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AFMC AFMC Solved Paper-2010

  • question_answer
    The Young's modulus of brass and steel are respectively \[1.0\times {{10}^{11}}N{{m}^{-2}}\] and \[2.0\times {{10}^{11}}N{{m}^{-2}}.\].A brass wire and a steel wire of the same length are extended by 1 mm each under the same force. If radii of brass and steel wires are RB and RS respectively, then

    A) Rs = \[\sqrt{2}\]RB                          

    B) RS = \[\frac{{{R}_{B}}}{2}\]

    C) Rs = 4RB                               

    D) Rs = \[\frac{{{R}_{B}}}{2}\]

    Correct Answer: B

    Solution :

    Increase in length,\[\Delta L=\frac{FL}{Y.A}=\frac{FL}{Y.\pi {{R}^{2}}}\]As F, L and \[\Delta L\]are same hence,\[Y.{{R}^{2}}=a\,cons\tan t\] \[\therefore \]   \[2.0\times {{10}^{11}}R_{s}^{2}=1.0\times {{10}^{11}}R_{B}^{2}\]\[\Rightarrow \]   \[{{R}_{s}}=\frac{{{R}_{B}}}{\sqrt{2}}\]


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