Question

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 $

Answer 1

Answer: (C)

$ {{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 $