Question

A circular disc $A$ of radius $r$ is made from an iron plate of thickness $t$ and another circular disc $B$ of radius $4r$ also made from an iron plate of thickness $t/4$ . The relation between moment of inertias $I_A$ and $I_B$ about the same axis is

• A. $I_A>I_B$
• B. $I_A=I_B$
• C. $I_A<I_B$
• D. depends on the actual value of $t$ and $r$ .

Answer 1

Answer: (C)

$I_A=\frac{M_A{r}^2}{2},M_A=\rho \pi r^{2}t$
$I_A=\frac{\left(\rho \pi r^{2}t\right)r^2}{2}=\frac{\rho \pi r^{4}t}{2}$
$I_B=\frac{M_B{\left(4r\right)}^2}{2},M_B=\rho \pi {\left(4r\right)}^2\left(\frac{t}{4}\right)=4\rho \pi r^{2}t$
$I_B=\frac{\left(4\rho \pi r^{2}t\right)\left(16r^2\right)}{2}=\frac{64\rho \pi r^{4}t}{2}$
$I_A<I_B$ .