J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    When a uniform wire of resistance R is stretched uniformly to increase its length by \[10%,\]the new resistance value would

    A)  remain as R     

    B)  become \[1.21R\]

    C)  become \[1.10\text{ }R\]  

    D)  become \[1.20\text{ }R\]

    Correct Answer: B

    Solution :

    Volume remains constant. \[\therefore \] \[{{A}_{1}}{{l}_{1}}={{A}_{2}}{{l}_{2}}\] \[{{A}_{1}}l={{A}_{2}}(1.1\,\,l)\] \[{{A}_{1}}={{A}_{2}}(1.1)\] \[{{R}_{1}}=\frac{\rho {{l}_{1}}}{{{A}_{1}}}\] Similarly,     \[{{R}_{2}}=\frac{\rho {{l}_{2}}}{{{A}_{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\rho \frac{{{l}_{1}}}{{{A}_{1}}}}{\rho \frac{{{l}_{2}}}{{{A}_{2}}}}\] \[=\frac{{{l}_{1}}}{{{A}_{1}}}\times \frac{{{A}_{2}}}{{{l}_{2}}}\] \[=\frac{l}{(1.1){{A}_{2}}}\times \frac{{{A}_{2}}}{(1.1)l}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{1.21}\] \[\Rightarrow \] \[{{R}_{2}}=1.21{{R}_{1}}\]


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