Two discs of same mass and thickness are made of materials having different densities. Which one of them will have larger M.I.?
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Since mass of both the discs is same $\therefore\pi\text{r}^2_1\text{t}\rho_1=\pi\text{r}_2^2\text{t}\rho^2$ [Mass = Vloume × Density = Area × Thickness × Density] $\therefore\frac{\rho_1}{\rho_2}=\frac{\text{r}^2_2}{\text{r}_1}$ $\therefore\frac{\text{I}_1}{\text{I}_2}=\frac{\frac{1}{2}\text{Mr}_1^2}{\frac{1}{2}\text{Mr}^2_2}=\frac{\text{r}^2_1}{\text{r}_2^2}$ $\frac{\rho_2}{\rho_1}\text{i.e., I}\propto\frac{\text{I}}{\rho}$This shows that M.I. of the disc having less density will be larger.
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