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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A spring of constant 100 N/m100 \mathrm{~N} / \mathrm{m} is stretched by applying equal forces each of magnitude FF at the two ends. The energy stored in the spring is 200 J200 \mathrm{~J}. Now, spring is cut into two equal parts and one of the part is stretched by applying equal forces each of magnitude FF at the two ends. The energy stored is
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Correct option: B
100 J100 \mathrm{~J}
Option B (100 J100 \mathrm{~J}) is correct because when the spring is cut in half, the spring constant doubles, and the energy stored is halved when the same force is applied.

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