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 is made from an iron plate of thickness t/4. The relation between the moments of inertia I_A and I_B is
(a) I_A > I_B
(b) I_A = I_B
(c) I_A < I_B
(d) depends on the actual values of t and r.

Answer 1

(c) I_A < I_B

Moment of inertia of circular disc of radius r:
I = $\frac{1}{2}m{r}^2$
Mass = Volume × Density
Volume of disc = $\mathrm{\pi }{r}^{2}t$
Here, t is the thickness of the disc.
As density is same for both the rods, we have:
Moment of inertia, $I\propto \mathrm{thickness}\times {\left(\mathrm{radius}\right)}^4$

$$\begin{gathered} \frac{I_A}{I_B}=\frac{t.{\left(r\right)}^4}{{\displaystyle \frac{t}{4}}{\left(4r\right)}^4}<1 \\ \Rightarrow \frac{I_A}{I_B}<1 \end{gathered}$$
$\Rightarrow I_{\mathrm{A}}<I_{\mathrm{B}}$