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IIT JEE Advanced 2007 Paper 1
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© examsnet.com
Question : 6
Total: 66
Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the centre P (as shown in the figure). Now, the midpoint of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is
F
2
m
a
√
a
2
−
x
2
F
2
m
x
√
a
2
−
x
2
F
2
m
x
a
F
2
m
√
a
2
−
x
2
x
Validate
Solution:
Since F is very small, we have the acceleration as 0 in the vertical direction.
Now, F = 2Tsinθ (1)
Also, Tcosθ =
m
a
x
(2)
Therefore, dividing Eq. (1) by Eq. (2), we get
2 tan θ =
F
max
where tan θ =
√
a
2
−
x
2
x
Therefore,
a
x
=
F
2
m
(
x
√
a
2
−
x
2
)
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