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NEET AIPMT SOLVED PAPER 2001

  • question_answer
                    Two springs A and B have force constants \[{{k}_{A}}\] and \[{{k}_{B}}\] such that \[{{k}_{B}}=2{{k}_{A}}\]. The four ends of the springs are stretched by the same force, If energy stored in spring A is E, then energy stored in spring B is:

    A)                                                                                                                                                                                            E/2        

    B)                 2 E                         

    C)                 E                             

    D)                 4 E

    Correct Answer: A

    Solution :

                              Key Idea: When a string is stretched, then work done in stretching it through a distance x is the potential energy stored in it.                 Potential energy stored = Work done is stretching \[or U=\frac{1}{2}k\,{{x}^{2}}\] \[Also F=k\,x\] \[or x=\frac{F}{x}\] \[So, U=\frac{1}{2}k\,{{\left( \frac{F}{k} \right)}^{2}}\] \[i.e., U\propto \,\,\frac{1}{k}(\text{for}\,\text{constant}\,\text{force})\] \[\therefore \frac{{{U}_{B}}}{{{U}_{A}}}=\frac{{{k}_{A}}}{{{k}_{B}}}\] but         \[{{k}_{B}}=2{{k}_{A}}\] \[\therefore {{U}_{B}}={{U}_{A}}\times \frac{{{k}_{A}}}{2\,{{k}_{A}}}=\frac{{{U}_{A}}}{2}=\frac{E}{2}\]


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