Q85 / 180 · AIIMS 2010 · Physics

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 $z$-axis is Figure

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

Solution

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

Hence, the correct answer is option: 'D'.

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