RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    A mass M is suspended by two springs of force constants\[{{k}_{1}}\]and\[{{k}_{2}}\]respectively as shown in the diagram. The total elongation (stretch) of the two springs is

    A)  \[\left( \frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}} \right)\]

    B)  \[\frac{2({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}mg\]

    C)  \[\frac{2({{k}_{1}}+{{k}_{2}})}{{{k}_{1}}{{k}_{2}}}mg\]              

    D)  None of the above          

    Correct Answer: A

    Solution :

     Effective force constant \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]


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