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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    A metal wire of circular cross-section has a resistance \[{{R}_{1}}.\]The wire is now stretched without breaking, so that its length is   doubled and the density is assumed to remain the same. If the resistance of the wire now becomes \[{{R}_{2}},\]then \[{{R}_{2}}:{{R}_{1}}\]is

    A)  1 : 1   

    B)  1 : 2   

    C)  4:1   

    D)  1 : 4

    Correct Answer: C

    Solution :

     As we know that, \[{{R}_{1}}=\rho \frac{l}{a}=\rho \frac{{{l}^{2}}}{V}\] where,\[l=\] length of wire a = area of cross-section of the       wire and V = volume of the wire \[{{R}_{1}}\propto {{l}^{2}}\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{2}}\] \[\Rightarrow \] \[{{R}_{2}}:{{R}_{1}}=4:1\]


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