Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    The force constants of two springs are \[{{k}_{1}}\] and\[{{k}_{2}}\] respectively. Both are stretched till their potential energies are equal. The forces \[{{f}_{1}}\] and \[{{f}_{2}}\] applied on them are in the ratio

    A)  \[{{k}_{1}}:{{k}_{2}}\]                                   

    B)  \[{{k}_{2}}:{{k}_{1}}\]

    C) \[\sqrt{{{k}_{1}}}:\sqrt{{{k}_{2}}}\]                         

    D) \[\sqrt{{{k}_{2}}}:\sqrt{{{k}_{1}}}\]

    Correct Answer: C

    Solution :

    Energy of spring \[U=\frac{1}{2}k{{x}^{2}}\] and force      \[f=kx\] From these two, we get                 \[U=\frac{1}{2}k\frac{{{f}^{2}}}{{{k}^{2}}}=\frac{1}{2}\frac{{{f}^{2}}}{k}\] \[\because \]Energies are equal, therefore                 \[\frac{f_{1}^{2}}{{{k}_{1}}}=\frac{f_{2}^{2}}{{{k}_{2}}}\] \[\Rightarrow \]               \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{\sqrt{{{k}_{1}}}}{\sqrt{{{k}_{2}}}}\]


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