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 \[{{d}_{A}}>{{d}_{g}}\].
\[\therefore \] \[{{I}_{B}}>{{I}_{A}}\]
You need to login to perform this action.
You will be redirected in
3 sec