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)

$I_{A}=\frac{M_{A} r^{2}}{2},M_{A}=\rho \pi r^{2}t$
$I_{A}=\frac{\left(\rho \pi r^{2} t\right) r^{2}}{2}=\frac{\rho \pi r^{4} t}{2}$
$I_{B}=\frac{M_{B} \left(4 r\right)^{2}}{2},M_{B}=\rho \pi \left(4 r\right)^{2}\left(\frac{t}{4}\right)=4\rho \pi r^{2}t$
$I_{B}=\frac{\left(4 \rho \pi r^{2} t\right) \left(16 r^{2}\right)}{2}=\frac{64 \rho \pi r^{4} t}{2}$
$I_{A} < I_{B}$ .