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

  • question_answer
    A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be                                              [EAMCET 1986]

    A) 8             

    B)                 4

    C)                 2             

    D)                 1

    Correct Answer: A

    Solution :

        \[l=\frac{FL}{\pi {{r}^{2}}r}\]Þ \[l\propto \frac{F}{{{r}^{2}}}\]          (Y and L are constant)             \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{F}_{2}}}{{{F}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times {{(2)}^{2}}=8\] \ \[{{l}_{2}}=8{{l}_{1}}=8\times 1=8mm\]


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