Q. A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$, and another disc $Y$ of radius 4$R\ $is made from an iron plate of thickness $t$/4. Then the relation between the moment of inertia $I_{X\ }$and $I_{Y}$ is
A) $I_{Y} = \ {32\ I}_{X}$
B) $I_{Y} = \ {16\ I}_{X}$
C) $I_{Y} = \ I_{X}$
D) $I_{Y} = \ {64\ I}_{X}$
Explanation
(d) Mass of disc $(X),\ m_{X} = \pi R^{2}t\rho$
Where, $\rho =$ density of material of disc
$\therefore$ $I_{X} = \frac{1}{2}m_{X}R^{2} = \frac{1}{2}\pi R^{2}t\rho R^{2}$
$I_{X} = \frac{1}{2}\pi\rho tR^{4}$ …(i)
Mass of disc $(Y)$
$m_{Y} = \pi(4R)^{2}\frac{t}{4}\rho = 4\pi R^{2}t\rho$
and $I_{Y} = \frac{1}{2}m_{Y}(4R)^{2} = \frac{1}{2}4\pi R^{2}\rho t.16R^{2}$
$\Rightarrow$ $I_{Y} = 32\pi t\rho R^{4}$ …(ii)
$\therefore$ $\frac{I_{Y}}{I_{X}} = \frac{32\pi t\rho R^{4}}{\frac{1}{2}\pi\rho tR^{4}} = 64$
$\therefore$ $I_{Y} = 64\ I_{X}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.