AMU Medical AMU Solved Paper-1999

  • question_answer
    A wire of resistance\[1\,\Omega \]is stretched to double its length. The resistance of the wire will become

    A)  \[1\,\Omega \]                

    B)  \[\frac{1}{4}\,\Omega \]

    C)  \[2\,\Omega \]               

    D)  \[4\,\Omega \]

    Correct Answer: D

    Solution :

    : Resistance of wire \[R=\rho \frac{l}{S}\] Volume of wire \[V=S\times l\] \[\therefore \] \[S=\frac{V/l}{{}}\] \[\therefore \] \[R=\frac{\rho {{l}^{2}}}{V}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\rho l_{2}^{2}}{V}\times \frac{V}{\rho l_{2}^{2}}\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{2{{l}_{1}}} \right)}^{2}}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{4}\] Or \[{{R}_{2}}=4{{R}_{1}}=4\times 1=4\] \[\therefore \] \[{{R}_{2}}=4\,\Omega \]


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