2. Solved Papers
  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    When a spring is stretched by 10 cm, the potential energy stored is E. When the spring is stretched by 10 cm more, the potential energy stored in the spring becomes

    A)  2E              

    B)  4 E

    C)  6E              

    D)  10 E

    Correct Answer: B

    Solution :

     For 1st situation, \[E=\frac{1}{2}k{{(10\,\times \,{{10}^{-2}})}^{2}}\] ?(i) For 2nd situation, \[E=\frac{1}{2}k{{(20\times {{10}^{-2}})}^{2}}\] ?(ii) On comparing Eqs. (i) and (ii), we get \[E=4E\]


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