JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Mechanical Properties of Solids

  • question_answer
    The length of elastic string, obeying Hooke's law is \[{{\ell }_{1}}\]metres when the tension 4N and\[{{\ell }_{2}}\]metres when the tension is 5N. The length in metres when the tension is 9N is -

    A) \[5{{\ell }_{1}}-4{{\ell }_{2}}\]

    B) \[5{{\ell }_{2}}-4{{\ell }_{1}}\]

    C) \[9{{\ell }_{1}}-8{{\ell }_{2}}\]

    D) \[9{{\ell }_{2}}-8{{\ell }_{1}}\]

    Correct Answer: B

    Solution :

    [b] Let \[{{\ell }_{0}}\] be the unstretched length and \[{{\ell }_{3}}\] be the length under a tension of 9 N. Then \[Y=\frac{4{{\ell }_{0}}}{A({{\ell }_{1}}-{{\ell }_{0}})}=\frac{5{{\ell }_{0}}}{A({{\ell }_{2}}-{{\ell }_{0}})}=\frac{9{{\ell }_{0}}}{A({{\ell }_{3}}-{{\ell }_{0}})}\] These give \[\frac{4}{{{\ell }_{1}}-{{\ell }_{0}}}=\frac{5}{{{\ell }_{2}}-{{\ell }_{0}}}\Rightarrow {{\ell }_{0}}=5{{\ell }_{1}}-4{{\ell }_{2}}\] Further, \[\frac{4}{{{\ell }_{1}}-{{\ell }_{0}}}=\frac{9}{{{\ell }_{2}}-{{\ell }_{0}}}\] Substituting the value of \[{{\ell }_{0}}\]and solving, we get \[{{\ell }_{3}}=5{{\ell }_{2}}-4{{\ell }_{1}}\]


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