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 co - Physics MCQ NEET (UG) 2017 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Force constant k \propto \frac{1}{L} .Let the original length be 6 L .Lengths are L, 2 L, 3 L .Corresponding k values: k_{1} = 6 k, k_{2} = 3 k, k_{3} = 2 k .Series connection (k^{'}): Re-joining them in series gives back the original spring, so k^{'} = k .Parallel connection (k^{''}): k^{''} = k_{1} + k_{2} + k_{3} = 6 k + 3 k + 2 k = 11 k .Ratio k^{'} : k^{''} = k : 11 k = 1 : 11 .

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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 : 9

1 : 11

1 : 14

1 : 6

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