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WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    Two massless springs of force constants \[{{k}_{1}}\]and \[{{k}_{2}}\]are joined end to end. The resultant force constant k of the system is

    A)  \[k=\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}}\]                         

    B)  \[k=\frac{{{k}_{1}}-{{k}_{2}}}{{{k}_{1}}{{k}_{2}}}\]

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

    D)  \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}-{{k}_{2}}}\]

    Correct Answer: C

    Solution :

                     In series, resultant force constant is given as \[\frac{1}{{{k}_{eq}}}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}\]                 \[\Rightarrow \]               \[{{k}_{eq}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]


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