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

Answer 1

Answer: (D)

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