Question:


As shown in the figure, a sector of 30° is removed from a circular disc of mass M and radius R. The moment of inertia of the remaining portion of disc about an axis passing through the centre and perpendicular to plane of disc is

  • \(\frac{5}{12}MR^2\)
  • \(\frac{17}{2}MR^2\)
  • \(\frac{11}{24}MR^2\)
  • \(\frac{1}{6}MR^2\)
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The Correct Option is C

Solution and Explanation

The correct option is (C): \(\frac{11}{24}MR^2\)
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