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NEET Sample Paper NEET Sample Test Paper-71

  • question_answer
    A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about

    A) 0.1 joule                       

    B) 0.2 joule

    C) 0.3 joule                       

    D) 0.5 joule

    Correct Answer: A

    Solution :

    \[\Delta PE=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})=\frac{1}{2}\times 10\left[ {{(0.25)}^{2}}-{{(0.20)}^{2}} \right]\] \[=5\,\,\times \,\,0.45\,\,\times \,\,0.05\,\,=\,\,0.1 J\]


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