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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. if the young?s modulii of copper and steel are respectively \[1.0\times {{10}^{11}}{{\operatorname{Nm}}^{-2}}\]and\[2.0\times {{10}^{11}}\,N{{m}^{-2}}\], the total extension of the composite wire is:     [JEE Main Online Paper ( Held On 23  April 2013 )

    A)                   1.75 mm                            

    B)                                          2.0 mm

    C)                                          1.50mm                             

    D)                                          1.25 mm

    Correct Answer: D

    Solution :

                      \[{{Y}_{c}}\times (\Delta {{L}_{c}}/{{L}_{c}})={{Y}_{s}}\times (\Delta {{L}_{s}}/{{L}_{s}})\] \[\Rightarrow \]\[1{{\times }^{11}}\times \left( \frac{1\times {{10}^{-3}}}{1}=2\times {{10}^{11}} \right)\times \left( \frac{\Delta {{L}_{s}}}{0.5} \right)\] \[\therefore \]\[\Delta {{L}_{s}}=\frac{0.5\times {{10}^{-3}}}{2}=0.25mm\] Therefore, total extension of the composite wire\[=\Delta {{L}_{c}}+\Delta {{L}_{s}}\]     \[=1\,mm+0.25m=1.25m\]


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