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 :

1. 1 : 9

2. 1 : 11

3. 1 : 14

4. 1 : 6

Doubtly Asked 30 Oct 2026 NEET UG 2017 Practice test neet Save WhatsApp Report

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Answer: 2. 1 : 11

Key idea: spring constant is inversely proportional to length.

Pieces have lengths \(\tfrac L6,\tfrac L3,\tfrac L2\), so constants \(6k,3k,2k\).

Series: \(\dfrac1{k'}=\dfrac1{6k}+\dfrac1{3k}+\dfrac1{2k}=\dfrac1k\Rightarrow k'=k\).

Parallel: \(k''=6k+3k+2k=11k\).

\(k':k''=1:11\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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