question_answer

Two circular discs A and B of the same mass and same thickness are made of two different metals whose densities are\[{{d}_{A}}\] and\[{{d}_{B}}\left( {{d}_{A}}>{{d}_{B}} \right)\] . Their moments of inertia about eh axes passing through their centres of gravity and perpendicular to their planes are\[{{I}_{A}}\] and\[{{I}_{B}}\]. Which is greater, \[{{I}_{A}}\]or\[{{I}_{B}}\]?

Answer:

                Let \[{{r}_{A}}\]and \[{{r}_{B}}\]be the radii of discs A & B. As their mass (m) and thickness (t) are same, therefore, \[m=\left( \pi r_{A}^{2} \right)t\times {{d}_{A}}=\left( \pi r_{B}^{2} \right)\times t\times {{d}_{B}}\]                      \[\therefore \]  \[\frac{r_{A}^{2}}{r_{B}^{2}}=\frac{{{d}_{B}}}{{{d}_{A}}}\] Now \[\frac{{{I}_{A}}}{{{I}_{B}}}=\frac{\frac{1}{2}mr_{A}^{2}}{\frac{1}{2}mr_{B}^{2}}=\frac{r_{A}^{2}}{r_{B}^{2}}=\frac{{{d}_{B}}}{{{d}_{A}}}\] As           \[{{d}_{A}}>{{d}_{g}}\].                                 \[\therefore \]  \[{{I}_{B}}>{{I}_{A}}\]