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JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • 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 [EAMCET 1986]

    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 :

        Initial length (circumference) of the ring = 2pr Final length (circumference) of the ring = 2pR Change in length = 2pR ? 2pr.             \[\text{strain}=\frac{\text{change in length}}{\text{original length}}\]\[=\frac{2\pi (R-r)}{2\pi r}\]\[=\frac{R-r}{r}\] Now Young's modulus \[E=\frac{F/A}{l/L}=\frac{F/A}{(R-r)/r}\] \\[F=AE\left( \frac{R-r}{r} \right)\]


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