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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    A conductor wire having\[{{10}^{29}}\]free electrons/ \[{{m}^{3}}\]carries a current of 20A. If the cross-section of the wire is\[1m{{m}^{2}}\], then the drift velocity of electrons will be \[\left( e=1.6\times {{1}^{-19}}C \right)\]

    A)  \[1.25\times {{10}^{-4}}m{{s}^{-1}}~\]

    B)  \[1.25\times {{10}^{-3}}m{{s}^{-1}}\]

    C)  \[1.25\times {{10}^{-5}}m{{s}^{-1}}\]

    D)  \[6.25\times {{10}^{-3}}m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    Current, \[I=nAe{{v}_{d}}\] \[\therefore \]Drift velocity \[{{v}_{d}}=\frac{I}{nAe}\] \[=\frac{20}{{{10}^{29}}\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}}\] \[{{v}_{d}}=1.25\times {{10}^{-3}}m{{s}^{-1}}\]


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