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JEE Main 11 April 2015 Paper
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© examsnet.com
Question : 6
Total: 90
A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is
12
rad
s
−
1
the magnitude of its angular momentum about a point on the ground right under the centre of the circle is :
8.64
k
g
m
2
s
−
1
11.52
k
g
m
2
s
−
1
14.4
k
g
m
2
s
−
1
20.16
k
g
m
2
s
−
1
Validate
Solution:
L
0
=
m
v
r
sin
90
°
=
m
(
0.6
ω
)
r
=
2
×
0.6
×
12
×
1
=
14.4
k
g
m
2
/
s
© examsnet.com
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