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Two disc having mass ratio $1 / 2$ and diameter ratio $2 / 1$ then find ratio of moment of inertia.
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Verified Answer
The correct answer is:
$2: 1$
Given, mass ratio of two disc
$\mathrm{m}_1: \mathrm{m}_2=1: 2 \text { i.e. } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{2}$
and diameter ratio $\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{2}{1} \Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2}{1}$
$\therefore$ Ratio of their moment of inertia,
$\begin{array}{ll}
\quad \frac{\mathrm{I}_1}{\mathrm{I}_2}= & \frac{\frac{\mathrm{m}_1 \mathrm{r}_1^2}{2}}{\frac{\mathrm{m}_2 \mathrm{r}_2^2}{2}}=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \cdot\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\frac{1}{2}\left(\frac{2}{1}\right)^2=\frac{2}{1} \\
\therefore \quad \mathrm{I}_1: \mathrm{I}_2=2: 1
\end{array}$
$\mathrm{m}_1: \mathrm{m}_2=1: 2 \text { i.e. } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{1}{2}$
and diameter ratio $\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{2}{1} \Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2}{1}$
$\therefore$ Ratio of their moment of inertia,
$\begin{array}{ll}
\quad \frac{\mathrm{I}_1}{\mathrm{I}_2}= & \frac{\frac{\mathrm{m}_1 \mathrm{r}_1^2}{2}}{\frac{\mathrm{m}_2 \mathrm{r}_2^2}{2}}=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \cdot\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\frac{1}{2}\left(\frac{2}{1}\right)^2=\frac{2}{1} \\
\therefore \quad \mathrm{I}_1: \mathrm{I}_2=2: 1
\end{array}$
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