Q145 / 150 · JEE Main 2006 · Physics
Four point masses, each of value $\mathrm{m}$, are placed at the corners of a square $A B C D$ of side $\ell$. The moment of inertia through $A$ and parallel to $B D$ is
- A$\mathrm{m} \ell^2$
- B$2 \mathrm{~m} \ell^2$
- C$3 m \ell^2$
- Correct answerD$3 \mathrm{~m} \ell^2$
Solution
$\mathrm{I}=2 \mathrm{~m}(\ell / \sqrt{2})^2=3 \mathrm{~m} \ell^2$
Hence, the correct answer is option: 'D'.