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IIT JEE Advanced 2006 Paper 1
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© examsnet.com
Question : 7
Total: 120
A massless rod is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to x. Further, it is observed that the frequency of 1st harmonic (fundamental frequency) in AB is equal to 2nd harmonic frequency in CD. Then, length of BO is
L/5
4L/5
3L/4
L/4
Validate
Solution:
We have
T
1
+
T
2
= mg (1)
T
1
x
=
T
2
(
L
–
x
)
(2)
Also, we have
1
2
l
√
T
1
µ
=
2
2
l
√
T
2
µ
(3)
where l is the length of AB and CD and for AB and CD, we have
µ =
Mass
Length
Therefore,
T
1
=
4
T
2
(4)
From Eq. (2), we get
4
T
2
x
=
T
2
(L - x)
⇒ 5x = L
⇒ x =
L
5
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