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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    A uniform circular disc of radius R lies in the X- Y plane with its centre coinciding with the origin of the co-ordinate system. Its moment of inertia about an axis, lying in the X-Y plane, parallel to the X-axis and passing through a point on the Y-axis at a distance y = 2R is \[{{I}_{1}}.\]Its moment of inertia about an axis lying in a plane perpendicular to X-Y plane passing through a point on the X-axis at a distance x = d is \[{{I}_{2}}.\]If \[{{I}_{1}}=\text{ }{{I}_{2}},\]the value of d is :

    A)  \[\frac{\sqrt{19}}{2}R\]                               

    B)  \[\frac{\sqrt{17}}{2}R\]

    C)   \[\frac{\sqrt{15}}{2}R\]                              

    D)  \[\frac{\sqrt{13}}{2}R\]

    Correct Answer: C

    Solution :

                     \[{{I}_{1}}=\frac{M{{R}^{2}}}{4}+M{{(2R)}^{2}}\] \[=\frac{M{{R}^{2}}}{4}+4M{{R}^{2}}\] \[=\frac{17\,M{{R}^{2}}}{4}\] \[{{I}_{2}}=\frac{M{{R}^{2}}}{2}+M{{d}^{2}}\]                 Given,                   \[{{I}_{1}}={{I}_{2}}\]                 \[\therefore \]  \[\frac{17\,M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}+M{{d}^{2}}\]                 \[\Rightarrow \]               \[M{{d}^{2}}=\frac{17}{4}M{{R}^{2}}-\frac{M{{R}^{2}}}{2}\]                 \[\Rightarrow \]               \[{{d}^{2}}=\frac{15}{4}{{R}^{2}}\] \[\therefore \]  \[d=\sqrt{\frac{15}{2}}R\]


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