2. Solved Papers
  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    A spring of force constant k is cut into two equal halves. The force constant of each half is

    A)  \[\frac{k}{\sqrt{2}}\]                                    

    B)  k

    C)  \[\frac{k}{2}\]                                  

    D)  2k

    Correct Answer: D

    Solution :

                                    For a spring system \[kl=\]constant \[\therefore \]  \[\frac{k}{k}=\frac{l}{l/2}=2\] \[\Rightarrow \]               \[k=2k\] Hence, force constant of each half spring is 2k.


You need to login to perform this action.
You will be redirected in 3 sec spinner