JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    A uniform wire (Young?s modulus\[2\times {{10}^{11}}{{\operatorname{Nm}}^{-2}}\]) is subjected to longitudinal tensile stress of\[5\times {{10}^{7}}{{\operatorname{Nm}}^{-2}}\]. If the  overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[1.0\times {{10}^{-4}}\]                   

    B)  \[1.5\times {{10}^{-4}}\]

    C)  \[0.25\times {{10}^{-4}}\]                 

    D)  \[5\times {{10}^{-4}}\]

    Correct Answer: C

    Solution :

     Given, \[y=2\times {{10}^{11}}\,N{{m}^{-2}}\] Stress \[\left( \frac{F}{A} \right)=5\times {{10}^{7}}N{{m}^{-2}}\] \[\Delta V=0.02%=2\times {{10}^{-4}}{{m}^{3}}\] \[\frac{\Delta r}{r}=?\] \[\gamma =\frac{stress}{strain}\Rightarrow strain\left( \frac{\Delta \ell }{{{\ell }_{0}}} \right)=\frac{\gamma }{stress}\] ?(i) \[\Delta V=2\pi {{\ell }_{0}}\Delta r-\pi {{r}^{2}}\Delta \ell \]                                        ?(ii) From eqns (i) and (ii) putting the value of \[\Delta \ell ,{{\ell }_{0}}\]and \[\Delta V\]and solving we get \[\frac{\Delta r}{r}=0.25\times {{10}^{-4}}\]


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