Five particles of masses 2 kg each 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

Five particles of masses 2 kg each 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
$1 kg - m^{2}$
B
$0.1 kg - m^{2}$
C
$2 kg \cdot m^{2}$
D
$0.2 kg - m^{2}$

Solution:

Moment of inertia of the system,

$I = 5 m R^{2} = 5 \times 2 \times (0.1)^{2} = 0.1 kg - m^{2}$

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Solution:

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