MCQ
Two rings of radius $R$ and $nR$ having different masses and made up of same wire have the ratio of moment of inertia about an axis passing through centre as $1 : 8$. The value of $n$ is
- ✓$2$
- B$2\sqrt 2$
- C$4$
- D$1/2$
✓
Answer
Correct option: A.
$2$
a
Ratio of moment of inertia of the rings
Ratio of moment of inertia of the rings
$\frac{I_{1}}{I_{2}}=\left(\frac{M_{1}}{M_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}=\left(\frac{\lambda L_{1}}{\lambda L_{2}}\right)^{2}\left(\frac{R_{1}}{R_{2}}\right)^{2}$
$=\left(\frac{2 \pi R}{2 \pi n R}\right)\left(\frac{R}{n R}\right)^{t}$
$[\lambda=\text { linear density of wire }=\text { constant }]$ $\Rightarrow \frac{L_{1}}{L_{2}}+\frac{1}{n_{3}}+\frac{1}{8}(\text { given })$
$\therefore n^{3}=8$
$\Rightarrow n=2$
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