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AIIMS 2010
Exam
AIIMS
Year
2010
Subject
Physics
Chapter
Rotational Motion

AIIMS 2010 · Physics · Rotational Motion

Four holes of radius R are cut from a thin square plate of side 4 R...

Four holes of radius RR are cut from a thin square plate of side 4R4 R and mass MM. The moment of inertia of the remaining portion about zz-axis is 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

Options

  1. Aπ12MR2\frac{\pi}{12} M R^2
  2. B(43−π4)MR2\left(\frac{4}{3}-\frac{\pi}{4}\right) M R^2
  3. C(43−π6)MR2\left(\frac{4}{3}-\frac{\pi}{6}\right) M R^2
  4. D(83−10π16)MR2\left(\frac{8}{3}-\frac{10 \pi}{16}\right) M R^2

Correct answer

D. (83−10π16)MR2\left(\frac{8}{3}-\frac{10 \pi}{16}\right) M R^2

Step-by-step solution

If MM is mass of the square plate before cutting the holes, then mass of portion of each hole, m=M16R2×πR2=Mπ16∴m=\frac{M}{16 R^2} \times \pi R^2=\frac{M \pi}{16}\therefore \quad Moment of inertia of remaining portion about ZZ axis I=Isquare −4Ihole =M12(16R2+16R2)−4[mR22+m(2R)2]=M12×32R2−10mR2=83MR2−1016MR2πI=(83−10π16)MR2.\begin{aligned}& I=I_{\text{square }}-4 I_{\text{hole }} \\ & =\frac{M}{12}\left(16 R²+16 R²\right)-4\left[\frac{m R^2}{2}+m(\sqrt{2}R)^2\right] \\ & =\frac{M}{12} \times32 R²-10 m R² \\ & =\frac{8}{3}M R²-\frac{10}{16}M R² \pi\\ & I=\left(\frac{8}{3}-\frac{10 \pi}{16}\right)M R² . \end{aligned}
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