JEE Main & Advanced Sample Paper JEE Main - Mock Test - 5

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

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

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

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

    Correct Answer: C

    Solution :

    [c] : Here, the springs are connected in series, their resultant force constant K is given by \[\frac{1}{K}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}=\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}{{k}_{2}}}\text{or}\,K=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]


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