CET Karnataka Medical CET - Karnataka Medical Solved Paper-2008

  • question_answer
    Two protons are kept at a separation of\[40{{A}^{o}}\]. \[{{F}_{n}}\] is the nuclear force and \[{{F}_{e}}\]is the electrostatic force between them. Then

    A)  \[{{F}_{n}}>>{{F}_{e}}\]

    B)  \[{{F}_{n}}={{F}_{e}}\]

    C)  \[{{F}_{n}}<<{{F}_{e}}\]

    D)  \[{{F}_{n}}={{F}_{e}}\]

    Correct Answer: C

    Solution :

    \[{{F}_{n}}\] is stronger than \[{{F}_{e}}.{{F}_{n}}\] operates at very short range inside the nucleus as little as \[{{10}^{-15}}m.\]As in the given case two protons are kept at a separation of \[40{\AA}.{{F}_{n}}<<{{F}_{e}}.\]


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