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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    In a conductor, if the number of conduction electrons per unit volume is \[8.5\times {{10}^{28}}\text{ }{{m}^{3}}\] and mean free time is 25?s (fem to second), it's approximate resistivity is :- \[\left( {{m}_{e}}=9.1\times {{10}^{31}}kg \right)\] [JEE Main 9-4-2019 Afternoon]

    A) \[{{10}^{-5}}\Omega m\]               

    B) \[{{10}^{-6}}\Omega m\]

    C) \[{{10}^{-7}}\Omega m\]               

    D) \[{{10}^{-8}}\Omega m\]

    Correct Answer: D

    Solution :

    \[\rho =\frac{m}{n{{e}^{2}}\tau }\]     \[=1.67\times {{10}^{-8}}\Omega m\]


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