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

  • question_answer
    At \[40{}^\text{o}C,\]a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from \[40{}^\text{o}C\]to \[20{}^\text{o}C\] it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young's modulus of brass are \[{{10}^{5}}/{}^\text{o}C\]and \[{{10}^{11}}N/{{m}^{2}},\] respectively; \[g=10\,m{{s}^{-2}}\])                                         [JEE Main Held on 12-4-2019 Morning]

    A) \[1.5\,kg\]                    

    B) \[9\text{ }kg\]

    C) \[0.9\text{ }kg\]                       

    D) \[0.5\text{ }kg\]

    Correct Answer: B

    Solution :

    \[Mg=\left( \frac{Ay}{\ell } \right)\Delta \ell \]  \[\frac{\Delta \ell }{\ell }=\alpha \Delta T\] \[Mg=(Ay)\alpha \Delta T=2\pi \] It is closest to 9.


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