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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 23
Total: 54
At time
t
=
0
, a disk of radius
1
m
starts to roll without slipping on a horizontal plane with an angular acceleration of
α
=
2
3
rad
s
−
2
. A small stone is stuck to the disk. At
t
=
0
, it is at the contact point of the disk and the plane. Later, at time
t
=
√
π
s
, the stone detaches itself and flies off tangentially from the disk. The maximum height (in
m
) reached by the stone measured from the plane is
1
2
+
x
10
. The value of
x
is____ , [Take
.
g
=
10
m
s
−
2
]
Your Answer:
Validate
Solution:
A
t
t
=
0
,
ω
=
0
at
t
=
√
π
,
ω
=
α
t
=
2
3
√
π
,
v
=
ω
r
=
2
3
√
π
θ
=
1
2
α
t
2
θ
=
1
2
×
2
3
×
π
=
π
3
θ
=
60
∘
v
y
=
v
sin
60
=
√
3
2
V
h
=
u
y
2
2
g
=
3
4
v
2
2
g
h
=
3
4
×
4
9
π
2
g
h
=
3
π
9
×
2
g
=
π
6
g
Maximum height from plane,
H
=
R
2
+
h
H
=
1
2
+
π
6
×
10
x
=
π
6
;
x
=
0.52
© examsnet.com
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