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JEE Advanced 2017 Paper 2
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© examsnet.com
Question : 13
Total: 54
A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which of the following statements about its motion is/are correct?
[JEE Adv 2017 P2]
The trajectory of the point A is a parabola
Instantaneous torque about the point in contact with the floor is proportional to sin θ
The midpoint of the bar will fall vertically downward
When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to (1 - cosθ)
Validate
Solution:
(A)
x
=
L
2
s
i
n
θ
y
=
L
c
o
s
θ
Hence
X
2
(
L
2
)
2
+
y
2
L
2
=
1
Hence NOT parabola
(B)
(
Torque about
)
B
=
(
m
g
L
2
s
i
n
θ
)
Hence correct
(C) As there is no external horizontal force and zero initial velocity Hence it will move downwards
(D) Displacement of mid point
=
(
L
2
−
L
2
c
o
s
θ
)
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