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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 7
Total: 69
A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is
9
18
27
36
Validate
Solution:
The centripetal force is provided by the tension on the string is
T =
m
r
ω
2
Now,
T
max
= 324 N, m = 0.5 kg and r = 0.5 m. Therefore,
(
ω
max
)
2
=
T
max
m
r
=
324
0.5
×
0.5
= 1296 ⇒
ω
max
= 36 rad/s
© examsnet.com
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