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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Two springs with spring constants k1=1200 N/mk_1=1200 \mathrm{~N} / \mathrm{m} and k2=500 N/mk_2=500 \mathrm{~N} / \mathrm{m} are stretched by the same force. The ratio of potential energy in the springs will be
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Correct option: C
510\frac{5}{10}
The correct answer is C (510\frac{5}{10}), which simplifies to 5/12, because the potential energy is inversely proportional to the spring constant when the force is the same.

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