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NEET AIPMT SOLVED PAPER SCREENING 2006

  • question_answer
                    The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:                                                                                                                 

    A)                  4U                        

    B)                  8U                        

    C)                  16U  

    D)                 U/4

    Correct Answer: C

    Solution :

                       Let extension produced in a spring be x initially. In stretched condition spring will have potential energy                 \[U=\frac{1}{2}k{{x}^{2}}\]                 where k is spring constant or force constant.                 \[\therefore \frac{{{U}_{1}}}{{{U}_{2}}}=\frac{x_{1}^{2}}{x_{2}^{2}}....(i)\]                 Given, \[{{U}_{1}}=U,\,{{x}_{1}}=2\,cm,\,{{x}_{2}}=8\,cm\]                 putting these values in Eq. (i), we have                 \[\frac{U}{{{U}_{2}}}=\frac{{{(2)}^{2}}}{{{(8)}^{2}}}=\frac{4}{64}=\frac{1}{16}\]                 \[\therefore {{U}_{2}}=16U\]


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