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JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Mock Test - Gravitation and Mechanical Properties of Solid

  • question_answer
    Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is 0.5 cm, the elongation (\[l\]) of each wireis\[{{Y}_{s}}(steel)=2.0\times {{10}^{11}}N/{{m}^{2}}\] \[{{Y}_{c}}(copper)=1.2\times {{10}^{11}}N/{{m}^{2}}\]

    A) \[{{l}_{s}}=0.75\,cm,\,{{l}_{c}}=1.25\,cm\]

    B)  \[{{l}_{s}}=1.25\,cm,\,{{l}_{c}}=0.75\,cm\]

    C)  \[{{l}_{s}}=0.25\,cm,\,{{l}_{c}}=0.75\,cm\]

    D)  \[{{l}_{s}}=0.75\,cm,\,{{l}_{c}}=0.25\,cm\]

    Correct Answer: A

    Solution :

    [a] \[l\propto \frac{1}{Y}\Rightarrow \frac{{{Y}_{s}}}{{{Y}_{c}}}=\frac{{{l}_{c}}}{{{l}_{s}}}\Rightarrow \frac{{{l}_{c}}}{{{l}_{s}}}=\frac{2\times {{10}^{11}}}{1.2\times {{10}^{11}}}=\frac{5}{3}\](i) Also \[{{l}_{c}}-{{l}_{s}}=0.5\]                                                (ii) On solving (i) and (ii), we get \[{{l}_{c}}=0.5\]cm and \[{{l}_{s}}=0.75\]cm


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