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NEET 2017 · Spring Systems

Spring Systems question from NEET 2017

Q86NEET 2017Difficult

Spring Systems

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

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Answer: (b)

$k \propto \frac{1}{l}$ , so pieces of lengths $\frac{l}{6}, \frac{2l}{6}, \frac{3l}{6}$ have constants $6k, 3k, 2k$ .
In series they rebuild the original spring: $k' = k$ .
In parallel: $k'' = 6k + 3k + 2k = 11k$ .
So $k' : k'' = 1 : 11$ , answer (b).

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