Q.
Four holes of radius $R$ are cut from a thin square plate of side 4 $R$ and mass $M$. The moment of inertia of the remaining portion about $\mathcal{z}$-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) $\left( \frac{8}{3} - \frac{10\pi}{16} \right)MR^{2}$
Explanation
(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
$I = I_{square} - 4I_{hole}$
$= \frac{M}{12}\left( 16R^{2} + 16R^{2} \right) - 4\left\lbrack \frac{mR^{2}}{2} + m\left( \sqrt{2}R \right)^{2} \right\rbrack$
$= \frac{M}{12} \times 32R^{2} - 10mR^{2}$
$= \frac{8}{3}MR^{2} - \frac{10\pi}{16}MR^{2}\ I = \left( \frac{8}{3} - \frac{10\pi}{16} \right)MR^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.