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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of \[\frac{7M}{8}\]and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let \[{{I}_{1}}\] be the moment of inertia of the disc about its axis and \[{{I}_{2}}\]be the moment of inertia of the new sphere about its axis. The ratio \[{{I}_{1}}/{{I}_{2}}\] is given by: [JEE Main 10-4-2019 Afternoon]

    A) 185                 

    B) 65

    C) 285                 

    D) 140

    Correct Answer: D

    Solution :

    \[{{I}_{1}}=\frac{\left( \frac{7M}{8} \right){{(2R)}^{2}}}{2}=\left( \frac{7}{16}\times 4 \right)M{{R}^{2}}=\frac{7}{4}M{{R}^{2}}\] \[{{I}_{2}}=\frac{2}{5}\left( \frac{M}{8} \right)R_{1}^{2}=\frac{2}{5}\left( \frac{M}{8} \right)\frac{{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{80}\] \[\frac{4}{3}\pi {{R}^{3}}=8\left( \frac{4}{3}\pi R_{1}^{3} \right)\] \[{{R}^{3}}=8R_{1}^{3}\] \[R=2{{R}_{1}}\] \[\therefore \]\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{7/4M{{R}^{2}}}{\frac{M{{R}^{2}}}{80}}=\frac{7}{4}\times 80=140\]


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