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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    Charge is distributed within a sphere of radius R with a volume charge density\[\rho (r)=\frac{A}{{{r}^{2}}}{{e}^{{}^{-2r}/{}_{a}}}\], where A and a are constants. If Q is the total charge of this charge distribution, the radius R is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[\frac{a}{2}\,\log \,\left( 1-\frac{Q}{2\,\pi \,a\,A} \right)\] 

    B) \[\frac{a}{2}\,\log \,\left( 1-\frac{1}{\frac{Q}{2\,\pi \,a\,A}} \right)\]

    C) \[a\,\,\log \,\,\left( \frac{1}{1-\frac{Q}{2\,\pi \,a\,A}} \right)\]

    D) \[a\,\,\log \,\,\left( 1-\frac{Q}{2\,\pi \,a\,A} \right)\]

    Correct Answer: B

    Solution :

    \[\int{\rho (r)\,4\,\pi {{r}^{2}}\,dr\,\,=\,\,Q}\] \[\Rightarrow \,\,\int\limits_{0}^{R}{\frac{A}{{{r}^{2}}}{{e}^{-2r/a}}\,4\pi {{r}^{2}}\,dr\,\,=\,\,Q}\] \[\frac{4\pi A\,{{\left[ {{e}^{-2r/a}} \right]}^{R}}_{0}}{\left( \frac{-2}{a} \right)}\,\,=\,\,Q\] \[2a\pi A\,[1-{{e}^{-2R/a}}]\,\,=\,\,Q\] \[R=\frac{a}{2}\,\,ln\,\,\left[ \frac{2\pi aA}{2\pi aA-Q} \right]=\frac{a}{2}\,\,ln\,\,\left( 1-\frac{1}{\frac{Q}{2\pi aA}} \right)\] Option (b)


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