Answer:
Here, \[{{m}_{1}}={{m}_{2}}\] \[\therefore
\] \[{{V}_{1}}{{\rho }_{1}}={{V}_{2}}{{\rho }_{2}}\]
\[\left(
\pi r_{1}^{2} \right)t.{{\rho }_{1}}=\left( \pi r_{2}^{2} \right)t{{\rho
}_{2}}\]
\[\frac{{{\rho
}_{1}}}{{{\rho }_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] ??..
(i)
Now, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\frac{1}{2}{{m}_{1}}r_{1}^{2}}{\frac{1}{2}{{m}_{2}}r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] \[\left(
\because \,\,\,{{m}_{1}}={{m}_{2}} \right)\]
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{\rho
}_{2}}}{{{\rho }_{1}}}\] ???
using (i)
\[\therefore \] \[I\propto
\frac{1}{\rho }\]
The disc of material
with lower density will have larger moment of inertia.
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