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 answer
    D
    $3 \mathrm{~m} \ell^2$

Solution

$\mathrm{I}=2 \mathrm{~m}(\ell / \sqrt{2})^2=3 \mathrm{~m} \ell^2$

Hence, the correct answer is option: 'D'.

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