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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    If \[{{g}_{E}}\] and \[{{g}_{M}}\] are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikans oil drop experiment could be performed on the two surfaces, one will find the ratio\[\frac{\text{electronic}\,\text{charge}\,\text{on}\,\text{the}\,\text{moon}}{\text{electronic}\,\text{charge}\,\text{on}\,\text{the}\,\text{earth}}\] to be

    A)  \[1\]                                    

    B)  zero

    C)  \[\frac{{{g}_{E}}}{{{g}_{M}}}\]                                  

    D)  \[\frac{{{g}_{M}}}{{{g}_{E}}}\]

    Correct Answer: A

    Solution :

    According to Millikans oil drop experiment, electronic charge is given by                 \[q=\frac{6\pi nr({{v}_{1}}+{{v}_{2}})}{E}\] which is independent of g. So,\[\frac{\text{electronic}\,\,\text{charge}\,\,\text{on}\,\,\text{the}\,\,\text{moon}}{\text{electronic}\,\,\text{charge}\,\,\text{on}\,\,\text{the}\,\,\text{earth}}\text{=1}\]


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