Q95 / 160 · JIPMER 2005 · Physics

Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

  • A
    $1 \mathrm{~kg} \mathrm{~m}^2$
  • Correct answer
    B
    $0.1 \mathrm{~kg} \mathrm{~m}^2$
  • C
    $2 \mathrm{~kg} \mathrm{~m}^2$
  • D
    $0.2 \mathrm{~kg} \mathrm{~m}^2$

Solution

The moment of inertia of the given system that contains 5 particles each of mass $=2 \mathrm{~kg}$ on the rim of circular disc of radius 0.1 m and of negligible mass is given by $=\text { MI of disc }+ \text { MI of particle }$ Since the mass of the disc is negligible therefore, MI of the system $=$ MI of particle $=5 \times 2 \times(0.1)^2=0.1 \mathrm{~kg} \mathrm{~m}^2$

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

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