BCECE Medical BCECE Medical Solved Papers-2001

  • question_answer
    The masses of three wires of copper are in the ratio 1:2:3 and their lengths are in the ratio 3 : 2 : 1. The ratio of their resistances is :

    A)  3 : 2 : 1       

    B)  1 : 2 : 3

    C)  27 : 6 : 1      

    D)  1 : 6 : 27

    Correct Answer: C

    Solution :

    Resistance of wire                 \[R=\frac{\rho l}{A}\] or            \[R\propto \frac{{{l}^{2}}}{Al}\] or            \[R\propto \frac{{{l}^{2}}}{V}\] or            \[R\propto \frac{{{l}^{2}}d}{m}\] (where d is density and m is mass) or            \[R\propto \frac{{{l}^{2}}}{m}\] Here,     \[{{l}_{1}}:{{l}_{2}}:{{l}_{3}}=3:2:1\]                 \[{{m}_{1}}:{{m}_{2}}:{{m}_{3}}=1:2:3\] \[\therefore \] \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}:\,\,:\frac{{{l}_{1}}^{2}}{{{m}_{1}}}:\frac{{{l}_{2}}^{2}}{{{m}_{2}}}:\frac{{{l}_{3}}^{2}}{{{m}_{3}}}\] or            \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}:\,\,:\frac{9}{1}:\frac{4}{2}:\frac{1}{3}\] or \[{{R}_{1}}:{{R}_{2}}:{{R}_{3}}:\,\,:\,27:6:1\]


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