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(B) Since the mass of the disc is negligible,the moment of inertia of the system is determined solely by the five particles attached to the rim.Each p

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$0.1$

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(B) Since the mass of the disc is negligible,the moment of inertia of the system is determined solely by the five particles attached to the rim.Each particle has a mass $m = 2 \ kg$ and is at a distance $r = 0.1 \ m$ from the axis of rotation.The moment of inertia $I$ for a system of particles is given by $I = \sum m_i r_i^2$.Since all $5$ particles are at the same distance $r$ from the center,$I = 5 	imes m 	imes r^2$.Substituting the values: $I = 5 	imes 2 	imes (0.1)^2 = 10 	imes 0.01 = 0.1 \ kg \cdot m^2$.

$5$ particles,each of mass $2 \ kg$,are attached to the rim of a circular disc of radius $0.1 \ m$ and negligible mass. The moment of inertia of this system about an axis passing through the center and perpendicular to its plane is ........... $kg \cdot m^{2}$.

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