
1
.
Then
[JEE Adv 2020 P1]
The particle will hit T if projected at an angle
45
°
from the horizontal
The particle will hit T if projected either at an angle
30
°
or
60
°
from the horizontal
Time taken by the particle to hit
T
could be
√
5
6
µ
s
as well as
√
5
2
µ
s
Time taken by the particle to hit
T
is
√
5
3
µ
s
Validate
Solution:
a
=
q
E
m
=
10
10
×
400
√
3
∴
R
=
u
2
‌
sin
(
2
θ
)
a
⇒
5
=
4
×
10
×
10
12
×
sin
(
2
θ
)
10
10
×
400
√
3
⇒
sin
(
2
θ
)
=
√
3
2
⇒
2
θ
=
60
°
or
120
°
⇒
θ
=
30
°
or
60
°
∴
Particle hits the target
if
θ
=
30
°
or
θ
=
60
°
∴
T
1
=
2
u
‌
sin
‌
θ
a
=
2
×
2
√
10
×
1
2
×
10
6
10
10
×
400
√
3
=
√
5
6
µ
s
and,
T
2
=
2
×
2
√
10
×
10
6
×
√
3
2
10
10
×
400
√
3
=
√
5
2
µ
s
∴ Answer (B, C)
© examsnet.com
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