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JIPMER Jipmer Medical Solved Paper-2003

  • question_answer
    In a neon discharge tube \[2.9\times {{10}^{18}}\] \[N{{e}^{+}}\] ions move to be the right per second while \[1.2\times {{10}^{8}}\] electron moves to the left per second electric charge is \[1.6\times {{10}^{-19}}C\]. The current in discharge tube is :

    A)  0.66 A towards left        

    B)  0.66 A towards right

    C)  1 A towards right

    D)  zero

    Correct Answer: B

    Solution :

    Current due to both types of ions is in the same direction towards right so \[i={{i}_{1}}+{{i}_{2}}_{,}\] \[i=2.9\times {{10}^{18}}\times 1.6\times {{10}^{-19}}+1.2\times {{10}^{18}}\times 1.6\times {{10}^{-19}}\] \[i=0.66\,amp\]


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