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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 34
Total: 54
List I describes four systems, each with two particles
A
and
B
in relative motion as shown in figure. List II gives possible magnitudes of then relative velocities (in
m
s
−
1
) at time
t
=
π
3
s
.
List-I
List-II
(I)
A
and
B
are moving on a horizontal circle of radius
1
m
with uniform angular speed
ω
=
1
rad
s
−
1
. The initial angular positions of
A
and
B
at time
t
=
0
are
θ
=
0
and
θ
=
π
2
respectively.
(P)
√
3
+
1
2
(II)
Projectiles
A
and
B
are fired (in the same vertical plane) at
t
=
0
and
t
=
0.1
s
respectively, with the same speed
v
=
5
π
√
2
m
s
−
1
and at
45
∘
from the horizontal plane. The initial separation between
A
and
B
is large enough so that they do not collide,
(
g
=
10
m
s
−
2
)
.
(Q)
(
√
3
−
1
)
√
2
(III)
Two harmonic oscillators
A
and
B
moving in the
x
direction according to
x
A
=
x
0
sin
t
t
0
and
x
B
=
x
0
sin
(
t
t
0
+
π
2
)
respectively, starting from
t
=
0
. Take
x
0
=
1
m
,
t
0
=
1
s
.
(R)
√
10
(IV)
Particle
A
is rotating in a horizontal circular path of radius
1
m
on the
x
y
plane, with constant angular speed
ω
=
1
rad
s
−
1
. Particle
B
is moving up at a constant speed
3
m
s
−
1
in the vertical direction as shown in the figure. (Ignore gravity.)
(S)
√
2
(T)
√
25
π
2
+
1
Which one of the following options is correct?
I
→
R
,
II
→
T
,
III
→
P
,
IV
→
S
I
→
S
,
II
→
P
,
III
→
Q
,
IV
→
R
I
→
S
,
II
→
T
,
III
→
P
,
IV
→
R
I
→
T
,
II
→
P
,
III
→
R
,
IV
→
S
Validate
Solution:
(I)
v
BA
2
=
v
A
2
+
v
B
2
−
2
v
AB
cos
θ
As
ω
A
=
ω
B
,
θ
=
90
∘
remains constant.
Also,
v
A
=
v
B
=
1
m
∕
s
So,
v
BA
=
√
2
m
∕
s
(II)
→
u
A
=
5
π
2
∧
i
+
5
π
2
∧
j
→
v
A
=
5
π
2
∧
i
+
(
5
π
2
−
10
⋅
π
3
)
∧
j
=
5
π
2
∧
i
−
5
π
6
∧
j
→
u
B
=
−
5
π
2
∧
i
+
5
π
2
∧
j
→
u
B
=
−
5
π
2
∧
i
−
(
5
π
6
+
1
)
∧
j
→
v
B
,
A
=
−
5
π
∧
i
−
∧
j
v
BA
=
√
25
π
2
+
1
(III)
x
A
=
sin
t
v
A
=
cos
t
=
1
2
m
∕
s
x
B
=
cost
v
B
=
−
sin
t
=
−
√
3
2
m
∕
s
v
BA
=
−
√
3
2
−
1
2
(IV)
→
v
A
&
→
v
B
are always perpendicular
So,
|
→
v
BA
|
=
√
v
A
2
+
v
B
2
=
√
10
m
∕
s
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