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

  • question_answer
    A steel ring of radius r and cross-section area 'A? is fitted on to a wooden disc of radius R(R > r). If Young's modulus be E, then the force with which the steel ring is expanded is

    A) \[AE\frac{R}{r}\]

    B) \[AE\left( \frac{R-r}{r} \right)\]

    C) \[\frac{E}{A}\left( \frac{R-r}{A} \right)\]

    D) \[\frac{Er}{Ar}\]

    Correct Answer: B

    Solution :

    [b] Initial length (circumference) of the ring \[=2\pi r\] Final length (circumference) of the ring \[=2\pi R\] Change in length \[=2\pi R-2\pi r\] \[Strain=\frac{2\pi (R-r)}{2\pi r}=\frac{R-r}{r}\] Young?s modulus \[E=\frac{F/A}{l/L}=\frac{F/A}{(R-r)/r}\] \[\therefore \,\,F=AE\left( \frac{R-r}{r} \right)\]


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