Question bank/JEE Main/Physics

Find momentum of inertia about an axis passing through COM of
system and it is perpendicular to theplane of 3 rods.

A${mL}^{2}\left( \frac{7}{3} \right)$B${mL}^{2}\left( \frac{2}{3} \right)$C${mL}^{2}\left( \frac{5}{2} \right)$D${mL}^{2}\left( \frac{1}{2} \right)$

Show answer and solution▾

$\because I_{2} = I_{3}$
$= \frac{(2m)(2L)^{2}}{12} + 2\left\lbrack \frac{{ML}^{2}}{12} + {Md}^{2} \right\rbrack$
$= {mL}^{2}\left( \frac{2 \times 2 \times 2}{12} \right) + 2\left( \frac{1}{12} + \frac{3}{4} \right){mL}^{2}$
${mL}^{2}\left( \frac{2}{3} \right) + 2\left( \frac{1 + 9}{12} \right){mL}^{2}$
${mL}^{2}\left( \frac{2}{3} \right) + 2\left( \frac{1 + 9}{12} \right){mL}^{2}$
$= {mL}^{2}\lbrack\left( \frac{2}{3} \right) + \frac{20}{12}\rbrack$
$I_{net\ } = {mL}^{2}\left( \frac{7}{3} \right)$

Source: Grass-G / P-07 Physics paper 17 October.docx