Consider a uniform wire of mass M and length L.

Question:

Consider a uniform wire of mass $\mathrm{M}$ and length $\mathrm{L}$. It is bent into a

semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :

  1. $\frac{1}{4} \frac{\mathrm{ML}^{2}}{\pi^{2}}$

  2. $\frac{2}{5} \frac{\mathrm{ML}^{2}}{\pi^{2}}$

  3. $\frac{\mathrm{ML}^{2}}{\pi^{2}}$

  4. $\frac{1}{2} \frac{\mathrm{ML}^{2}}{\pi^{2}}$


Correct Option: , 3

Solution:

(3)

$\pi \mathrm{r}=\mathrm{L} \Rightarrow \mathrm{r}=\frac{\mathrm{L}}{\pi}$

$\mathrm{I}=\mathrm{Mr}^{2}=\frac{\mathrm{ML}^{2}}{\pi^{2}}$

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