Question:medium

The close-coiled helical springs 'A' and 'B' are of same material, same coil diameter, same wire diameter and subjected to same load. If the number of turns of spring 'A' is half that of spring 'B', the ratio of deflection of spring 'A' to spring 'B' is

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Deflection ($\delta$) and stiffness ($k$) behaviors in helical springs: - Deflection is directly proportional to the number of turns: \(\delta \propto n\) - Axial spring stiffness is inversely proportional to the number of active turns: \(k = \frac{W}{\delta} = \frac{G d^4}{8 D^3 n} \implies k \propto \frac{1}{n}\) Cutting a spring in half reduces the turns by half, which doubles its stiffness and halves its deflection under the same load.
Updated On: Jul 9, 2026
  • \(\frac{1}{2}\)
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The Correct Option is A

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