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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Mechanical Properties of Solids

  • question_answer
    Two, spring P and Q of force constants \[{{k}_{p}}\] and \[kQ\left( kQ=\frac{{{k}_{p}}}{2} \right)\] 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) E/2

    D) E/4

    Correct Answer: C

    Solution :

    [c] Here, \[{{k}_{Q}}=\frac{{{k}_{p}}}{2}\] According to Hooke's law \[\therefore {{F}_{p}}=-{{k}_{p}}{{x}_{p}}\] \[{{F}_{Q}}=-{{k}_{Q}}{{x}_{Q}}\Rightarrow \frac{{{F}_{p}}}{{{F}_{Q}}}=\frac{{{k}_{p}}}{{{k}_{Q}}}\frac{{{x}_{p}}}{{{x}_{Q}}}\] \[{{F}_{p}}={{F}_{Q}}\]  [Given] \[\therefore \,\,\frac{{{x}_{p}}}{{{x}_{Q}}}=\frac{{{k}_{Q}}}{{{k}_{p}}}\]    ?. (i) Energy stored in a spring is\[U=\frac{1}{2}k{{x}^{2}}\] \[\therefore \,\,\frac{{{U}_{p}}}{{{U}_{Q}}}=\frac{{{k}_{p}}x_{p}^{2}}{{{k}_{Q}}x_{Q}^{2}}=\frac{{{k}_{p}}}{{{k}_{Q}}}\times \frac{k_{Q}^{^{2}}}{k_{^{p}}^{2}}=\frac{1}{2}\]     \[\left[ \therefore {{k}_{Q}}=\frac{{{k}_{p}}}{2} \right]\] \[\Rightarrow \,\,\,{{U}_{p}}=\frac{{{U}_{Q}}}{2}=\frac{E}{2}\]             \[[\therefore \,\,\,\,{{U}_{Q}}=E]\]


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