JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Mass per unit area of a circular disc of radius a depends on the distance r from its centre as\[\sigma \,(r)=A+Br\]. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is: [JEE MAIN Held on 07-01-2020 Evening]

    A)  \[2\pi {{a}^{4}}\,\left( \frac{A}{4}+\frac{B}{5} \right)\]

    B) \[\pi {{a}^{4}}\,\left( \frac{A}{4}+\frac{aB}{5} \right)\]

    C) \[2\pi {{a}^{4}}\,\left( \frac{A}{4}+\frac{aB}{5} \right)\]

    D) \[2\pi {{a}^{4}}\,\left( \frac{aA}{4}+\frac{B}{5} \right)\]

    Correct Answer: C

    Solution :

    [c] \[I=\int{dm{{r}^{2}}}=\int{\sigma \,2\,\pi \,rdr\cdot {{r}^{2}}}\] \[\Rightarrow \,\,\,I=2\pi \int\limits_{0}^{a}{(A+Br)}\,{{r}^{3}}dr=2\pi \left[ \frac{A{{a}^{4}}}{4}+\frac{B{{a}^{5}}}{5} \right]\] \[\Rightarrow \,\,\,I=2\pi {{a}^{4}}\left[ \frac{A}{4}+\frac{Ba}{5} \right]\]

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