A) \[1\text{ }kg\text{ }{{m}^{2}}\]
B) \[0.1\text{ }kg\text{ }{{m}^{2}}\]
C) \[2\text{ }kg\text{ }{{m}^{2}}\]
D) \[0.2\text{ }kg\text{ }{{m}^{2}}\]
Correct Answer: B
Solution :
The moment of inertia of the given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by = Moment of inertia of disc + Moment of inertia of disc of particle Since, the mass of the disc is negligible therefore, M.I. of the system = Moment of inertia of disc of particle \[=5\times 2\times {{(0.1)}^{2}}=0.1\text{ }kg\text{ }{{m}^{2}}\]
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