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Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    A long spring, when stretched by a distance\[x,\]has potential energy U. On increasing the stretching to\[nx,\]the potential energy of the spring will be

    A)  \[\frac{U}{n}\]

    B)  \[nU\]

    C)  \[{{n}^{2}}U\]

    D)  \[\frac{U}{{{n}^{2}}}\]

    Correct Answer: C

    Solution :

     Potential energy of the spring \[U=\frac{1}{2}k{{x}^{2}}\]...(i)   and \[U=\frac{1}{2}k{{(nx)}^{2}}\] \[\Rightarrow \]\[U={{n}^{2}}\frac{1}{2}k\,x\] \[\therefore \]\[U={{n}^{2}}U\]             [From Eq. (i)]


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