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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Two springs P and Q of force constants\[{{k}_{p}}\]and\[{{k}_{Q}}\] \[\left( {{k}_{Q}}=\frac{{{k}_{p}}}{2} \right)\]are stretched by applying forces of equal magnitude. If the energy stored in Q is E, then the energy stored in P is

    A)  \[E\]                    

    B)         \[2E\]

    C)  \[\frac{E}{8}\]                  

    D)         \[\frac{E}{4}\]

    E)  \[\frac{E}{2}\]

    Correct Answer: E

    Solution :

    Given: \[{{k}_{p}}=2{{k}_{Q}}\] By stretching spring energy given                 \[E=\frac{1}{2}k{{x}^{2}}\]                           \[[\because F=kx]\] Or           \[E=\frac{1}{2}\frac{{{F}^{2}}}{k}\] For spring P                 \[{{E}_{p}}=\frac{1}{2}\frac{{{F}^{2}}}{{{k}_{p}}}\] For spring Q                 \[{{E}_{Q}}=\frac{1}{2}\frac{{{F}^{2}}}{{{k}_{Q}}}\] \[\therefore \]  \[\frac{{{E}_{P}}}{{{E}_{Q}}}=\frac{{{k}_{Q}}}{{{k}_{P}}}=\frac{1}{2}\] Or           \[{{E}_{P}}=\frac{{{k}_{Q}}}{2}=\frac{E}{2}\]


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