Question

A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness r/4. Then the relation between Ac moments of inertia $ {{I}_{x}} $ and $ {{I}_{\gamma }} $ is

• A. $ {{I}_{\gamma }}=32{{I}_{x}} $
• B. $ {{I}_{\gamma }}=16{{I}_{x}} $
• C. $ {{I}_{\gamma }}={{I}_{x}} $
• D. $ {{I}_{\gamma }}=64{{I}_{x}} $

Answer 1

Answer: (D)

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}} $