Q160 / 200 · Manipal (MET) 2013 · Physics

The moment of inertia of a disc of mass $m$ and radius $R$ about an axis, which is tangential to the circumference of the disc and parallel to its diameter is

  • A
    $\frac{3}{2} m R^2$
  • B
    $\frac{2}{3} m R^2$
  • Correct answer
    C
    $\frac{5}{4} m R^2$
  • D
    $\frac{4}{5} m R^2$

Solution

Use parallel axis theorem, from the theorem of parallel axis, the moment of inertia ( $/$ ) of a body about given axis is equal to its moment of inertia/about its diameter, plus the product of the mass $M$ of the body and square of perpendicular distanće between the two axes. That is $I=I_d+M R^2=\frac{1}{4} M R^2+M R^2=\frac{5}{4} M R^2$

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

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