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Question

A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is k. Then they are connected in parallel and force constant is k′′. Then k:k′′ is

A
1 : 9
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B
1 : 11
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C
1 : 14
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D
1 : 16
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Solution

The correct option is B 1 : 11
Let the original length of the spring be l.
Then length of the spring segments will be l6, l3, l2
As we know k1l
So spring constants for spring segments will be k1=6k, k2=3k, k3=2k
For series combination,
1k=16k+12k+13kk=k
For parallel combination,
k′′=6k+3k+2k=11k
kk′′=k11k = 111

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