2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Wave Mechanics

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

  • question_answer
    A metal wire of length \[{{L}_{1}}\]and area of cross-section A is attached to a rigid support. Another metal wire of length \[{{L}_{2}}\] and of the same cross-sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire. If \[{{Y}_{1}}\]and \[{{Y}_{2}}\] are the young's moduli of the wires respectively, the effective force constant of the system of two wires is                 

    A) \[\frac{({{Y}_{1}}{{Y}_{2}})A}{2({{Y}_{1}}{{L}_{2}}+{{Y}_{2}}{{L}_{1}})}\]

    B) \[\frac{({{Y}_{1}}{{Y}_{2}})A}{{{({{L}_{1}}{{L}_{2}})}^{1/2}}}\]

    C) \[\frac{({{Y}_{1}}{{Y}_{2}})A}{{{Y}_{1}}{{L}_{2}}+{{Y}_{2}}{{L}_{1}}}\]

    D) \[\frac{{{({{Y}_{1}}{{Y}_{2}})}^{1/2}}A}{{{({{L}_{2}}{{L}_{1}})}^{1/2}}}\]

    Correct Answer: C

    Solution :

    [c] Using the usual expression for the Young's modulus, the force constant for the wire can be written as \[k=\frac{F}{\Delta l}=\frac{YA}{L}\]where the symbols have their usual meanings. Now the two wires together will have an effective force constant \[\left[ \frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \right]\] Substituting the corresponding lengths and the Young's moduli we get the answer.


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