Question

Four holes of radius $R$ are cut from a thin square plate of side $4R$ and mass $M$ . The moment of inertia of the remaining portion about the $x$ -axis is

• A. $\frac{\pi }{12}MR^{2}$
• B. $\left(\frac{4}{3} - \frac{\pi }{4}\right)MR^{2}$
• C. $\left(\frac{4}{3} - \frac{\pi }{6}\right)MR^{2}$
• D. $\left(\frac{8}{3} - \frac{10 \pi }{16}\right)MR^{2}$

Answer 1

Answer: (D)

If $M$ mass of the square plate before cutting the holes, then mass of portion of each hole,
$m=\frac{M}{16 R^{2}}\times \pi R^{2}=\frac{\pi }{16} \, M$
$\therefore $ Moment of inertia of remaining portion
$I=I_{s q u a r e}-4I_{h o l e}$
$=\frac{M}{12}\left(16 R^{2} + 16 R^{2}\right)-4\left[\frac{m R^{2}}{2} + m \left(\sqrt{2} R\right)^{2}\right]$
$=\frac{M}{12}\times 32R^{2}-10mR^{2}$
$=\frac{8}{3}MR^{2}-\frac{10 \pi }{16}MR^{2}=\left(\frac{8}{3} - \frac{10 \pi }{16}\right)MR^{2}$