A) Both disc have same moment of inertia
B) disc having less density have larger moment of inertia
C) disc having more density have larger moment of inertia
D) None of the above
Correct Answer: B
Solution :
Since mass of both the disc is same \[\therefore \] \[\pi r_{1}^{2}t{{\rho }_{1}}=\pi r_{2}^{2}t{{\rho }_{2}}\] (\[\because \] Mass = volume\[\times \]density = area\[\times \]thickness\[\times \]density) \[\Rightarrow \] \[\frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\frac{1}{2}Mr_{1}^{2}}{\frac{1}{2}Mr_{2}^{2}}\] \[=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\] i.e., \[I\propto \frac{1}{\rho }\] So, the disc having less density have larger moment of inertia.
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