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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The potential energy of spring when stretched by a distance x is E. The energy of the spring when stretched by x/2 is

    A)  E                                            

    B)  E/2

    C)  E/4                       

    D)         E/6

    Correct Answer: C

    Solution :

    \[\frac{E}{E}=\frac{\frac{1}{2}k{{x}^{2}}}{\frac{1}{2}k{{\left( \frac{x}{2} \right)}^{2}}}\]                 \[\frac{E}{E}=\frac{4}{1}\] \[\Rightarrow \]               \[E=\frac{E}{4}\]


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