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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    A spring whose unstretched length is \[l\] has a force constant k. The spring is cut into two pieces of un stretched lengths \[{{l}_{1}}\]and \[{{l}_{2}}\] where, \[{{l}_{1}}=n{{l}_{2}}\]and n is an integer. The ratio \[{{k}_{1}}/{{k}_{2}}\]of the corresponding force constants, \[{{k}_{1}}\]and \[{{k}_{2}}\]will be : [JEE Main 12-4-2019 Afternoon]

    A) \[\frac{1}{{{n}^{2}}}\]                              

    B) \[{{n}^{2}}\]

    C) \[\frac{1}{n}\]

    D)   \[n\]

    Correct Answer: C

    Solution :

    \[{{k}_{1}}=\frac{C}{{{\ell }_{1}}}\] \[{{k}_{2}}=\frac{C}{{{\ell }_{2}}}\]           \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{C{{\ell }_{2}}}{{{\ell }_{1}}C}{{\ell }_{2}}=\frac{{{\ell }_{2}}}{n{{\ell }_{2}}}=\frac{1}{n}\]             


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