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Two similar springs P and Q have spring constants $K_{p}$ and $K_{Q}$ , such that $K_{P} > K_{Q}$ . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs $W_{P} and W_{Q}$ are related as, in case (a) and case (b) respectively

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Wp>WQ;WQ>WP

WP=WQ;WP>WQ

WP=WQ;WP=WQ

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detailed solution

Correct option is A

Here, Kp>KQ Case (a): Elongation (x) in each spring is same. WP=12Kpx2,WQ=12KQx2∴ WP>WQ Case (b): Force of elongation is same. So, x1=FKP and x2=FKQWP=12KPx12=12F2KPWQ=12KQx22=12F2KQ ∴ WP

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Two similar springs P and Q have spring constants Kp and KQ, such that KP>KQ. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b) respectively

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Two similar springs P and Q have spring constants $K_{p}$ and $K_{Q}$ , such that $K_{P} > K_{Q}$ . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs $W_{P} and W_{Q}$ are related as, in case (a) and case (b) respectively