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