A) \[{{I}_{\gamma }}=32{{I}_{x}}\]
B) \[{{I}_{\gamma }}=16{{I}_{x}}\]
C) \[{{I}_{\gamma }}={{I}_{x}}\]
D) \[{{I}_{\gamma }}=64{{I}_{x}}\]
Correct Answer: D
Solution :
Mass of disc \[(X),{{m}_{x}}=\pi {{R}^{2}}t\rho \] where, \[\rho =\]density of material of disc \[\therefore \] \[{{I}_{x}}=\frac{1}{2}{{m}_{x}}{{R}^{2}}=\frac{1}{2}\pi {{R}^{2}}t\rho {{R}^{2}}\] \[{{I}_{x}}=\frac{1}{2}\pi \rho t{{R}^{2}}\] ...(i) Mass of disc (Y) \[{{m}_{Y}}=\pi {{(4R)}^{2}}\frac{t}{4}\rho =4\pi {{R}^{2}}t\rho \] and \[{{I}_{Y}}=\frac{1}{2}{{m}_{y}}{{(4R)}^{2}}=\frac{1}{2}4\pi {{R}^{2}}t\rho .16{{R}^{2}}\] \[\Rightarrow \] \[{{I}_{Y}}=32\pi t\rho {{R}^{2}}\] ...(ii) \[\therefore \] \[\frac{{{I}_{Y}}}{{{I}_{X}}}=\frac{32\pi t\rho {{R}^{2}}}{\frac{1}{2}\pi \rho t{{R}^{4}}}=64\] \[\therefore \] \[{{I}_{Y}}=64{{I}_{X}}\]
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