masses $=\mathrm{m}_{1}=\mathrm{m}_{2}$
density $\left(\rho_{1} / \rho_{2}\right)=(1 / 3)$
$\mathrm{I}=\mathrm{Ml}$ of disc $=\left[\left(\mathrm{MR}^{2}\right) / 2\right]$
also $\rho=[(\text { mass }) / \text { (volume) }]$
$=[(\text { mass }) /(\text { area } \times \text { thickness) }]$
$=\left[\mathrm{m} /\left(\pi \mathrm{R}^{2} \mathrm{t}\right)\right]$
$\mathrm{R}^{2}=[\mathrm{m} /(\pi \mathrm{t} \rho)]$
from $( 1)$
$\mathrm{I}=(\mathrm{M} / 2) \mathrm{R}^{2}$
$=(M / 2) \cdot[M /(\pi t \rho)]$
$=\left[M^{2} /(2 \pi T \rho)\right]$
given: thickness $\&$ masses same.
hence I $\propto(1 / p)$
$\left(l_{1} / l_{2}\right)=\left(\rho_{2} / \rho_{1}\right)$ as $t_{1}=t_{2} m_{1}=m_{2}$
$\left(l_{1} / l_{2}\right)=(3 / 1)$
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