Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    One-fourth length of a spring of force constant k is cut away. The force constant of the remaining spring will be

    A)  \[\frac{3}{4}k\]                               

    B) \[\frac{4}{3}k\]

    C)  k                                            

    D)  4k

    Correct Answer: B

    Solution :

    Spring constant \[k\propto \frac{1}{l}\] Therefore,  \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] Given,       \[{{l}_{1}}=l,\]\[{{l}_{2}}=3/4l,\] \[{{k}_{1}}=k\] \[\therefore \]  \[\frac{{{k}_{2}}}{k}=\frac{l}{3/4l}\] or \[{{k}_{2}}=\frac{4}{3}k\]


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