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Question

The springs in fig. A and B are identical but the length of spring A is three times the length of each spring in B. The ratio of period TA/TB is

A
3
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B
1/3
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C
3
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D
1/3
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Solution

The correct option is C 3
Length of each spring in B is L and spring constant is K
Keq=3K
K1LL1K
For B:LB1KB;(Keq)B=3K
For A: Length of A is three times each of that in B.

KA1LA13LBK3
(Keq)A=K3
(Keq)A=K/3,KB=3KTATB=KBKATATB=KBKA=31

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