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JEE Advanced 2019 Paper 2
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© examsnet.com
Question : 4
Total: 54
A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical ?
[g is the acceleration due to gravity]
The normal reaction force from the floor on the rod will be
M
g
16
The angular acceleration of the rod will be
2
g
L
The angular speed of the rod will be
√
3
g
2
L
The radial acceleration of the rod’s center of mass will be
3
g
4
Validate
Solution:
→
m
g
(
l
2
sin
60
)
=
m
l
2
3
α
√
3
g
4
=
l
3
α
⇒
α
=
3
√
3
g
4
l
⇒
m
g
l
2
(
1
−
cos
60
)
=
1
2
m
l
2
3
ω
2
g
4
=
l
6
ω
2
ω
2
=
3
g
2
l
→ radial acceleration
a
=
1
2
ω
2
a
=
3
g
4
→
M
g
−
N
=
M
(
3
g
4
×
1
2
+
3
√
3
g
4
l
×
√
3
2
l
)
M
g
−
N
=
M
g
(
3
8
+
9
16
)
N
=
M
g
−
15
M
g
16
=
M
g
16
© examsnet.com
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